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# What are the kinematic formulas?

Here are the main equations you can use to analyze situations with constant acceleration.

## What are the kinematic formulas?

The kinematic formulas are a set of formulas that relate the five kinematic variables listed below.
delta, x, start text, D, i, s, p, l, a, c, e, m, e, n, t, end text
t, start text, T, i, m, e, space, i, n, t, e, r, v, a, l, end text, space
v, start subscript, 0, end subscript, space, space, start text, I, n, i, t, i, a, l, space, v, e, l, o, c, i, t, y, end text, space
v, space, space, space, start text, F, i, n, a, l, space, v, e, l, o, c, i, t, y, end text, space
a, space, space, start text, space, C, o, n, s, t, a, n, t, space, a, c, c, e, l, e, r, a, t, i, o, n, end text, space
If we know three of these five kinematic variables—delta, x, comma, t, comma, v, start subscript, 0, end subscript, comma, v, comma, a—for an object under constant acceleration, we can use a kinematic formula, see below, to solve for one of the unknown variables.
The kinematic formulas are often written as the following four equations.
1, point, v, equals, v, start subscript, 0, end subscript, plus, a, t
2, point, delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t
3, point, delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared
4, point, v, squared, equals, v, start subscript, 0, end subscript, squared, plus, 2, a, delta, x
Since the kinematic formulas are only accurate if the acceleration is constant during the time interval considered, we have to be careful to not use them when the acceleration is changing. Also, the kinematic formulas assume all variables are referring to the same direction: horizontal x, vertical y, etc.

## What is a freely flying object—i.e., a projectile?

It might seem like the fact that the kinematic formulas only work for time intervals of constant acceleration would severely limit the applicability of these formulas. However one of the most common forms of motion, free fall, just happens to be constant acceleration.
All freely flying objects—also called projectiles—on Earth, regardless of their mass, have a constant downward acceleration due to gravity of magnitude g, equals, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction.
g, equals, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction, start text, left parenthesis, M, a, g, n, i, t, u, d, e, space, o, f, space, a, c, c, e, l, e, r, a, t, i, o, n, space, d, u, e, space, t, o, space, g, r, a, v, i, t, y, right parenthesis, end text
A freely flying object is defined as any object that is accelerating only due to the influence of gravity. We typically assume the effect of air resistance is small enough to ignore, which means any object that is dropped, thrown, or otherwise flying freely through the air is typically assumed to be a freely flying projectile with a constant downward acceleration of magnitude g, equals, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction.
This is both strange and lucky if we think about it. It's strange since this means that a large boulder will accelerate downwards with the same acceleration as a small pebble, and if dropped from the same height, they would strike the ground at the same time.
It's lucky since we don't need to know the mass of the projectile when solving kinematic formulas since the freely flying object will have the same magnitude of acceleration, g, equals, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction, no matter what mass it has—as long as air resistance is negligible.
Note that g, equals, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction is just the magnitude of the acceleration due to gravity. If upward is selected as positive, we must make the acceleration due to gravity negative a, start subscript, y, end subscript, equals, minus, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction for a projectile when we plug into the kinematic formulas.
Warning: Forgetting to include a negative sign is one of the most common sources of error when using kinematic formulas.

## How do you select and use a kinematic formula?

We choose the kinematic formula that includes both the unknown variable we're looking for and three of the kinematic variables we already know. This way, we can solve for the unknown we want to find, which will be the only unknown in the formula.
For instance, say we knew a book on the ground was kicked forward with an initial velocity of v, start subscript, 0, end subscript, equals, 5, start text, space, m, slash, s, end text, after which it took a time interval t, equals, 3, start text, space, s, end text for the book to slide a displacement of delta, x, equals, 8, start text, space, m, end text. We could use the kinematic formula delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared to algebraically solve for the unknown acceleration a of the book—assuming the acceleration was constant—since we know every other variable in the formula besides adelta, x, comma, v, start subscript, 0, end subscript, comma, t.
Problem solving tip: Note that each kinematic formula is missing one of the five kinematic variables—delta, x, comma, t, comma, v, start subscript, 0, end subscript, comma, v, comma, a.
1, point, v, equals, v, start subscript, 0, end subscript, plus, a, t, start text, left parenthesis, T, h, i, s, space, f, o, r, m, u, l, a, space, i, s, space, m, i, s, s, i, n, g, space, delta, x, point, right parenthesis, end text
2, point, delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t, start text, left parenthesis, T, h, i, s, space, f, o, r, m, u, l, a, space, i, s, space, m, i, s, s, i, n, g, space, a, point, right parenthesis, end text
3, point, delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared, start text, left parenthesis, T, h, i, s, space, f, o, r, m, u, l, a, space, i, s, space, m, i, s, s, i, n, g, space, v, point, right parenthesis, end text
4, point, v, squared, equals, v, start subscript, 0, end subscript, squared, plus, 2, a, delta, x, start text, left parenthesis, T, h, i, s, space, f, o, r, m, u, l, a, space, i, s, space, m, i, s, s, i, n, g, space, t, point, right parenthesis, end text
To choose the kinematic formula that's right for your problem, figure out which variable you are not given and not asked to find. For example, in the problem given above, the final velocity v of the book was neither given nor asked for, so we should choose a formula that does not include v at all. The kinematic formula delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared is missing v, so it's the right choice in this case to solve for the acceleration a.

## How do you derive the first kinematic formula, $v=v_0+at$v, equals, v, start subscript, 0, end subscript, plus, a, t ?

This kinematic formula is probably the easiest to derive since it is really just a rearranged version of the definition of acceleration. We can start with the definition of acceleration,
a, equals, start fraction, delta, v, divided by, delta, t, end fraction $\quad$
Now we can replace delta, v with the definition of change in velocity v, minus, v, start subscript, 0, end subscript.
a, equals, start fraction, v, minus, v, start subscript, 0, end subscript, divided by, delta, t, end fraction
Finally if we just solve for v we get
v, equals, v, start subscript, 0, end subscript, plus, a, delta, t
And if we agree to just use t for delta, t, this becomes the first kinematic formula.
v, equals, v, start subscript, 0, end subscript, plus, a, t

## How do you derive the second kinematic formula, ${\Delta x}=(\dfrac{v+v_0}{2})t$delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t?

