soligem
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My brother's having trouble with his homework and my mechanic's really rusty, so help's really appreciated 😅 Thank you 😊

Question:

A car is moving along a straight horizontal road. The speed of the car as it passes the point A is 25m/s and the car maintains this speed for 30 s. The car then decelerates uniformly to a speed of 10 m/s. The speed of 10 m/s is then maintained until the car passes the point B.
The time taken to travel from A to B is 90 s and AB = 1410 m.

a) Sketch a speed-time graph to show the motion of the car from A to B.
(He's already done that no problem)

b) Calculate the deceleration of the car as it decelerates from 25 m/s to 10 m/s.
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FoxNuke
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There are essentially 5 different kinematic equations of constant acceleration to remember.
1) v = u + at
2) v^2 = u^2 + 2as
3) s = ut + \frac{at^2}{2}
4)  s = vt - \frac{at^2}{2}
5) s = \frac{(u+v)t}{2}
where v = final velocity, u = initial velocity, s = displacement, a = acceleration and t = time taken

Besides that here are some other hints:
- The area under a velocity-time graph is the displacement, and the distance under a speed-time graph is the total distance moved.
- The gradient of a velocity-time graph is the acceleration, and the gradient of a speed-time graph is the magnitude of the acceleration (direction is not considered).

I think you can find the deceleration by considering the slope of the 25m/s to 10m/s line. You do need to solve for time though and that's where AB=1410m and time from A to B = 90s comes in.
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soligem
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(Original post by FoxNuke)
There are essentially 5 different kinematic equations of constant acceleration to remember.
1) v = u + at
2) v^2 = u^2 + 2as
3) s = ut + \frac{at^2}{2}
4)  s = vt - \frac{at^2}{2}
5) s = \frac{(u+v)t}{2}
where v = final velocity, u = initial velocity, s = displacement, a = acceleration and t = time taken

Besides that here are some other hints:
- The area under a velocity-time graph is the displacement, and the distance under a speed-time graph is the total distance moved.
- The gradient of a velocity-time graph is the acceleration, and the gradient of a speed-time graph is the magnitude of the acceleration (direction is not considered).

I think you can find the deceleration by considering the slope of the 25m/s to 10m/s line. You do need to solve for time though and that's where AB=1410m and time from A to B = 90s comes in.
I'm aware of all the things mentioned but I'm still sorta stuck and can't figure out what to apply to get the missing time 😅
How do I use the total time and distance?
We figured out the distance covered in the first 30 s which is equal to 750 m, which makes the rest of the distance covered between unknown time t till the end of the 90 s = 660 m, but I still really don't know how this helps :confused:
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timif2
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It takes 30s so use suvat, given u = 25, V = 10, a = ? T=30

Solve for a, you get ans < 0 so deceleration
You can also find gradient of line from 25 to 10 on the graph
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Muttley79
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(Original post by soligem)
I'm aware of all the things mentioned but I'm still sorta stuck and can't figure out what to apply to get the missing time 😅
How do I use the total time and distance?
We figured out the distance covered in the first 30 s which is equal to 750 m, which makes the rest of the distance covered between unknown time t till the end of the 90 s = 660 m, but I still really don't know how this helps :confused:
Have you used the sketch at all? The area under the graph is the same as the distance gone.
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soligem
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(Original post by timif2)
It takes 30s so use suvat, given u = 25, V = 10, a = ? T=30

Solve for a, you get ans < 0 so deceleration
You can also find gradient of line from 25 to 10 on the graph
The 30 s is the time for the initial phase of constant velocity not that of the deceleration, so we still have the time variable for the deceleration part missing and that's the whole problem 😅
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soligem
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(Original post by Muttley79)
Have you used the sketch at all? The area under the graph is the same as the distance gone.
Yes we have, that's what I just described when I said we figured out the distance in the first 30 s which is equal to 750 m (area from graph), but I'm still missing out on what we could do now with the new values.
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Muttley79
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(Original post by soligem)
Yes we have, that's what I just described when I said we figured out the distance in the first 30 s which is equal to 750 m (area from graph), but I'm still missing out on what we could do now with the new values.
The time for the decel and 10 bit of the motion is how many seconds?

Split that into two parts and you should be able to find the time by calling one bit T and the other bit 'above answer - T'.

When they ask you to sketch first it's usually a hint that this will help.
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soligem
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(Original post by Muttley79)
The time for the decel and 10 bit of the motion is how many seconds?

Split that into two parts and you should be able to find the time by calling one bit T and the other bit 'above answer - T'.

When they ask you to sketch first it's usually a hint that this will help.
Thank you so much! 😍
I finally figured it out thanks to your help :flower:
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Muttley79
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(Original post by soligem)
Thank you so much! 😍
I finally figured it out thanks to your help :flower:
You are welcome - without the sketch it is harder to 'see'
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