# Physics GCSE

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Any help MUCH appreciated

A car drove past a speed camera at a constant speed

2 images were recorded 0.70s apart

the car traveled 12m between the 2 images taken

the maximum deceleration of the car is 6.25m/s squared

calculate the minimum braking distance for the car at the speed it passed the speed cameras.

A car drove past a speed camera at a constant speed

2 images were recorded 0.70s apart

the car traveled 12m between the 2 images taken

the maximum deceleration of the car is 6.25m/s squared

calculate the minimum braking distance for the car at the speed it passed the speed cameras.

Last edited by Scarlett.sully; 1 month ago

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#2

(Original post by

Any help MUCH appreciated

A car drove past a speed camera at a constant speed

2 images were recorded 0.70s apart

the car traveled 12m between the 2 images taken

the maximum deceleration of the car is 6.25m/s squared

calculate the minimum braking distance for the car at the speed it passed the speed cameras.

**Scarlett.sully**)Any help MUCH appreciated

A car drove past a speed camera at a constant speed

2 images were recorded 0.70s apart

the car traveled 12m between the 2 images taken

the maximum deceleration of the car is 6.25m/s squared

calculate the minimum braking distance for the car at the speed it passed the speed cameras.

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(Original post by

What have you tried so far? (I am a third-year physics student and have solved the problem)

**Nagromicous**)What have you tried so far? (I am a third-year physics student and have solved the problem)

I had a look in my books and i could only find an equation that is

distance x force = 0.5 x mass x velocity squared

i have no clue if that is along the right tracks other than that im goosed

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#4

(Original post by

heyyy

I had a look in my books and i could only find an equation that is

distance x force = 0.5 x mass x velocity squared

i have no clue if that is along the right tracks other than that im goosed

**Scarlett.sully**)heyyy

I had a look in my books and i could only find an equation that is

distance x force = 0.5 x mass x velocity squared

i have no clue if that is along the right tracks other than that im goosed

Given that, what you know is the following:

Velocity = 12/0.7 m/s

Max deceleration = 6.25 m/s^2,

and you need to work out braking distance.

Is there an equation you know which relates velocity, acceleration and distance? One of the suvat equations perhaps?

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#5

(Original post by

Do you mean the equation

(final velocity)2 – (initial velocity)2 = 2 × acceleration × distance?

**Scarlett.sully**)Do you mean the equation

(final velocity)2 – (initial velocity)2 = 2 × acceleration × distance?

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(Original post by

That doesn't help you because you are not given any forces or masses - you don't have anything to substitute into that equation. You are given a distance and a time which you can use to work out a velocity.

Given that, what you know is the following:

Velocity = 12/0.7 m/s

Max deceleration = 6.25 m/s^2,

and you need to work out braking distance.

Is there an equation you know which relates velocity, acceleration and distance? One of the suvat equations perhaps?

**Nagromicous**)That doesn't help you because you are not given any forces or masses - you don't have anything to substitute into that equation. You are given a distance and a time which you can use to work out a velocity.

Given that, what you know is the following:

Velocity = 12/0.7 m/s

Max deceleration = 6.25 m/s^2,

and you need to work out braking distance.

Is there an equation you know which relates velocity, acceleration and distance? One of the suvat equations perhaps?

(Original post by

That's the one. Remember that since the car is supposed to be braking and coming to a stop, the final velocity is zero.

**Nagromicous**)That's the one. Remember that since the car is supposed to be braking and coming to a stop, the final velocity is zero.

Ok so like that is

(12)2 – (0)2 =144

but im kinda confused on how that whole calculation is rearranged into working out the braking distance

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#7

(Original post by

Rightt

Ok so like that is

(12)2 – (0)2 =144

but im kinda confused on how that whole calculation is rearranged into working out the braking distance

**Scarlett.sully**)Rightt

Ok so like that is

(12)2 – (0)2 =144

but im kinda confused on how that whole calculation is rearranged into working out the braking distance

v^2 - u^2 = 2as

and get it in the form:

s = ...

No offence, but this is a very simple thing to rearrange.

The velocity isn't 12 btw - it's 12/0.7. Speed is distance divided by time. It travels 12 metres in a time of 0.7 seconds.

In summary, the variables in the equation are:

s, u, v, a

You know u = 12/0.7, v = 0, a = -6.25

Just rearrange the suvat equation to make s the subject, and substitute in the values of u, v and a. There is nothing more to it. This is as far as I can take you, unless there are any more specific questions (not "I don't understand"!)

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(Original post by

All you are trying to do is take

v^2 - u^2 = 2as

and get it in the form:

s = ...

No offence, but this is a very simple thing to rearrange.

The velocity isn't 12 btw - it's 12/0.7. Speed is distance divided by time. It travels 12 metres in a time of 0.7 seconds.

In summary, the variables in the equation are:

s, u, v, a

You know u = 12/0.7, v = 0, a = -6.25

Just rearrange the suvat equation to make s the subject, and substitute in the values of u, v and a. There is nothing more to it. This is as far as I can take you, unless there are any more specific questions (not "I don't understand"!)

**Nagromicous**)All you are trying to do is take

v^2 - u^2 = 2as

and get it in the form:

s = ...

No offence, but this is a very simple thing to rearrange.

The velocity isn't 12 btw - it's 12/0.7. Speed is distance divided by time. It travels 12 metres in a time of 0.7 seconds.