A cool way to visually derive this kinematic formula is by considering the velocity graph for an object with constant acceleration—in other words, a constant slope—and starts with initial velocity v, start subscript, 0, end subscript as seen in the graph below.
The area under any velocity graph gives the displacement delta, x. So, the area under this velocity graph will be the displacement delta, x of the object.
delta, x, equals, start text, space, t, o, t, a, l, space, a, r, e, a, end text
We can conveniently break this area into a blue rectangle and a red triangle as seen in the graph above.
The height of the blue rectangle is v, start subscript, 0, end subscript and the width is t, so the area of the blue rectangle is v, start subscript, 0, end subscript, t.
The base of the red triangle is t and the height is v, minus, v, start subscript, 0, end subscript, so the area of the red triangle is start fraction, 1, divided by, 2, end fraction, t, left parenthesis, v, minus, v, start subscript, 0, end subscript, right parenthesis.
The total area will be the sum of the areas of the blue rectangle and the red triangle.
delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, t, left parenthesis, v, minus, v, start subscript, 0, end subscript, right parenthesis
If we distribute the factor of start fraction, 1, divided by, 2, end fraction, t we get
delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, v, t, minus, start fraction, 1, divided by, 2, end fraction, v, start subscript, 0, end subscript, t
We can simplify by combining the v, start subscript, 0, end subscript terms to get
delta, x, equals, start fraction, 1, divided by, 2, end fraction, v, t, plus, start fraction, 1, divided by, 2, end fraction, v, start subscript, 0, end subscript, t
And finally we can rewrite the right hand side to get the second kinematic formula.
delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t
This formula is interesting since if you divide both sides by t, you get start fraction, delta, x, divided by, t, end fraction, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis. This shows that the average velocity start fraction, delta, x, divided by, t, end fraction equals the average of the final and initial velocities start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction. However, this is only true assuming the acceleration is constant since we derived this formula from a velocity graph with constant slope/acceleration.

## How do you derive the third kinematic formula, $\Delta x=v_0 t+\dfrac{1}{2}at^2$delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared?

There are a couple ways to derive the equation delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared. There's a cool geometric derivation and a less exciting plugging-and-chugging derivation. We'll do the cool geometric derivation first.
Consider an object that starts with a velocity v, start subscript, 0, end subscript and maintains constant acceleration to a final velocity of v as seen in the graph below.
Since the area under a velocity graph gives the displacement delta, x, each term on the right hand side of the formula delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared represents an area in the graph above.
The term v, start subscript, 0, end subscript, t represents the area of the blue rectangle since A, start subscript, r, e, c, t, a, n, g, l, e, end subscript, equals, h, w.
The term start fraction, 1, divided by, 2, end fraction, a, t, squared represents the area of the red triangle since A, start subscript, t, r, i, a, n, g, l, e, end subscript, equals, start fraction, 1, divided by, 2, end fraction, b, h.
That's it. The formula delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared has to be true since the displacement must be given by the total area under the curve. We did assume the velocity graph was a nice diagonal line so that we could use the triangle formula, so this kinematic formula—like all the rest of the kinematic formulas—is only true under the assumption that the acceleration is constant.

Here's the alternative plugging-and-chugging derivation. The third kinematic formula can be derived by plugging in the first kinematic formula, v, equals, v, start subscript, 0, end subscript, plus, a, t, into the second kinematic formula, start fraction, delta, x, divided by, t, end fraction, equals, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction.
start fraction, delta, x, divided by, t, end fraction, equals, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction
and we use v, equals, v, start subscript, 0, end subscript, plus, a, t to plug in for v, we get
start fraction, delta, x, divided by, t, end fraction, equals, start fraction, left parenthesis, v, start subscript, 0, end subscript, plus, a, t, right parenthesis, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction
We can expand the right hand side and get
start fraction, delta, x, divided by, t, end fraction, equals, start fraction, v, start subscript, 0, end subscript, divided by, 2, end fraction, plus, start fraction, a, t, divided by, 2, end fraction, plus, start fraction, v, start subscript, 0, end subscript, divided by, 2, end fraction
Combining the start fraction, v, start subscript, 0, end subscript, divided by, 2, end fraction terms on the right hand side gives us
start fraction, delta, x, divided by, t, end fraction, equals, v, start subscript, 0, end subscript, plus, start fraction, a, t, divided by, 2, end fraction
And finally multiplying both sides by the time t gives us the third kinematic formula.
delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared
Again, we used other kinematic formulas, which have a requirement of constant acceleration, so this third kinematic formula is also only true under the assumption that the acceleration is constant.

## How do you derive the fourth kinematic formula, $v^2=v_0^2+2a\Delta x$v, squared, equals, v, start subscript, 0, end subscript, squared, plus, 2, a, delta, x?

