Mechanics Modelling Problem of Two Sprinters Urgent Help!

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Alexandramartis
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Hello, I have been practising mechanics questions and have found the question below as a revision exercise. I have answered the first part of the question but I am really struggling with the second part, which consequently leads to the third.
I would greatly appreciate if anyone could comment upon what I have calculated so far and could suggest any improvements to point me in the right direction.

Two sprinters compete in a 100m race, crossing the finishing line together after 12 seconds. The two models, A and B are models for the motions of the two sprinters:

Model A: The sprinter accelerates from rest at a constant rate for 4 seconds and then travels at a constant speed for the rest of the race.

Model B: The sprinter accelerates from rest at a constant rate until reaching a speed 9ms^-1 and then travels at this speed for the rest of the race.

1. Model A; find the maximum speed and the initial acceleration of the sprinter.

The information given is s=100m, u=0ms^-1,v=?,a=?, t=12s

v=final and maximum velocity
Use the equation v=u+at
Therefore, in the first four seconds:
v=0+a*4
v=4a
Distance travelled during the first four seconds;
s=ut+1/2at^2
s=0*4+1/2*a*4^2
s=8a
Distance travelled in the final eight seconds;
s=ut+1/2at^2
s=0*4+1/2*a*8^2
s=32a
Total distance travelled = 100m
100m=8a+32a=40a
a=100/40=2.5ms^-2

Use the value obtained for the acceleration to find the maximum velocity which was previously shown to equal; v=4a
v=2.5*4
v=10ms^-1

2. Model B; Find the time taken to reach the maximum speed and the initial acceleration of the sprinter.

This is where I am struggling, I was thinking of using the suvat equation
t=2s/u+v to find the time taken to reach 9ms^-1 but realised that without knowing the distance travelled to get there I could not do so.
Could I try to find the distance travelled using s=u^2+v^2/2a
s=0^2+9^2/2a=81/2a
Which can be rearranged that a=2s/81
Then substitute the value of a to find the time taken to reach 9ms^-1?

My apologies but I have really confused myself here.

3. Sketch distance-time graphs for each of the sprinters on the same axis. Describe how the distance between the sprinters varies in the race.

I have began to draw the graph for Model A, which I have attached, but I have excluded including Model B until I have solved part 2.

I would be grateful of any help 😁
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mqb2766
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(Original post by Alexandramartis)
Hello, I have been practising mechanics questions and have found the question below as a revision exercise. I have answered the first part of the question but I am really struggling with the second part, which consequently leads to the third.
I would greatly appreciate if anyone could comment upon what I have calculated so far and could suggest any improvements to point me in the right direction.

Two sprinters compete in a 100m race, crossing the finishing line together after 12 seconds. The two models, A and B are models for the motions of the two sprinters:

Model A: The sprinter accelerates from rest at a constant rate for 4 seconds and then travels at a constant speed for the rest of the race.

Model B: The sprinter accelerates from rest at a constant rate until reaching a speed 9ms^-1 and then travels at this speed for the rest of the race.

1. Model A; find the maximum speed and the initial acceleration of the sprinter.

The information given is s=100m, u=0ms^-1,v=?,a=?, t=12s

v=final and maximum velocity
Use the equation v=u+at
Therefore, in the first four seconds:
v=0+a*4
v=4a
Distance travelled during the first four seconds;
s=ut+1/2at^2
s=0*4+1/2*a*4^2
s=8a
Distance travelled in the final eight seconds;
s=ut+1/2at^2
s=0*4+1/2*a*8^2
s=32a
Total distance travelled = 100m
100m=8a+32a=40a
a=100/40=2.5ms^-2

Use the value obtained for the acceleration to find the maximum velocity which was previously shown to equal; v=4a
v=2.5*4
v=10ms^-1

2. Model B; Find the time taken to reach the maximum speed and the initial acceleration of the sprinter.

This is where I am struggling, I was thinking of using the suvat equation
t=2s/u+v to find the time taken to reach 9ms^-1 but realised that without knowing the distance travelled to get there I could not do so.
Could I try to find the distance travelled using s=u^2+v^2/2a
s=0^2+9^2/2a=81/2a
Which can be rearranged that a=2s/81
Then substitute the value of a to find the time taken to reach 9ms^-1?

My apologies but I have really confused myself here.

3. Sketch distance-time graphs for each of the sprinters on the same axis. Describe how the distance between the sprinters varies in the race.

I have began to draw the graph for Model A, which I have attached, but I have excluded including Model B until I have solved part 2.

I would be grateful of any help 😁
Your A is not correct.
For the final 8 seconds, the initial speed is the same as the previous final speed and acceleration is zero.
Last edited by mqb2766; 1 month ago
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Alexandramartis
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(Original post by mqb2766)
Your A is not correct.
For the final 8 seconds, the initial speed is the same as the previous final speed and acceleration is zero.
Thank you for your reply. Oh should it be that in the final eight seconds the initial velocity is 4a?

