# Further maths mechanics help - complicated SUVAT Watch

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Q) Two trains start together from rest at a station. They travel for 3 minutes and come to rest together at the next station. One train, A, accelerates uniformly at the rate of 1 m/s for 30s, continues at that speed for the next 2 minutes and decelerates uniformly for the last 30 seconds. The other train, B, has a uniform acceleration for 90 seconds and uniform deceleration for the remaining time.

Distance between stations= 4500m

Find the largest distance between the trains.

How do you go about this? I used simultaneous Suvats but keep getting the wrong answer Correct answer is 360m according to textbook.

Thanks

Distance between stations= 4500m

Find the largest distance between the trains.

How do you go about this? I used simultaneous Suvats but keep getting the wrong answer Correct answer is 360m according to textbook.

Thanks

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(Original post by

Q) Two trains start together from rest at a station. They travel for 3 minutes and come to rest together at the next station. One train, A, accelerates uniformly at the rate of 1 m/s for 30s, continues at that speed for the next 2 minutes and decelerates uniformly for the last 30 seconds. The other train, B, has a uniform acceleration for 90 seconds and uniform deceleration for the remaining time.

Distance between stations= 4500m

Find the largest distance between the trains.

How do you go about this? I used simultaneous Suvats but keep getting the wrong answer Correct answer is 360m according to textbook.

Thanks

**theeetimdoherty**)Q) Two trains start together from rest at a station. They travel for 3 minutes and come to rest together at the next station. One train, A, accelerates uniformly at the rate of 1 m/s for 30s, continues at that speed for the next 2 minutes and decelerates uniformly for the last 30 seconds. The other train, B, has a uniform acceleration for 90 seconds and uniform deceleration for the remaining time.

Distance between stations= 4500m

Find the largest distance between the trains.

How do you go about this? I used simultaneous Suvats but keep getting the wrong answer Correct answer is 360m according to textbook.

Thanks

You need to use a bit of thought as to the shape, and deciding what feature of the graph represents when the two trains are a max. distance apart (there are two occurrences - I'd use the first.). Then it's a case of doing the necessary calculations. In order: What's the acceleration of B? When are the two trains at the "feature"? How far has each of them gone? And find the difference between the two.

Can't see it being done easily with just suvat due to the changing accelerations.

Post working/diagram if you need further assistance.

**Edit:**PS. Agree with 360m

Last edited by ghostwalker; 3 months ago

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I suggest using a velocity-time graph.

You need to use a bit of thought as to the shape, and deciding what feature of the graph represents when the two trains are a max. distance apart (there are two occurrences - I'd use the first.). Then it's a case of doing the necessary calculations. In order: What's the acceleration of B? When are the two trains at the "feature"? How far has each of them gone? And find the difference between the two.

Can't see it being done easily with just suvat due to the changing accelerations.

Post working/diagram if you need further assistance.

**ghostwalker**)I suggest using a velocity-time graph.

You need to use a bit of thought as to the shape, and deciding what feature of the graph represents when the two trains are a max. distance apart (there are two occurrences - I'd use the first.). Then it's a case of doing the necessary calculations. In order: What's the acceleration of B? When are the two trains at the "feature"? How far has each of them gone? And find the difference between the two.

Can't see it being done easily with just suvat due to the changing accelerations.

Post working/diagram if you need further assistance.

**Edit:**PS. Agree with 360mMy.method: Using the velcocity-time graph, I determined that the times at whjch distance between the trains is at a maximum both occur whilst train A is at a constant speed. Thus, I set the SUVAT for point A to be (S=S, U=30, V=30, A=0, T=T1 or T2). For Train B, acceleration becomes negative after 90 seconds and so there are two SUVATs, which I found to be (S=S-x, U= 0, V=50, A= 5/9, T=T1) and (S=S x, U=50, V=0, A=-5/8, T=T2), where x is the distance between Train A and Trajn B. I believe I then used the equation S= 0.5(u v)t and then v=u at to find that x=15t, and finding t to be 58s (off the top of my head) but that'd mean x=870 (maybe I have to subtract the actual value of S, tho shouldn't if I used simultaneous equation to cancel S). Regardless, I got the wrong answer. Given there's two T values it'd be a lot easier if there was a polynomial.

Thanks in advance tho, you're surprisingly helpful for a stranger on TSR.

Last edited by theeetimdoherty; 3 months ago

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(Original post by

Its the equatjons bit at the end that Idk how to do.