To derive the fourth kinematic formula, we'll start with the second kinematic formula:
delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t
We want to eliminate the time t from this formula. To do this, we'll solve the first kinematic formula, v, equals, v, start subscript, 0, end subscript, plus, a, t, for time to get t, equals, start fraction, v, minus, v, start subscript, 0, end subscript, divided by, a, end fraction. If we plug this expression for time t into the second kinematic formula we'll get
delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, left parenthesis, start fraction, v, minus, v, start subscript, 0, end subscript, divided by, a, end fraction, right parenthesis
Multiplying the fractions on the right hand side gives
delta, x, equals, left parenthesis, start fraction, v, squared, minus, v, start subscript, 0, end subscript, squared, divided by, 2, a, end fraction, right parenthesis
And now solving for v, squared we get the fourth kinematic formula.
v, squared, equals, v, start subscript, 0, end subscript, squared, plus, 2, a, delta, x

## What's confusing about the kinematic formulas?

People often forget that the kinematic formulas are only true assuming the acceleration is constant during the time interval considered.
Sometimes a known variable will not be explicitly given in a problem, but rather implied with codewords. For instance, "starts from rest" means v, start subscript, 0, end subscript, equals, 0, "dropped" often means v, start subscript, 0, end subscript, equals, 0, and "comes to a stop" means v, equals, 0. Also, the magnitude of the acceleration due to gravity on all freely flying projectiles is assumed to be g, equals, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction, so this acceleration will usually not be given explicitly in a problem but will just be implied for a freely flying object.
People forget that all the kinematic variables—delta, x, comma, v, start subscript, o, end subscript, comma, v, comma, a—except for t can be negative. A missing negative sign is a very common source of error. If upward is assumed to be positive, then the acceleration due to gravity for a freely flying object must be negative: a, start subscript, g, end subscript, equals, minus, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction.
The third kinematic formula, delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared, might require the use of the quadratic formula, see solved example 3 below.
People forget that even though you can choose any time interval during the constant acceleration, the kinematic variables you plug into a kinematic formula must be consistent with that time interval. In other words, the initial velocity v, start subscript, 0, end subscript has to be the velocity of the object at the initial position and start of the time interval t. Similarly, the final velocity v must be the velocity at the final position and end of the time interval t being analyzed.

## What do solved examples involving the kinematic formulas look like?

### Example 1: First kinematic formula, $v=v_0+at$v, equals, v, start subscript, 0, end subscript, plus, a, t

A water balloon filled with Kool-Aid is dropped from the top of a very tall building.
What is the velocity of the water balloon after falling for t, equals, 2, point, 35, start text, space, s, end text?
Assuming upward is the positive direction, our known variables are
v, start subscript, 0, end subscript, equals, 0 (Since the water balloon was dropped, it started at rest.)
t, equals, 2, point, 35, start text, space, s, end text (This is the time interval after which we want to find the velocity.)
a, start subscript, g, end subscript, equals, minus, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction(This is implied since the water balloon is a freely falling object.)
The motion is vertical in this situation, so we'll use y as our position variable instead of x. The symbol we choose doesn't really matter as long as we're consistent, but people typically use y to indicate vertical motion.
Since we don't know the displacement delta, y and we weren't asked for the displacement delta, y, we'll use the first kinematic formula v, equals, v, start subscript, 0, end subscript, plus, a, t, which is missing delta, y.
v, equals, v, start subscript, 0, end subscript, plus, a, t, start text, left parenthesis, U, s, e, space, t, h, e, space, f, i, r, s, t, space, k, i, n, e, m, a, t, i, c, space, f, o, r, m, u, l, a, space, s, i, n, c, e, space, i, t, apostrophe, s, space, m, i, s, s, i, n, g, space, delta, y, point, right parenthesis, end text
v, equals, 0, start text, space, m, slash, s, end text, plus, left parenthesis, minus, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction, right parenthesis, left parenthesis, 2, point, 35, start text, space, s, end text, right parenthesis, start text, left parenthesis, P, l, u, g, space, i, n, space, k, n, o, w, n, space, v, a, l, u, e, s, point, right parenthesis, end text
v, equals, minus, 23, point, 1, start text, space, m, slash, s, end text, start text, left parenthesis, C, a, l, c, u, l, a, t, e, space, a, n, d, space, c, e, l, e, b, r, a, t, e, !, right parenthesis, end text
Note: The final velocity was negative since the water balloon was heading downward.

### Example 2: Second kinematic formula, ${\Delta x}=(\dfrac{v+v_0}{2})t$delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t

A leopard is running at 6.20 m/s and after seeing a mirage that's taken the form of an ice cream truck; the leopard then speeds up to 23.1 m/s in a time of 3.3 s.
How much ground did the leopard cover in going from 6.20 m/s to 23.1 m/s?
Assuming the initial direction of travel is the positive direction, our known variables are
v, start subscript, 0, end subscript, equals, 6, point, 20, start text, space, m, slash, s, end text (The initial speed of the leopard)
v, equals, 23, point, 1, start text, space, m, slash, s, end text (The final speed of the leopard)
t, equals, 3, point, 30, start text, space, s, end text (The time it took for the leopard to speed up)
Since we do not know the acceleration a and were not asked for the acceleration, we'll use the second kinematic formula for the horizontal direction delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t, which is missing a.
delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t, start text, left parenthesis, U, s, e, space, t, h, e, space, s, e, c, o, n, d, space, k, i, n, e, m, a, t, i, c, space, f, o, r, m, u, l, a, space, s, i, n, c, e, space, i, t, apostrophe, s, space, m, i, s, s, i, n, g, space, a, point, right parenthesis, end text
delta, x, equals, left parenthesis, start fraction, 23, point, 1, start text, space, m, slash, s, end text, plus, 6, point, 20, start text, space, m, slash, s, end text, divided by, 2, end fraction, right parenthesis, left parenthesis, 3, point, 30, start text, space, s, end text, right parenthesis, start text, left parenthesis, P, l, u, g, space, i, n, space, k, n, o, w, n, space, v, a, l, u, e, s, point, right parenthesis, end text
delta, x, equals, 48, point, 3, start text, space, m, end text, start text, left parenthesis, C, a, l, c, u, l, a, t, e, space, a, n, d, space, c, e, l, e, b, r, a, t, e, !, right parenthesis, end text

### Example 3: Third kinematic formula, $\Delta x=v_0 t+\dfrac{1}{2}at^2$delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared

