Need help!!

[madoscar65]

Hi People,

A dodgem car of mass 200kg is driven south into a rigid barrier at an initial speed of 5.0m/s and the dodgem rebounds at a speed of 2.0m/s, it is in contact with the barrier for 0.2s. Calculate the average acceleration of the car during its interaction with the barrier.

Thanks !!

[kamil9876]

Just use the definition of average acceleration.

Let us take south as positive.



So that means the direction is north as we have taken south as positive.

[QuantumJG]

Ok so u = 5m/s & v = -2m/s (depending on which way you define as positive and negative)

v = u + at

So,

-2 = 5 + 0.2a

-7 = 0.2a

a = -35ms^-2

So what was the problem?

I just want to add:

Kamil is more precise in how she? took on the problem as that is the definition of average. In VCE physics though, our suvat (distance, initial velocity, final velocity, acceleration & time) formulas deal with acceleration being constant and in a realistic case they give you the average. If you want to deal with non-constant acceleration you need to deal with calculus (which you can easily use to derive the suvat formulas).

[madoscar65]

I also got the same answer as you which is 35m/s^2, but the answer says 3.5m/s^2... , I thought I must have done something wrong. But it appear to be the back of the book that is wrong.. Problem all solved now. Thanks

[bomb]

Yeah I remember doing that question last year haha, the back of the book is wrong