Mechanics A Level Maths Problem

Please could I have some help with this question?

A tram starts from rest at station A and accelerates uniformly for t1 seconds covering a distance of 1750m. It then travels at a constant speedvms-1 for t2 seconds covering a distance of 17500m. The tram then decelerates for t3 seconds and comes to rest at station B. Given that the total time for the journey is 7minutes and 3t1 = 4t3, find t1, t2 and t3 and the distance between station and B

Reply 1

mqb2766

21

Original post by aflowers

Please could I have some help with this question?

A tram starts from rest at station A and accelerates uniformly for t1 seconds covering a distance of 1750m. It then travels at a constant speedvms-1 for t2 seconds covering a distance of 17500m. The tram then decelerates for t3 seconds and comes to rest at station B. Given that the total time for the journey is 7minutes and 3t1 = 4t3, find t1, t2 and t3 and the distance between station and B

Its composed of 3 suvat phases where acceleration is constant in each phase. Id sketch the velocity-time curve, and mark on what you know and how the different phases are joined together. You can get the distances using area under the velocity time curve or using the usual suvat equations.

Think about how the unknowns (times, velocity v, and distances) can be used to start by finding one of them.

(edited 7 months ago)

Reply 2

aflowers

OP

9

Which suvat equations would you suggest I use in which order

Reply 3

mqb2766

21

Original post by aflowers

Which suvat equations would you suggest I use in which order

Can you upload your diagram with the info/appropriate suvats marked on? Without working it through, Id guess use the total time info to find the velocity in phase 2, then the rest should fall out.

(edited 7 months ago)

OP

OP

Would you mind talking it through

Reply 6

mqb2766

21

Original post by aflowers

Assuming that the equation you want to solve is
t1 + t2 + t3 = 7
and you want to get the left hand side in terms of v. Note the rhs could be 420 s if you want to work in sec rather than min

How could you write t2 in terms of v using the given info. Its a simple speed-distance-time. Then do something similar for phase 1 where the area of the triangle is equal to the distance (s = (u+v)*t/2), then use the t3<->t1 info for phase 3.

(edited 7 months ago)

Reply 7

Muttley79

20

Original post by aflowers

Would you mind talking it through

No, that's against the forum rules - we'll give hints

Reply 8

Muttley79

20

Original post by aflowers

Keep going ...

Reply 9

aflowers

OP

9

Am I on the right track

Reply 10

tiny hobbit

15

Remember that the 7 is in minutes so the total time is 7 X 60 = 420 seconds - beaten to it

(edited 7 months ago)

OP

I have written 420

OP

Please can I have another hint

Reply 13

mqb2766

21

Original post by aflowers

Am I on the right track

Sure so in terms of v
t1 = ...
t2 = ...
and use 3t1 = 4t3 to get t3 in terms of v. Then sub into the total time equation and solve for v.

Even if you dont solve them fully, annotating your sketch with the basic ideas/suvat for each phase really helps you see how to do the problem.