A student is fed up with doing her kinematic formula homework, so she throws her pencil straight upward at 18.3 m/s.
How long does it take the pencil to first reach a point 12.2 m higher than where it was thrown?
Assuming upward is the positive direction, our known variables are
v, start subscript, 0, end subscript, equals, 18, point, 3, start text, space, m, slash, s, end text (The initial upward velocity of the pencil)
delta, y, equals, 12, point, 2, start text, space, m, end text (We want to know the time when the pencil moves through this displacement.)
a, equals, minus, 9, point, 81, start fraction, start text, space, m, end text, divided by, start text, space, s, end text, squared, end fraction (The pencil is a freely flying projectile.)
Since we don't know the final velocity v and we weren't asked to find the final velocity, we will use the third kinematic formula for the vertical direction delta, y, equals, v, start subscript, 0, y, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, start subscript, y, end subscript, t, squared, which is missing v.
delta, y, equals, v, start subscript, 0, y, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, start subscript, y, end subscript, t, squared, start text, left parenthesis, S, t, a, r, t, space, w, i, t, h, space, t, h, e, space, t, h, i, r, d, space, k, i, n, e, m, a, t, i, c, space, f, o, r, m, u, l, a, point, right parenthesis, end text
Normally we would just solve our expression algebraically for the variable we want to find, but this kinematic formula can not be solved algebraically for time if none of the terms are zero. That's because when none of the terms are zero and t is the unknown variable, this equation becomes a quadratic equation. We can see this by plugging in known values.
12, point, 2, start text, space, m, end text, equals, left parenthesis, 18, point, 3, start text, space, m, slash, s, end text, right parenthesis, t, plus, start fraction, 1, divided by, 2, end fraction, left parenthesis, minus, 9, point, 81, start fraction, start text, space, m, end text, divided by, start text, space, s, end text, squared, end fraction, right parenthesis, t, squared, start text, left parenthesis, P, l, u, g, space, i, n, space, k, n, o, w, n, space, v, a, l, u, e, s, point, right parenthesis, end text
To put this into a more solvable form of the quadratic equation, we move everything onto one side of the equation. Subtracting 12.2 m from both sides we get
0, equals, start fraction, 1, divided by, 2, end fraction, left parenthesis, minus, 9, point, 81, start fraction, start text, space, m, end text, divided by, start text, space, s, end text, squared, end fraction, right parenthesis, t, squared, plus, left parenthesis, 18, point, 3, start text, space, m, slash, s, end text, right parenthesis, t, minus, 12, point, 2, start text, space, m, end text, start text, left parenthesis, P, u, t, space, i, t, space, i, n, t, o, space, t, h, e, space, f, o, r, m, space, o, f, space, t, h, e, space, q, u, a, d, r, a, t, i, c, space, e, q, u, a, t, i, o, n, point, right parenthesis, end text
At this point, we solve the quadratic equation for time t. The solutions of a quadratic equation in the form of a, t, squared, plus, b, t, plus, c, equals, 0 are found by using the quadratic formula t, equals, start fraction, minus, b, plus minus, square root of, b, squared, minus, 4, a, c, end square root, divided by, 2, a, end fraction. For our kinematic equation a, equals, start fraction, 1, divided by, 2, end fraction, left parenthesis, minus, 9, point, 81, start fraction, start text, space, m, end text, divided by, start text, space, s, end text, squared, end fraction, right parenthesis, b, equals, 18, point, 3, start text, space, m, slash, s, end text, and c, equals, minus, 12, point, 2, start text, space, m, end text.
So, plugging into the quadratic formula, we get
t, equals, start fraction, minus, 18, point, 3, start text, space, m, slash, s, end text, plus minus, square root of, left parenthesis, 18, point, 3, start text, space, m, slash, s, end text, right parenthesis, squared, minus, 4, open bracket, start fraction, 1, divided by, 2, end fraction, left parenthesis, minus, 9, point, 81, start fraction, start text, space, m, end text, divided by, start text, space, s, end text, squared, end fraction, right parenthesis, left parenthesis, minus, 12, point, 2, start text, space, m, end text, right parenthesis, close bracket, end square root, divided by, 2, open bracket, start fraction, 1, divided by, 2, end fraction, left parenthesis, minus, 9, point, 81, start fraction, start text, space, m, end text, divided by, start text, space, s, end text, squared, end fraction, right parenthesis, close bracket, end fraction
Since there is a plus or minus sign in the quadratic formula, we get two answers for the time t: one when using the plus and one when using the minus. Solving the quadratic formula above gives these two times:
t, equals, 0, point, 869, start text, space, s, end text and t, equals, 2, point, 86, start text, space, s, end text
There are two positive solutions since there are two times when the pencil was 12.2 m high. The smaller time refers to the time required to go upward and first reach the displacement of 12.2 m high. The larger time refers to the time required to move upward, pass through 12.2 m high, reach a maximum height, and then fall back down to a point 12.2 m high.
So, to find the answer to our question of "How long does it take the pencil to first reach a point 12.2 m higher than where it was thrown?" we would choose the smaller time t, equals, 0, point, 869, start text, space, s, end text.

### Example 4: Fourth kinematic formula, $v^2=v_0^2+2a\Delta x$v, squared, equals, v, start subscript, 0, end subscript, squared, plus, 2, a, delta, x

