# Physics GCSE

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Thread starter 1 month ago
#1
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.
Last edited by Scarlett.sully; 1 month ago
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1 month ago
#2
(Original post by 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.
What have you tried so far? (I am a third-year physics student and have solved the problem)
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Thread starter 1 month ago
#3
(Original post by Nagromicous)
What have you tried so far? (I am a third-year physics student and have solved the problem)
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
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1 month ago
#4
(Original post by 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
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?
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1 month ago
#5
(Original post by Scarlett.sully)
Do you mean the equation
(final velocity)2 – (initial velocity)2 = 2 × acceleration × distance?
That's the one. Remember that since the car is supposed to be braking and coming to a stop, the final velocity is zero.
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Thread starter 1 month ago
#6
(Original post by 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 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.
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
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1 month ago
#7
(Original post by 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
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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Thread starter 1 month ago
#8
(Original post by 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"!)
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
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1 month ago
#9
(Original post by 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
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
Last edited by Incede; 1 month ago
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Thread starter 1 month ago
#10
(Original post by 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
Even tho the car is decelerating?
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1 month ago
#11
(Original post by Scarlett.sully)
Even tho the car is decelerating?
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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Thread starter 1 month ago
#12
(Original post by 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.
ohhhhh yaaa thank you very much, you've been a blessing : )
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4 weeks ago
#13
would it be 32m?
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4 weeks ago
#14
(Original post by 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.
would it be 32m? and by the way my question was 14m instead of 12m
Last edited by ur123; 4 weeks ago
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4 weeks ago
#15
(Original post by ur123)
would it be 32m? and by the way my question was 14m instead of 12m
No. It was 23.5m
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4 weeks ago
#16
(Original post by Nagromicous)
No. It was 23.5m
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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4 weeks ago
#17
(Original post by Nagromicous)
No. It was 23.5m
Is it possible if you could help me with some other questions too?
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4 weeks ago
#18
(Original post by 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
Yes, in that case it would be 32. OP's answer was 23.5 though.
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4 weeks ago
#19
(Original post by Nagromicous)
Yes, in that case it would be 32. OP's answer was 23.5 though.
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)
Last edited by ur123; 4 weeks ago
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4 weeks ago
#20
(Original post by 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)
kinetic energy = 1/2 * m * v^2
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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