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Kinematics mechanics assignment - 2 objects, find initial displacement watch

1. 1. The problem statement, all variables and given/known data

A cheetah is estimated to be able to run at a maximum speed of whilst and antelope can run at a maximum speed of . A cheetah at rest sees an antelope at rest and starts running towards it. The antelope immediately starts running away. Both cheetah and antelope are assumed to move with constant acceleration and reach their maximum speeds in 4 seconds. Assuming that both run along the same straight line and that the cheetah catches the antelope in 15 seconds, find the distance between the animals when they first started moving.

2. Equations I have been given

I have been staring at this for a good hour now and am yet to come up with the answer which I know to be 126m. How do i find this value using those equations and the data given? Our mechanics tutor is away with an illness and we have been set work which we haven't been taught properly so any help would be greatly appreciated!

Kind regards, derRoboter
2. (Original post by derRoboter)
1. The problem statement, all variables and given/known data

A cheetah is estimated to be able to run at a maximum speed of whilst and antelope can run at a maximum speed of . A cheetah at rest sees an antelope at rest and starts running towards it. The antelope immediately starts running away. Both cheetah and antelope are assumed to move with constant acceleration and reach their maximum speeds in 4 seconds. Assuming that both run along the same straight line and that the cheetah catches the antelope in 15 seconds, find the distance between the animals when they first started moving.

2. Equations I have been given

I have been staring at this for a good hour now and am yet to come up with the answer which I know to be 126m. How do i find this value using those equations and the data given? Our mechanics tutor is away with an illness and we have been set work which we haven't been taught properly so any help would be greatly appreciated!

Kind regards, derRoboter
Perhaps I am missing something, but I can't see a solution with that information. The best I can get is 13(v-u) as the separation where v is the max. speed of the cheetah etc.
3. (Original post by ghostwalker)
Perhaps I am missing something, but I can't see a solution with that information. The best I can get is 13(v-u) as the separation where v is the max. speed of the cheetah etc.

yeah, fixed that, see my first post again!
4. (Original post by derRoboter)
yeah, fixed that, see my first post again!
The suvat equations are only valid when you are dealing with constant acceleration. In this case the first 4 seconds. So how far did each of them travel in that first 4 seconds? Then for the remaining 11 seconds they are travelling at constant speed, so how far did each travel. Subtract the two and there's your answer.

It's rather long winded, and to be honest if you use a velocity time graph instead, then it's only a couple of lines of working. Get to the formula I gave and just plug in the numbers.
5. (Original post by derRoboter)
I have been staring at this for a good hour now and am yet to come up with the answer which I know to be 126m. How do i find this value using those equations and the data given? Our mechanics tutor is away with an illness and we have been set work which we haven't been taught properly so any help would be greatly appreciated!

Kind regards, derRoboter
I've not done the working out, but 126m is definitely way too big. I'd suspect it should be more like 20-30m at most.
6. (Original post by ghostwalker)
The suvat equations are only valid when you are dealing with constant acceleration. In this case the first 4 seconds. So how far did each of them travel in that first 4 seconds? Then for the remaining 11 seconds they are travelling at constant speed, so how far did each travel. Subtract the two and there's your answer.

It's rather long winded, and to be honest if you use a velocity time graph instead, then it's only a couple of lines of working. Get to the formula I gave and just plug in the numbers.

cheetah-

antelope-

cheetah displacement -

first 4 secs

final 11 secs

total S =

antelope displacement -

first 4 secs

final 11 secs

total S =

Does that look correct? According to the answer i know to be correct (its on the back of the assignment paper) it is wrong :s

Thanks again for this!
7. (Original post by Rubgish)
I've not done the working out, but 126m is definitely way too big. I'd suspect it should be more like 20-30m at most.
Can you take a look at my workings and see if the are correct?
8. (Original post by derRoboter)
Can you take a look at my workings and see if the are correct?
Your distance travelled in the final 11 seconds is incorrect in each case. During this time period the velocity is constant.

You can use the formula s=ut where u is the maximum velocity.

The final answer is 126m to 3 sig.fig.
9. (Original post by derRoboter)
Can you take a look at my workings and see if the are correct?
Ghostwalker got it correct, your equation for s = 0.5 (u+v)t should have u as the same as v, as speed is a constant (so effectively the equation becomes s = ut.

And 126m is correct as well, when I was guessing in my head I forgot to times the top speed section by 11

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