A European motorcyclist starts with a speed of 23.4 m/s and, seeing traffic up ahead, decides to slow down over a length of 50.2 m with a constant deceleration of magnitude 3, point, 20, start fraction, start text, space, m, end text, divided by, start text, space, s, end text, squared, end fraction. Assume the motorcycle is moving forward for the entire trip.
What is the new velocity of the motorcyclist after slowing down through the 50.2 m?
Assuming the initial direction of travel is the positive direction, our known variables are
v, start subscript, 0, end subscript, equals, 23, point, 4, start text, space, m, slash, s, end text (The initial forward velocity of the motorcycle)
a, equals, minus, 3, point, 20, start fraction, start text, space, m, end text, divided by, start text, space, s, end text, squared, end fraction (Acceleration is negative since the motorcycle is slowing down and we assumed forward is positive.)
delta, x, equals, 50, point, 2, start text, space, m, end text (We want to know the velocity after the motorcycle moves through this displacement.)
Since we don't know the time t and we weren't asked to find the time, we will use the fourth kinematic formula for the horizontal direction v, start subscript, x, end subscript, squared, equals, v, start subscript, 0, x, end subscript, squared, plus, 2, a, start subscript, x, end subscript, delta, x, which is missing t.
v, start subscript, x, end subscript, squared, equals, v, start subscript, 0, x, end subscript, squared, plus, 2, a, start subscript, x, end subscript, delta, x, start text, left parenthesis, S, t, a, r, t, space, w, i, t, h, space, t, h, e, space, f, o, u, r, t, h, space, k, i, n, e, m, a, t, i, c, space, f, o, r, m, u, l, a, point, right parenthesis, end text
v, start subscript, x, end subscript, equals, plus minus, square root of, v, start subscript, 0, x, end subscript, squared, plus, 2, a, start subscript, x, end subscript, delta, x, end square root, start text, left parenthesis, A, l, g, e, b, r, a, i, c, a, l, l, y, space, s, o, l, v, e, space, f, o, r, space, t, h, e, space, f, i, n, a, l, space, v, e, l, o, c, i, t, y, point, right parenthesis, end text
Note that in taking a square root, you get two possible answers: positive or negative. Since our motorcyclist will still be going in the direction of motion it started with and we assumed that direction was positive, we'll choose the positive answer v, start subscript, x, end subscript, equals, plus, square root of, v, start subscript, 0, x, end subscript, squared, plus, 2, a, start subscript, x, end subscript, delta, x, end square root.
Now we can plug in values to get
v, start subscript, x, end subscript, equals, square root of, left parenthesis, 23, point, 4, start text, space, m, slash, s, end text, right parenthesis, squared, plus, 2, left parenthesis, minus, 3, point, 20, start fraction, start text, space, m, end text, divided by, start text, space, s, end text, squared, end fraction, right parenthesis, left parenthesis, 50, point, 2, start text, space, m, end text, right parenthesis, end square root, start text, left parenthesis, P, l, u, g, space, i, n, space, k, n, o, w, n, space, v, a, l, u, e, s, point, right parenthesis, end text
v, start subscript, x, end subscript, equals, 15, point, 0, start text, space, m, slash, s, end text, start text, left parenthesis, C, a, l, c, u, l, a, t, e, space, a, n, d, space, c, e, l, e, b, r, a, t, e, !, right parenthesis, end text

## Want to join the conversation?

• How to derive equations of motion by using calculus ?

We know that acceleration is approximately -9.8 m/s^2 (we're just going to use -9.8 so the math is easier) and we know that acceleration is the derivative of velocity, which is the derivative of position. We can use this knowledge (and our knowledge of integrals) to derive the kinematics equations.

First, we need to establish that acceleration is represented by the equation a(t) = -9.8.

Because velocity is the antiderivative of acceleration, that means that v'(t) = a(t) and v(t) = int[a(t)]. Simplifying the integral results in the equation v(t) = -9.8t + C_1, where C_1 is the initial velocity (in physics, this the initial velocity is v_0). This means that for every second, the velocity decreases by -9.8 m/s.

To find the position equation, simply repeat this step with velocity. Position is the antiderivative of velocity, so that means that x'(t) = v(t) and x(t) = int[v(t)]. Simplifying the integral results in the position equation x(t) = -4.9t^2 + (C_1)t + C_2, where C_1 is the initial velocity and C_2 is the initial position (in physics, C_2 is usually represented by x_0).

In order to make this equation more universal, the position equation can be generalized as x(t) = 1/2(at^2) + v_0 + x_0
• In example 3, where the pencil is being thrown upward, should g = +9.8 m/s^2 or -9.8 m/s^2 ? I thought it should be positive (upward), but here it is negative. Could someone explain this to me.
• For this situation (any most situations), any vector that point UP or to the RIGHT is taken as positive.

Whilst the initial vertical VELOCITY vector is upwards and therefore positive, the force of graviy is always downwards, and therefore (F=ma) the acceleration is always downward and negative.

(Note: when the velocity vector and acceleration vectors are in opposite direct this means that the object is slowing down. When the pencil starts to fall, the velocity and acceleration are in the same direction and so it is speeding up)

ok??
• Could someone please explain in a step by step fashion; how to solve for Vf in the first kinematic equation:
a=vf-v0/delta t.
I cannot seem to be able to wrap my mind around the manipulation.

Thank you.
• We want to get Vf = something, so we start by multiplying both sides of the equation:
A(delta t) = vf - v0.
The add v0 to both sides, and voila!
A(delta t) + v0 = vf
(1 vote)
• How can two objects with different masses experience the same acceleration (-9.8m/s^2)? I thought heavier things fall to the ground quicker.
• You're close, but not quite there. While it's true that there is more gravitational force acting on a heavier object, this doesn't correspond to an increase in acceleration. In fact, the opposite is true. A heavier object has more inertia, which is a resistance to a change in motion.
If you look at Newton's law of universal gravitation, you see that the force of attraction between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them. If we take this equation and frame it in terms of somebody standing on the earth, we get a force due to gravitational attraction that is the product of their masses divided by the square of the earth's radius.
From here, we can take Newton's second law of motion, f = ma. Here, we are looking for the force on the person from the earth. So, the mass on the left side of the following equation would refer to the person, which we can arbitrarily call m2. We can plug this in for the attraction force, giving us this:
m2 * a = G (m1 * m2) / (r^2)
Canceling out m2, we get a definition for gravitational acceleration:
a = G (M1) / R^2
From here, we have to use measurements to help us. Using measurements of the earth's mass and radius, as well as Newton's constant of gravitation, we can determine that the average value of a on the Earth's surface is about 9.81 m/s^2

