# Acceleration of a car required to reach point B with velocity D from point A with velocity C in time T

How do I get the acceleration of a car required to reach point B with velocity D from point A with velocity C in time T?

I understand the acceleration is not constant but how can I calculate the required acceleration as a function with time as its parameter.

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Can you give a bit more detail? Do you need to get from A to B in a specified time and reach B with a specified velocity? If so does it matter how the acceleration varies with time? – John Rennie Jul 27 '12 at 10:47
Yes, in a specified time and then the car's velocity should be D when it reaches the point B.  I am interested in the acceleration function and how it varies with time. – RickyTar Jul 27 '12 at 11:05
Very simple answer. Apply an immediate infinite acceleration (okay, deceleration) so as to bring the car to a complete stop. Accelerate to speed $|B-A|/T$ in the direction of point B. Upon arrival at point B, apply infinite deceleration to stop car, and infinite acceleration to give the desired final velocity. [Other solutions may be easier on the car and driver.] – Carl Brannen Jul 27 '12 at 11:40
You need some decision about the functional form of the acceleration, otherwise the problem is too poorly constrained and there are an infinite number of solutions. One possibility would be to travel at speed C for time $t_1$ then accelerate to speed D with some acceleration $a$ and finally travel at speed D until you reach point B. You can adjust $t_1$ and $a$ to make the travel time equal to your required time T. Is this the sort of solution you're looking for? – John Rennie Jul 27 '12 at 13:55

## 1 Answer

Let:

$a(t) = a + Kt$

It is given that:

$v(0) = C, \ v(T) = D$

$x(0) = A, \ x(T) = B$

Then,

$v(T) = C + aT + \dfrac{1}{2}KT^2 = D$

$x(T) = A + CT +\dfrac{1}{2}aT^2 + \dfrac{1}{6}KT^3 = B$

You have two equations and two unknowns, $a$ and $K$.

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