# Thread:

1. ## Physics1

A hockey puck hits the board with a velocity of 10m/s at an angle of 20deg to the board. It is deflected with a velocity of 8m/s at 24deg to the board. If the time of impact is 0.03s, what is the average acceleration of the puck?
$(2.3 x 10^2m/s^2$ out from the board at an angle of 107deg)

Thank you!

2. Acceleration is the rate of change of velocity in time. It asks for average acceleration. The formula of average acceleration is:

$a = \frac{\Delta V}{\Delta t}$

$\text{Attention !}:$ you can use this formula only for average acceleration, don't even think of using it for instantaneous acceleration.

So, we have to find $\Delta V$, which means the change of velocity. Remember that velocity is a vector. We have to find it using vector subtraction. You can't say $\Delta V = V_1 - V_0 = 8 - 10$. This only would be true if the velocities were in the same direction. In here, they have different angles.

O.K, here are the velocity vectors.

I leave you the subtraction work. After doing that, find the velocity of the subtracted vector, this will be our $\Delta V$. You know $\Delta t = 0.03 sec$. So use them in the average acceleration formula to finish.

3. It turns out to be the exact same calculation, but I'll bet you were supposed to use the "Impulse-Momentum" theorem, though I agree that simply using the acceleration definition is easier.

$\bar{F} \Delta t = m \Delta v$

The average force can be assumed to be a constant net force over the time interval so $\bar{F} = ma$

$ma \Delta t = m \Delta v$

$a \Delta t = \Delta v$

$a = \frac{\Delta v}{\Delta t}$

-Dan
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