My.method: Using the velcocity-time graph, I determined that the times at whjch distance between the trains is at a maximum both occur whilst train A is at a constant speed.

**theeetimdoherty**)Its the equatjons bit at the end that Idk how to do.

My.method: Using the velcocity-time graph, I determined that the times at whjch distance between the trains is at a maximum both occur whilst train A is at a constant speed.

(Original post by

Thus, I set the SUVAT for point A to be (S=S, U=30, V=30, A=0, T=T1 or T2).

**theeetimdoherty;undefined**)Thus, I set the SUVAT for point A to be (S=S, U=30, V=30, A=0, T=T1 or T2).

But in your equations for train B, it's clear you're using the start of the journey for t=0. So, these two items don't tie up.

Can't really follow the rest of your working. Some of it looks as if it could be right, but it's impossible to tell.

There's no requirement for a polynomial

Last edited by ghostwalker; 3 months ago

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**theeetimdoherty**)

Q) Two trains start together from rest at a station. They travel for 3 minutes and come to rest together at the next station. One train, A, accelerates uniformly at the rate of 1 m/s for 30s, continues at that speed for the next 2 minutes and decelerates uniformly for the last 30 seconds. The other train, B, has a uniform acceleration for 90 seconds and uniform deceleration for the remaining time.

Distance between stations= 4500m

Find the largest distance between the trains.

How do you go about this? I used simultaneous Suvats but keep getting the wrong answer Correct answer is 360m according to textbook.

Thanks

Since it's symmetric, we need only really consider the first half of the journey.

In order to obtain the distance-time graph, we may then simply integrate the piecewise function as shown.

The distance between the two trains as a function of time between the times 30 and 90 is then given by d_A-d_B=30t-450-5t^2/18 which is maximised at t=54 (this is shown by setting the first derivative to zero).

Assuming that the maximum distance between the trains occurs during this period (this is easily shown), the maximum separation between the trains is obtained by evaluating d_A-d_B at t=54, whereupon we obtain 360 as required.

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Although apparantly complex, with a bit of reasoning, this can be turned into quite a simple question that doesn't require polynomials, or integration & differentation in all their glory.

As noted previously, and diagrammed. Our graphs are both symmetrical about t=90, hence the trains meet at the start, at the end, and at t=90 (half way through their journey - the line of symmetry).

As noted previously, and diagrammed. Our graphs are both symmetrical about t=90, hence the trains meet at the start, at the end, and at t=90 (half way through their journey - the line of symmetry).

Spoiler:

Train A accelerates faster than B, so they are separating initially.

Train A stops accelerating, but is still travelling faster than B (so separating), until the speed of B equals that of A. Thereafter B is travelling faster than A, and they are coming together to meet again at t=90.

So, first greatest separation occurs when speed of A = speed of B.

For the first half (t<90), acceleration of B = 5/9, and it reaches the speed of A, after 30/(5/9) = 54 seconds.

We now work out the distrances travelled by A and B, and subtract one from t'other to get the desired result.

And by symmetry the greatest separation as B is decelerating is the same.

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Train A accelerates faster than B, so they are separating initially.

Train A stops accelerating, but is still travelling faster than B (so separating), until the speed of B equals that of A. Thereafter B is travelling faster than A, and they are coming together to meet again at t=90.

So, first greatest separation occurs when speed of A = speed of B.

For the first half (t<90), acceleration of B = 5/9, and it reaches the speed of A, after 30/(5/9) = 54 seconds.

We now work out the distrances travelled by A and B, and subtract one from t'other to get the desired result.

And by symmetry the greatest separation as B is decelerating is the same.

Last edited by ghostwalker; 3 months ago

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#7

The distance between a and b will contine increasing as long as b is traveling faster than a. Since the b will be slower than a at 30m/s you can setup a suvat equation for b (s=s, u=50, v=30, a=-5/9, t=t) using this equation the distance is 1440, but this is not the total distance, you need to add the distance from the first part of the journey of b, 1440+2250= 3690. You now need to find how far a has traveled at that point, this means you need to find the time where b has traveled 2690, use the suvat equation (s=2690, u=50, v=30, a=-5/9, t=t) to find t=36, then add 36 to the first part to find the total time (90+36=126). Now we need to do a suvat for a at the time of 126, to do this we do 126-30=96 because the first part of train a lasts for 30 seconds. Now do a suvat for this (s=s, u=30, v=30, a=0, t=96) with this we find s=2880. You add up the distance both parts of journey a, 2880+450=3330. You then take this away from the distance of b, 3690-3330=360

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