Distance travelled in the final eight seconds;
s=ut+1/2at^2
s=4a*4+1/2*a*8^2
s=16a+32a=48a
Total distance travelled = 100m
100m=8a+48a=56a
a=100/40=1.76ms^-2 (3.s.f)


v=4a
v=4*1.76
v=7.14 ms^-2 3.s.f.


I do not think that this is correct though? Sorry I have become acutely confused 😳
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mqb2766
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(Original post by Alexandramartis)
Thank you for your reply. Oh should it be that in the final eight seconds the initial velocity is 4a?

Distance travelled in the final eight seconds;
s=ut+1/2at^2
s=4a*4+1/2*a*8^2
s=16a+32a=48a
Total distance travelled = 100m
100m=8a+48a=56a
a=100/40=1.76ms^-2 (3.s.f)


v=4a
v=4*1.76
v=7.14 ms^-2 3.s.f.


I do not think that this is correct though? Sorry I have become acutely confused 😳
Your first line is correct, but if there is no acceleration
Distance = speed * time
Don't over think it, a=0 in the final 8 secs.
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Alexandramartis
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(Original post by mqb2766)
Your first line is correct, but if there is no acceleration
Distance = speed * time
Don't over think it, a=0 in the final 8 secs.
Oh, ok so a=0 in the final 8 seconds. Do you mean the first line of my calculations or of the graph? Thank you for your help, sorry I just keep stumbling on this problem, like you say I may be overthinking it! 😳😁
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(Original post by Alexandramartis)
Oh, ok so a=0 in the final 8 seconds. Do you mean the first line of my calculations or of the graph? Thank you for your help, sorry I just keep stumbling on this problem, like you say I may be overthinking it! 😳😁
The first line of the previous post. The rest was wrong.
You should realise that luckily the numbers (8s = 2*4s) are such that the answer in the original post is correct, but the working is wrong.

The second is similar, work it the acceleration and time (or distance) for each phase and sum the distances and equate to 100 (again). The second phase is zero acceleration (again).
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Alexandramartis
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(Original post by mqb2766)
The first line of the previous post. The rest was wrong.
You should realise that luckily the numbers (8s = 2*4s) are such that the answer in the original post is correct, but the working is wrong.

The second is similar, work it the acceleration and time (or distance) for each phase and sum the distances and equate to 100 (again). The second phase is zero acceleration (again).
Thank you for your reply. I can understand what you are saying but now I am not sure how to begin correcting my answers.
So my workings for part 1 in the original post are incorrect?

Would it be that in the final eight seconds, u=4a and a=0ms^-2
Distance travelled in the final eight seconds;
s=ut+1/2at^2
s=4a*4+1/2*0*8^2
s=16a
Total distance travelled = 100m
100m=8a+16a=24a
a=100/24=4.17ms^-2 (3.s.f)


v=4a
v=16.7ms^-1 (3.s.f)
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mqb2766
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(Original post by Alexandramartis)
Thank you for your reply. I can understand what you are saying but now I am not sure how to begin correcting my answers.
So my workings for part 1 in the original post are incorrect?

Would it be that in the final eight seconds, u=4a and a=0ms^-2
Distance travelled in the final eight seconds;
s=ut+1/2at^2
s=4a*4+1/2*0*8^2
s=16a
Total distance travelled = 100m
100m=8a+16a=24a
a=100/24=4.17ms^-2 (3.s.f)


v=4a
v=16.7ms^-1 (3.s.f)
Sometimes, it seems your answers are too verbose and hence you make small but important mistakes. Pre gcse, you learnt that
s = ut = 4a*8 = ...
If the acceleration is zero, just use the speed - distance - time as referred to in a previous post. Hence the final answer is the same.
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Alexandramartis
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(Original post by mqb2766)
Sometimes, it seems your answers are too verbose and hence you make small but important mistakes. Pre gcse, you learnt that
s = ut = 4a*8 = ...
If the acceleration is zero, just use the speed - distance - time as referred to in a previous post. Hence the final answer is the same.
Thank you for your reply 👍Right, speed=distance/time
speed=80/8=10 ms^-1?

Sorry I put t=4 not 8 seconds didn't I.

s=ut+1/2at^2
s=4a*8+1/2*0*8^2
s=32a
Total distance travelled = 100m
100m=8a+32a=40a
a=100/40=2.5ms^-2 (3.s.f)


v=4a
v=10ms^-1 (3.s.f)
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mqb2766
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(Original post by Alexandramartis)
Thank you for your reply 👍Right, speed=distance/time
speed=80/8=10 ms^-1?

