STAV 2010 Question6 motion

[Cuntryboner]

I am unsure of how the answer is justified. Isn't the force after 8 up to 12 seconds purely the braking force? I thought the speed before he applied the brakes was from t=0 to t=8, as the force is constant, and there is also a uniform acceleration for this time, when the graph's gradient goes negative this must be deceleration yeah?. However the answer states it's t=0 up to t=12. So confused atm. Any help would be great

[fady_22]

I am unsure of how the answer is justified. Isn't the force after 8 up to 12 seconds purely the braking force? I thought the speed before he applied the brakes was from t=0 to t=8, as the force is constant, and there is also a uniform acceleration for this time, when the graph's gradient goes negative this must be deceleration yeah?. However the answer states it's t=0 up to t=12. So confused atm. Any help would be great

This is a force TIME graph. It does not indicate acceleration directly.

I would use impulse=area=change in momentum to find the initial velocity (keeping in mind that that final velocity is 0).

[m@tty]

For the first 8 seconds, yes, there is constant BRAKING force and thus constant deceleration(as this force is opposing motion). Then the braking force weans.

But this is pretty much irrelevant to the question. What you need to do here is determine the area beneath the graph, this is impulse as it is effectively F*t which is Ns or momentum. The loss of momentum implies what the original momentum must have been. Then solve for the change in speed, this is the original speed as the final speed is zero.

From memory the answer to this question is wrong, at least the answer which kenhung had.

Should be

[Cuntryboner]

Cheers m@tty, I understand