Hope this helped!
• In example 4 when plugging in the formulas there were no step by steps on how they got the answer.What are they someone please...
• Is quadratic formula necessary as used in the example with the third formula? I completed the problem without using the quadratic formula and I was wondering if this was an inferior strategy.
• I've seen a different list of five kinetic variables with position as one of them. Is there one standard list of "the" five kinetic variables that I should know? Or is it more subjective and situational?
(1 vote)
• I don't think you should look for such a list, try and understand the most important concepts like position, velocity (how position changes over time) and acceleration (how velocity changes over time). We can visually understand these three simply, you probably already have an accurate intuition of how each should look like. Position is merely an object's location in space; its also easy to notice when something has some sort of velocity; About acceleration, its noticeable whenever the object's velocity is changing somehow (becoming faster or changing direction).


text:
# Generic mesh readers under ParaView via the new Python interface

7 分钟阅读

Recently I discoved the Python VTKPythonAlgorithmBase interface introduced in the ParaView 5.6 version for defining user plugins with pure Python files. These plugins can then be loaded through Tools / Manage Plugins / Load New under ParaView and can be served as custom sources, readers, writers and filters.

Previously before 5.6 only plugins defined by a XML file (or compiled library files) can be loaded. For generating such XML files, I used a Python package pvpyfilter. With the newly introduced native VTKPythonAlgorithmBase interface, it is no longer necessary.

In this post, I will illustrate the development of such Python plugins via the example of a generic mesh format reader, with the help of meshio. The code source of this plugin is available on Github.

## meshio

meshio is a Python library that reads and writes various frequently-used mesh formats: XDMF, VTK, gmsh, Abaqus…Personnally since 2018 I’m actively involved in this project especially for the MED format. The finite element result produced by code_aster is a MED file and in general requires the rather fatty salome platform for visualization and post-processing. With the MED interface of meshio, these results files can now be converted to a more generic format readable directly by ParaView, like XDMF or VTU. Directly Python processing is also possible without the use of salome environment.

The API of meshio is minimalist. For reading or writing a mesh file foo, simply use

mesh = meshio.read(foo)  # read mesh from foo
meshio.write(foo, mesh)  # write mesh to foo

The actual mesh format used by foo is either automatically deduced from its extension, or in case of ambiguity, can be specified by the file_format argument

mesh = meshio.read(foo, file_format="...")
meshio.write(foo, mesh, file_format="...")

The mesh information is contained in a Mesh instance returned by meshio.read. The most useful attributes are defined as follows

mesh.points  # node coordinates
mesh.cells  # cell connectivities
mesh.point_data  # nodal fields
mesh.cell_data  # cell fields

For cells, mesh.cells is a dictionary that defines the connectivity information for each geomerical type of cells. In the case of a 2d mesh composed of triangles and quads, we will have

mesh.cells = {
"triangle": ...,  # a (n_triangles, 3) numpy array
}

Each line of these two numpy arrays specifies the node id’s of a particular cell: 3 points for triangles and 4 for quad’s. The id’s correspond to the node coordinates of the mesh defined in mesh.points.

The point data is a dictionary of numpy arrays with the keys as field names, for instance

mesh.point_data = {
"Displacement": ...,  # a (n_points, 3) numpy array
}

defines a 3d (with 3 spatial components) displacement field defined on each node of the mesh. As for cell data, an additional dictionary level is necessary to specify the cell type

mesh.cell_data = {
"triangle":  {
"Displacement": ...,  # a (n_triangles, 3) numpy array
},
"Displacement": ...,  # a (n_quads, 3) numpy array
},
}

In this case, a 3d displacement is defined on each triangular and quadrilateral cell.

## VTKPythonAlgorithmBase interface for defining ParaView plugins

For a more comprehensive introduction, interested readers can consult section 12.4 of the ParaView Guide and the PythonAlgorithmExamples.py file provided by ParaView.

Essentially, to develop a user ParaView plugin, we just need to define a subclass of VTKPythonAlgorithmBase in your Python file. In my generic mesh reader case, we have

from paraview.util.vtkAlgorithm import VTKPythonAlgorithmBase
def __init__(self):
VTKPythonAlgorithmBase.__init__(
self, nInputPorts=0, nOutputPorts=1, outputType="vtkUnstructuredGrid"
)
...

If you are familiar with ParaView or VTK, you know that post-processing of the input data is organized through pipeline. Each operation can be characterized by the number of inputs and the number of outputs. A “source” or “reader” do not require inputs, but produce some outputs; a “writer” take an or several inputs but do not produce additional data; a “filter” performs some operation to inputs and generates the results. This input / output meta-informaton is needed to initialize the class. In this case, as a generic mesh format reader, no inputs are required (so you do not need to select an existing object in ParaView before using this plugin to open a file), and it only returns 1 object as a vtkUnstructuredGrid, an unstructured mesh defined by points coordinates and cell connectivities (as meshio reads and writes).

Until now, only VTK objects are involved (VTKPythonAlgorithmBase is in fact defined by VTK). To declare this class as an effective reader usable under ParaView, it has to be decorated by ParaView functions.

from paraview.util.vtkAlgorithm import smdomain, smhint, smproperty, smproxy
extensions=meshio_supported_ext,  # [msh, med, inp, fem, ...]
file_description="meshio-supported files",
)
...

By doing so, the plugin will be registered as a reader and when you open files, the addtional file formats will be available as “meshio-supported files” in the dialog.

