# Determining average acceleration during point of impact

1. Feb 13, 2015

### miles davis

1. The problem statement, all variables and given/known data
Hello, new to the physics world, I am trying to self-teach through reading textbooks, videos etc and application.

I came across a problem I do not know how to begin to answer.

A dodgem car of mass 200g is driven due south into a rigid barrier at an initial speed of 5.0 ms^-1. The dodgem rebounds at a speed of 2.0m^-1. It is in contact with the barrier for 0.2s. Calculate:

a) the average acceleration of the car during its interaction with the barrier
b) the average net force applied to the car during its interaction with the barrier.

2. Relevant equations

I am unsure which equations to utilise: F=ma, a= Delta v/ Delta t or kinematic equations

3. The attempt at a solution

a) a = (-2-5)/(0.2) but for barrier unsure?
b) Fnet= 0.2 X unknown acceleration

2. Feb 13, 2015

### BvU

Hello M, welcome to PF :)

a= Delta v/ Delta t is a good one for average a. Delta v is v before - v afterwards. Note that v is a vector, so the before and after have different signs in this calculation. a is also a vector and it should be clear that it points opposite to the original direction of motion of the car !

And you will do fine with F = ma in part b. So not what you write in your attempt: the 0.2 seconds have no place there....
 Sorry, didn't see that it was a toy car. 0.2 kg is ok.

Again, F is a vector. Check the sign using your intuition!

3. Feb 13, 2015

### miles davis

Thank you BvU,

It is liberating receiving help from someone when navigating the minefield of physics.

Your guidance is perfectly clear, I am just unsure as to how to incorporate delta T here: if I were to divide the change in velocity (delta V) which appears to be -7 (in the opposite direction as you pointed out! :) ) by the change in time (0.2s?) (Final T - Initial T; 0.2 - 0) I end up with 35ms-2N, however this does not appear to be correct!

Apologies for my rudimentary understanding, I am still grappling with content.

4. Feb 13, 2015

### BvU

Perhaps they want a different direction in the answer: you found $-2 - 5$ m/s per $0.2$ s = $-35$ m/s2 If I am correct. This is taking the 5 m/s South as positive. So the answer is $-35$ m/s2 South which you can correctly report as $35$ m/s2 North.

But if the program insists on receiving the acceleration in a southerly direction, then entering $-35$ m/s2 (with or without the S, I don't know) is the only way to get it to approve your answer.

( I'm deducing all this from your " this does not appear to be correct" -- without knowing if there is a computer program involved at all...)

For me your result is impeccable. No need for apologies; on the contrary: kudos for your wanting to learn !