OCR MEI M1 Question

I've only recently started learning the suvat equations and I'm finding it hard to depict what to use with this question. It's only a 2 mark question so the answer must be simple but I cannot for the life of me solve it.

Cars A and B are travelling in the same direction along a straight road.
The time t is in seconds. At t = 0, car A is at rest. It accelerates at 3 ms-1 for 01t ≤≤ and then travels at a constant speed.
Car B travels at 15 ms-1 for 03t ≤≤ and then accelerates at 1 ms-2 until it reaches a speed of 25 ms-1, after which it continues at this constant speed.

Given that car A is 500 m behind car B at t = 0, at what value of t does car A catch up with car B?

Could anyone please point me in the right direction?

Reply 1

benwellsday

12

What is 01t and 03t?
At t=0, the car accelerates at 3ms-2, from rest, so you know initial velocity, acceleration, time . This can be used in one of the equations to find v at time 01t, for example v = u + at = 0 + 3(01t). This is useful, but you also need the distance it travelled, s = ut + 0.5at^2 should help.

The whole question requires you find equations for the displacement of each (don't forget the -500m for A!!) and these should have a t variable in them. You can then solve the equations simultaneously at the point where their displacements are equal.

Reply 2

73iso

17

saw this on edx
the question is as follows:
Cars A and B are travelling in the same direction along a straight road. The time t is in seconds. At t=0, car A is at rest. It accelerates at 3ms−2 for 0⩽t⩽10 and then travels at a constant speed. Car B travels at 15ms−1 for 0⩽t⩽30 and then accelerates at 1ms−2 until it reaches a speed of 25ms−1, after which it continues at this constant speed.
Given that car A is 500m behind car B when t=0, at what value of t does car A catch up with car B?
The answer is 60 but I can't figure out why?
For 60s:
displacement of a=-500+0.5x3x10^2+50x30
=-500+150+1500=1150m
displacement of b=450+450+50=950m

Reply 3

old_engineer

11

Original post by 73iso

saw this on edx
the question is as follows:
Cars A and B are travelling in the same direction along a straight road. The time t is in seconds. At t=0, car A is at rest. It accelerates at 3ms−2 for 0⩽t⩽10 and then travels at a constant speed. Car B travels at 15ms−1 for 0⩽t⩽30 and then accelerates at 1ms−2 until it reaches a speed of 25ms−1, after which it continues at this constant speed.
Given that car A is 500m behind car B when t=0, at what value of t does car A catch up with car B?
The answer is 60 but I can't figure out why?
For 60s:
displacement of a=-500+0.5x3x10^2+50x30
=-500+150+1500=1150m
displacement of b=450+450+50=950m

You need to look again at the displacement of B. I agree with the initial 450m (15m/s for 30s) but not with the other two distances. Please post your working for those two numbers.

(edited 3 years ago)

OCR MEI M1 Question

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