M1

Hi guys I'm really struggling on this Mechanics question. I would really appreciate it if you could help me out.

Two trains A and B run on parallel straight tracks. Initially both are at rest in a station and level with each other. At time t=0, A starts to move. It moves with constant acceleration for 12 s up to a speed of 30m/s, and then moves at a constant speed of 30m/s. Train B starts to move in the same direction as A when t=40, where t is measured in seconds. It accelerated with the same initial acceleration as A, up to a speed of 60m/s. It then moves at a constant speed of 60m/s. It thenmoes at a constant speed of 60m/s. Train B overtakes A after both trains have reached their maximum speed. Train B overtakes A when t=T.

a). Sketch, on the same diagram, the speed-time graphs of both trains for 0≤t≤T. (3 marks)

b). Find the value of T. (9 marks)

How far have you got?

The gradient of a speed-time graph gives the acceleration.

I've found that the acceleration is 2.5 m/s2

That's right.

a = (v-u)/t

Rearrange this for t, and use v=60 to find the time that B reaches 60m/s.

That's right.

a = (v-u)/t

Rearrange this for t, and use v=60 to find the time that B reaches 60m/s.

Is the graph I drew for part 1 right or not

What would u be in this situation.
Is that all that is needed because the question is worth 9 marks.

That was for part a). B started at rest, so u = 0.

So is T=24 seconds

No, that is still for part a). Train B reaches maximum speed 24 seconds after it started to accelerate.

So does train B start accelerating at 40 seconds and reaches top speed of 60m/s at 64 seconds

Yes! I think you need to mark that on your x-axis as it is required for part b).

What steps would you then need to do to find out the value of T.

Distance = speed * time = area under the graph.

At the time when B overtakes A, they will have travelled the same distance, so the area under line A will equal the area under line B. I suggest that you use the formula for the area of a trapezium, rather than triangles and rectangles.

I'm getting different areas for both A and B. Can you tell me how to get the area for both of them pls.

Find them in terms of T, e.g. the area under B = 1/2 * 60 * (T - 40 + T - 64), using 'area = half x distance between the parallel sides x sum of the parallel sides'.

... This simplifies to B = 30 * (2T - 104).

Repeat for the area under A, equate them, and then solve for T.

Is area of A = 30T-180

Area under line A = 1/2 * 30 * (T + T - 12)
= 15 * (2T - 12)
= 30T - 180.

... Yes.

So would T=98 seconds