### Data generation

The mesh reading and output to ParaView pipeline is performed in the RequestData method of VTKPythonAlgorithmBase. In my case, it can be summarized as follows

from vtkmodules.numpy_interface import dataset_adapter as dsa
from vtkmodules.vtkCommonDataModel import vtkUnstructuredGrid
def RequestData(self, request, inInfo, outInfo):
# Use meshio to read the mesh

# Produce ParaView output from mesh
output = dsa.WrapDataObject(vtkUnstructuredGrid.GetData(outInfo))
output.SetPoints(points)  # point coordinates
output.SetCells(cell_types, cell_offsets, cell_conn)  # cell connectivities

output.PointData.append(array, name)  # point data
output.CellData.append(array, name)  # cell data
output.FieldData.append(array, name)  # field data

The amazing part of this Python approach is that the meshio library can be directly used to do the dirtiest work (through meshio.read) of actually reading the mesh file. Later on we only need to carefully wrap the mesh object into the ParaView-desired output outInfo. The connection between ParaView VTK objects and pure Python is achieved by vtkmodules.numpy_interface.dataset_adapter. By doing

output = dsa.WrapDataObject(vtkUnstructuredGrid.GetData(outInfo))

The numpy-friendly object output can be used to construct the mesh for output. Numpy arrays can be directly used when defining node coordinates via SetPoints, cell connectivities via SetCells, etc.

Some GUI elements can be added to the plugin to provide additional parameters to the plugin. These parameters can be continuous (integers, floats) or discrete in nature (a list of choices). In this example, as suggested by one of the Github user, I added a file format (string) selector to for users to explicitly indicate the mesh format of the input file. It can be defined as additional methods inside the class

@smproperty.stringvector(name="StringInfo", information_only="1")
def GetStrings(self):
return meshio_input_filetypes

@smproperty.stringvector(name="FileFormat", number_of_elements="1")
@smdomain.xml(
"""<StringListDomain name="list">
<RequiredProperties>
<Property name="StringInfo" function="StringInfo"/>
</RequiredProperties>
</StringListDomain>
"""
)
def SetFileFormat(self, file_format):
...

In this case, the GUI (string list) must be defined through raw XML format. In other cases (floats, integers…), simple decorators can be used, see the documentation.

## Conclusions

The Python VTKPythonAlgorithmBase interface provides a powerful and easy-to-maintain approach of defining ParaView plugins. Since it is pure Python, additional technical Python libraries can be directly used to process the data. It can also be easily distributed to other users, since only one source file is involved.

format: plain


text:
# Problem in plotting complex valued data in Mathematica

I have a situation where a function sys yields solutions in complex values while I change its argument x from $-\pi$ to $+\pi$ as in the following:

Bands[sys[t1, t2, 3], {x, -\[Pi], \[Pi]}]

which gives me list of complex values with x as following:

{{-3.14159, -1.99705 + 0.0000290634 I}, {-3.02073, -1.99703 -
0.0000293392 I}, {-2.8897, -1.99698 -
0.0000302499 I}, {-2.76735, -1.99691 - 0.0000316411 I},...{3.14159, 1.99705
+0.0000290634 I}}

Now if I write

Re[Bands[sys[t1, t2, 3], {x, -\[Pi], \[Pi]}]]

it gives me correct list of real parts of the complex values which is:

{{-3.14159, -1.99705},{-3.02073, -1.99703},{-2.8897, -1.99698}...so on} but for imaginary parts:

Im[Bands[sys[t1, t2, 3], {x, -\[Pi], \[Pi]}]]

it gives me:

{{0,0.0000290634 I},{0,- 0.0000293392 I},{0,- 0.0000302499 I}...so on}

thus, not showing the correct values of variable x. As a result, the real values giving me correct plot but the imaginary values don't give me correct plot since the values of 'x' show all zero which is not true.

Someone (@Rom38) suggested me to use: {First@#, Im@Last@#} & /@ [Bands[...]. But in my case how do I implement it when the plot command I use to plot the imaginary part is:

PlotBands[Im[Bands[sys[t1, t2, 1.5], {x, -2 \[Pi], 2[Pi]}]], FrameLabel -> {"x", "Im(E)"}]

Any help will be much appreciated. Many thanks!

• It is clear that imaginary part of the real number is zero. Try something like {First@#,Im@Last@#}&/@Bands[....] – Rom38 Apr 16 '18 at 6:27
• Actually my problem is precisely as in this link: mathematica.stackexchange.com/questions/58549/… @ Rom38 I couldn't understand your point. Can you explain. – foi Apr 16 '18 at 6:48
• My point is following: applying the Im for both {x,y}, you will have {0,Im[y]} in case if x is already real. Mapping my pure function {First@#,Im@Last@#}& for each sublist of your list, you will apply the Im only to second element of each {x,y} pair. You should use the /@` command for this. As result, just copy-paste my code from previous comment and see the result – Rom38 Apr 16 '18 at 7:19
• @Sam don't use answers to comment. Even with low reputation you can comment under your questions. The problem seems to be that you have created two accounts, see: mathematica.stackexchange.com/help/merging-accounts to fix this. – Kuba Apr 16 '18 at 7:31


text:
# ISODD

Returns TRUE if number is odd, or FALSE if number is even.

## Syntax

ISODD(number)

### Parameters

Term Definition
number The value to test. If number is not an integer, it is truncated.

## Return value

Returns TRUE if number is odd, or FALSE if number is even.

## Remarks

• If number is nonnumeric, ISODD returns the #VALUE! error value.

• This function is not supported for use in DirectQuery mode when used in calculated columns or row-level security (RLS) rules.


text:
# Show that the function is not differentiable at the point z

#### mathmari

##### Well-known member
MHB Site Helper
Hi!! Could you help me with the following?
Let $$g: R \rightarrow R$$ a bounded function. There is a point $$z \epsilon R$$ for which the function $$h: R$$ \ $$\{z\} \rightarrow R$$, where $$h(x)=\frac{g(x)-g(z)}{x-z}$$ is not bounded. Show that the function $$g$$ is not differentiable at the point $$z$$.

My idea is the following:
Let the function $$g$$ be differentiable at the point $$z$$, so $$lim_{x \rightarrow z}{ \frac{g(x)-g(z)}{x-z}}=L$$. So there is a ε>0: $$|\frac{g(x)-g(z)}{x-z}-L|<ε$$.
$$|h(x)|=|h(x)-L+L| \leq |h(x)-L|+|L|=|\frac{g(x)-g(z)}{x-z}-L|+|L|<ε+|L|$$.
So the function $$h$$ is bounded. That cannot be true. So the function $$g$$ is not differentiable at the point $$z$$.