Sorry I put t=4 not 8 seconds didn't I.

s=ut+1/2at^2
s=4a*8+1/2*0*8^2
s=32a
Total distance travelled = 100m
100m=8a+32a=40a
a=100/40=2.5ms^-2 (3.s.f)


v=4a
v=10ms^-1 (3.s.f)
OK, so have a think about Q2. The second phase is zero acceleration, just like Q1 and you should be able to do something similar for the first phase to get a couple of simultaneous equations to solve.
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(Original post by mqb2766)
OK, so have a think about Q2. The second phase is zero acceleration, just like Q1 and you should be able to do something similar for the first phase to get a couple of simultaneous equations to solve.
So, during the constant acceleration in the second phase;
a=0 ms ^-2, u=9ms^-1 and v=9ms^-1

I am so sorry I just do not know how to progress. I really appeciate your help and do not want you to think that I am not trying.
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(Original post by Alexandramartis)
So, during the constant acceleration in the second phase;
a=0 ms ^-2, u=9ms^-1 and v=9ms^-1

I am so sorry I just do not know how to progress. I really appeciate your help and do not want you to think that I am not trying.
In the second phase (time length T)
s = 9T
So use something similar to the first phase to get a distance in terms of time (12-T), and sum the two distances to 100 and solve for T. You've got a few suvat equations to choose from which involve distance, time, acceleration, ... see which one(s) are the most suitable.
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(Original post by mqb2766)
In the second phase (time length T)
s = 9T
So use something similar to the first phase to get a distance in terms of time (12-T), and sum the two distances to 100 and solve for T. You've got a few suvat equations to choose from which involve distance, time, acceleration, ... see which one(s) are the most suitable.
Sorry how does s=9T
Is this using s=ut+1/2at^2?

I can see how you have states to find a distance in terms of time, 12-T, but I am uncertain how to sum the distances to solve for T?
Is this like saying 9T+12-T=s
s=100m
100=9T+12-T
100=8T+12
8T=88
T=11s?
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mqb2766
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(Original post by Alexandramartis)
Sorry how does s=9T
Is this using s=ut+1/2at^2?

I can see how you have states to find a distance in terms of time, 12-T, but I am uncertain how to sum the distances to solve for T?
Is this like saying 9T+12-T=s
s=100m
100=9T+12-T
100=8T+12
8T=88
T=11s?
This is the third time in this thread. When acceleration is zero
distance = speed * time
This is KS2 / KS3, you're doing A-level, you must know this?

For each phase, find the distance in terms of T. Then sum those two distances (functions of T) and equate to 100 and solve for T.
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(Original post by mqb2766)
This is the third time in this thread. When acceleration is zero
distance = speed * time
This is KS2 / KS3, you're doing A-level, you must know this?

For each phase, find the distance in terms of T. Then sum those two distances (functions of T) and equate to 100 and solve for T.
Oh of course, I see using d=s*t that s=9T
So the distance for the second phase is 9T
For the first phase, how would I find the distance using d=s*t as the speed is changing as the sprinter accelerates to reach 9ms^-1?
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(Original post by Alexandramartis)
Oh of course, I see using d=s*t that s=9T
So the distance for the second phase is 9T
For the first phase, how would I find the distance using d=s*t as the speed is changing as the sprinter accelerates to reach 9ms^-1?
In the first phase acceleration is constant but not zero, so you use suvat.
Have a think about which equations you could use and try?
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(Original post by mqb2766)
In the first phase acceleration is constant but not zero, so you use suvat.
Have a think about which equations you could use and try?
If the acceleration in the first phase is constant, could I use s=u^2+v^2/2a?
Or to find the distance of the first phase in terms of time s= t/2(u+v)
s=9t/2
Then, sum the distances and equate to 100;
9t/2+9t=100
27T=200
T=200/27=7.4 s

I do not think this is correct either though.
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(Original post by Alexandramartis)
If the acceleration in the first phase is constant, could I use s=u^2+v^2/2a?
Or to find the distance of the first phase in terms of time s= t/2(u+v)
s=9t/2
Then, sum the distances and equate to 100;
9t/2+9t=100
27T=200
T=200/27=7.4 s

I do not think this is correct either though.
The first equation is wrong. The second equation is correct.
When you try and use a suvat equation, write down which terms you know. The two times in the different phases are not the same time, which is what you've written.
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Alexandramartis
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Ok, in phase two; s=? u=9ms-1 v=9ms-1 a=0 t=?
distance=speed*time
s=9T

In phase one; s= ? u=0ms-1 v=9ms-1 a=? t=?
s= t/2(u+v)
s=t/2(0+9)
s=9t/2
This is equal to 12-T
12-T=9t/2

Sorry where should I go from here or am I still wrong?
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In phase 1, the displacement s is not the time 12-T. The time t = 12-T.
Do as suggested above.
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