In summary, the variables in the equation are:

s, u, v, a

You know u = 12/0.7, v = 0, a = -6.25

Just rearrange the suvat equation to make s the subject, and substitute in the values of u, v and a. There is nothing more to it. This is as far as I can take you, unless there are any more specific questions (not "I don't understand"!)

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#9

(Original post by

Thank you for your patience, i'm a bit slow sometimes but honestly physics is hard for me, but non the less i got -1.37142856m and i somehow feel that is extremely wrong

**Scarlett.sully**)Thank you for your patience, i'm a bit slow sometimes but honestly physics is hard for me, but non the less i got -1.37142856m and i somehow feel that is extremely wrong

You need to use this equation v^2 - u^2 = 2as.

You want to find s so this rearranges to s = (v^2-u^2)/2a

Input these values that you have worked out/been given into the equation above. u= 12/0.7, v = 0, a = -6.25.

Since you want to find distance, this should not be a negative number so try again

Last edited by Incede; 1 month ago

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(Original post by

The user above posted the values that you can input into the equation

You need to use this equation v^2 - u^2 = 2as.

You want to find s so this rearranges to s = (v^2-u^2)/2a

Input these values that you have worked out/been given into the equation above. u= 12/0.7, v = 0, a = -6.25.

Since you want to find distance, this should not be a negative number so try again

**Incede**)The user above posted the values that you can input into the equation

You need to use this equation v^2 - u^2 = 2as.

You want to find s so this rearranges to s = (v^2-u^2)/2a

Input these values that you have worked out/been given into the equation above. u= 12/0.7, v = 0, a = -6.25.

Since you want to find distance, this should not be a negative number so try again

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#11

(Original post by

Even tho the car is decelerating?

**Scarlett.sully**)Even tho the car is decelerating?

Think about a real situation. If you are driving a car and press the brakes, you will still be moving forward and covering some distance whilst you slow down to a stop.

Deceleration means that your velocity is reducing with time. You can still be moving forwards whilst your velocity decreases from say 10 metres per second to 0 metres per second and this will cover a distance.

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(Original post by

Yep.

Think about a real situation. If you are driving a car and press the brakes, you will still be moving forward and covering some distance whilst you slow down to a stop.

Deceleration means that your velocity is reducing with time. You can still be moving forwards whilst your velocity decreases from say 10 metres per second to 0 metres per second and this will cover a distance.

**Incede**)Yep.

Think about a real situation. If you are driving a car and press the brakes, you will still be moving forward and covering some distance whilst you slow down to a stop.

Deceleration means that your velocity is reducing with time. You can still be moving forwards whilst your velocity decreases from say 10 metres per second to 0 metres per second and this will cover a distance.

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#14

**Incede**)

Yep.

Think about a real situation. If you are driving a car and press the brakes, you will still be moving forward and covering some distance whilst you slow down to a stop.

Deceleration means that your velocity is reducing with time. You can still be moving forwards whilst your velocity decreases from say 10 metres per second to 0 metres per second and this will cover a distance.

Last edited by ur123; 4 weeks ago

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#15

(Original post by

would it be 32m? and by the way my question was 14m instead of 12m

**ur123**)would it be 32m? and by the way my question was 14m instead of 12m

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#16

(Original post by

No. It was 23.5m

**Nagromicous**)No. It was 23.5m

the car travelled 14m between the two images being taken.

the maximum deceleration of the car is 0.625m/ssquared

calculate the minimum breaking distance for the car at the speed it passed the speed camera

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#17

(Original post by

No. It was 23.5m

**Nagromicous**)No. It was 23.5m

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#18

(Original post by

wait my full question was the camera recorded two images of the car at 0.70s apart.

the car travelled 14m between the two images being taken.

the maximum deceleration of the car is 0.625m/ssquared

calculate the minimum breaking distance for the car at the speed it passed the speed camera

**ur123**)wait my full question was the camera recorded two images of the car at 0.70s apart.

the car travelled 14m between the two images being taken.

the maximum deceleration of the car is 0.625m/ssquared

calculate the minimum breaking distance for the car at the speed it passed the speed camera

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#19

(Original post by

Yes, in that case it would be 32. OP's answer was 23.5 though.

**Nagromicous**)Yes, in that case it would be 32. OP's answer was 23.5 though.

Figure 12 shows a delivery van full of packages.

The driver delivers all the packages.

The empty van has a shorter stopping distance than the full van when driven at the same speed.

Explain why (3 marks)

Last edited by ur123; 4 weeks ago

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#20

(Original post by

Thank you so much! and by the way ive got another question if you could help me out with if thats possible?

Figure 12 shows a delivery van full of packages.

The driver delivers all the packages.

The empty van has a shorter stopping distance than the full van when driven at the same speed.

Explain why (3 marks)

**ur123**)Thank you so much! and by the way ive got another question if you could help me out with if thats possible?

Figure 12 shows a delivery van full of packages.

The driver delivers all the packages.

The empty van has a shorter stopping distance than the full van when driven at the same speed.

Explain why (3 marks)

since the van is empty, it has less kinetic energy for the same speed.

If we assume that the braking force is the same in both scenarios,

the work done slowing the van down is Force * [braking] distance

equating the work done and kinetic energy

1/2 * m * v^2 = F * d

since the kinetic energy of the empty van is less and the braking force is constant, the stopping distance must be less.

I do find it rather a strange question because, in reality, the braking force would depend on the weight as long as the braking is limited by the traction of the wheels (and not the power of the brakes). But regardless, what I said should probably be enough to get three marks.

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