# Mechanics question help

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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.

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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#2

There are essentially 5 different kinematic equations of

1)

2)

3)

4)

5)

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.

*constant acceleration*to remember.1)

2)

3)

4)

5)

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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(Original post by

There are essentially 5 different kinematic equations of

1)

2)

3)

4)

5)

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.

**FoxNuke**)There are essentially 5 different kinematic equations of

*constant acceleration*to remember.1)

2)

3)

4)

5)

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.

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

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#4

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

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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#5

(Original post by

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

**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

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(Original post by

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

**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

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(Original post by

Have you used the sketch at all? The area under the graph is the same as the distance gone.

**Muttley79**)Have you used the sketch at all? The area under the graph is the same as the distance gone.

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#8

(Original post by

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.

**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.

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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(Original post by

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.

**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.

I finally figured it out thanks to your help

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