Is my idea right??

#### Opalg

##### MHB Oldtimer
Staff member
Hi!! Could you help me with the following?
Let $$g: R \rightarrow R$$ a bounded function. There is a point $$z \epsilon R$$ for which the function $$h: R$$ \ $$\{z\} \rightarrow R$$, where $$h(x)=\frac{g(x)-g(z)}{x-z}$$ is not bounded. Show that the function $$g$$ is not differentiable at the point $$z$$.

My idea is the following:
Let the function $$g$$ be differentiable at the point $$z$$, so $$lim_{x \rightarrow z}{ \frac{g(x)-g(z)}{x-z}}=L$$. So there is a ε>0: $$|\frac{g(x)-g(z)}{x-z}-L|<ε$$.
$$|h(x)|=|h(x)-L+L| \leq |h(x)-L|+|L|=|\frac{g(x)-g(z)}{x-z}-L|+|L|<ε+|L|$$.
So the function $$h$$ is bounded. That cannot be true. So the function $$g$$ is not differentiable at the point $$z$$.

Is my idea right??
Absolutely correct!

MHB Site Helper
Great!!!

#### mathmari

##### Well-known member
MHB Site Helper
Now I'm looking again the exercise..
There is a $ε>0$ such that $|\frac{g(x)-g(z)}{x-z}-L|<ε$ for $|x-z|<δ$, right? So $h$ is bounded for $x$ such that $|x-z|<δ$.
Do I have to show also that $h$ is bounded for $x$ such that $|x-z|>δ$, or is the above enough?

#### Opalg

##### MHB Oldtimer
Staff member
Now I'm looking again the exercise..
There is a $ε>0$ such that $|\frac{g(x)-g(z)}{x-z}-L|<ε$ for $|x-z|<δ$, right? So $h$ is bounded for $x$ such that $|x-z|<δ$.
Do I have to show also that $h$ is bounded for $x$ such that $|x-z|>δ$, or is the above enough?
Yes, you need to show that if $h$ is unbounded, then it must be unbounded near $z$, not anywhere else. But if $|x-z|\geqslant \delta$ then $\left|\dfrac1{x-z}\right|$ is bounded by $1/\delta$. Also, you know that $g$ is bounded. So it follows that $h$ must indeed be bounded outside the region $|x-z|< \delta$.

#### mathmari

##### Well-known member
MHB Site Helper
Yes, you need to show that if $h$ is unbounded, then it must be unbounded near $z$, not anywhere else. But if $|x-z|\geqslant \delta$ then $\left|\dfrac1{x-z}\right|$ is bounded by $1/\delta$. Also, you know that $g$ is bounded. So it follows that $h$ must indeed be bounded outside the region $|x-z|< \delta$.
So do I have to show both cases $|x-z|< \delta$ and $|x-z|\geq \delta$ ??


text:
# 0.9 Logarithms

 Page 3 / 4

At t =0, the switch is closed. Because the voltage across the capacitor cannot change instantaneously, the value of the voltage across the resistor immediately after the closing of the swith ( v (0 + ) will be 10 V. After the switch has been closed, current will flow throughout the circuit and the voltage across the resistor will diminish exponentially. Because the decay can be described by the use of an exponential function, this is an example of exponential decay .

We can write an expression for the transient response, that is, the voltage across the resistor for positive time

$v\left(t\right)=\text{10}{e}^{-t/\text{RC}}$

Here, the units of v ( t ) are Volts.

The product of the resistance and the capacitance is called the time constant for the circuit and is often denoted as (τ). The time constant is measured in seconds. For this example, $\tau =\left(\text{100}×{\text{10}}^{3}\right)\left(8×{\text{10}}^{-6}\right)=\text{800}×{\text{10}}^{-3}=0\text{.}\text{800}s$

So the transient response of the circuit becomes

Question : At what instant of time will the transent response be equal to 5 Volts?

Solution : We have been asked to find the value of t that satisfies the following equation.

$5=\text{10}{e}^{-t/0\text{.}8}$

Let us divide each side of the equation by (10) and interchange the sides

${e}^{-t/0\text{.}8}=0\text{.}5$

We now take the natural logarithm of each side

$\frac{-t}{0\text{.}8}=-0\text{.}\text{693}$
$t=\left(0\text{.}8\right)\left(0\text{.}\text{693}\right)=0\text{.}\text{555}$

So we conclude that at the time 0.555 seconds after the switch closes, the value of the voltage across the resistor will be 5.0 volts.

## Compound interest

It is always essential to consider the financial aspects of an engineering project. The field of industrial engineering provides us with the analytical tools necessary to weigh the merits of competing designs.

Most situations surrounding the financial aspects of an engineering project involve the determination of what is most economical in the long run. That is, engineers must be aware of the costs and benefits of a project over a considerable period of time. In situations as these, it is important to consider the time value of money. Because of the existence of interest, the current value of a dollar is worth more than the value of a dollar some time in the future.

Interest can be defined as money that is paid for the use of money that has been borrowed. The rate of interest is the ratio between the interest chargeable or payable at the end of a period of time. This period of time is typically yearly, quarterly or monthly. In this module, we will restrict our attention to interest that is paid yearly or per annum .

As an example, a sum of money is invested at an annual rate of interest of 4%. One year later, the interest that would be paid on the investment would be $40 (4% of$1,000). So after one year, the initial sum would grow to a value of $1,040 one year later. Suppose that the$1,040 were invested for a second year at the end of the first year. At the end of the second year, the amount of interest that would be payable would be 4% of $1,040 or$41.60. The amount of interest earned in the second year exceeds the amount earned in the first year because of a phenomenon known as compound interest .

how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Do somebody tell me a best nano engineering book for beginners?
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
how did you get the value of 2000N.What calculations are needed to arrive at it
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Can someone give me problems that involes radical expressions like area,volume or motion of pendulum with solution