# Stopping Distances! Help!

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#1
So, I’ve got a question from the May 2016 Physics paper. Asks me to calculate Stopping Distance for a car. I know the formula is Thinking + Braking. I also know Thinking is Speed x Reaction time. Don’t understand the braking distance part. Will try and attach question and mark scheme below.

Thanks for any help given in advance!

dreamerz11 :-)
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#2
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#3
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#4
Just to clarify, I’m fine with part b (i)
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#5
It’s the 0.08 bit that I’m not sure about
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3 years ago
#6
In part b (i) you found that the braking distance x = 0.08u^2.

For u = 30, the braking distance x = 0.08(30)^2 = 72 and the thinking distance = 0.6 * 30

Add the braking distance and the thinking distance to get the stopping distance which is 90m
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#7
(Original post by BobbJo)
In part b (i) you found that the braking distance x = 0.08u^2.

For u = 30, the braking distance x = 0.08(30)^2 = 72 and the thinking distance = 0.6 * 30

Add the braking distance and the thinking distance to get the stopping distance which is 90m
I get the thinking bit. I’m still unsure about the 0.08 part. How’d you get that from the first bit? It get x is directly proportional to u^2.
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#8
I got 64m from doing:

Thinking = 30 x 0.60 = 18m (C1)
Braking =
u= 30
v= 0
a= 9.81
So, v^2 = u^2 + 2as (rearranged for “s”) = (u^2-v^2)/2xa. (0^2-30^2)/2x9.81 = -45.871....
45.871.....+18 = (rounded to 2sf) 64m
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#9
I know 64m is wrong because of the mark scheme. But I don’t understand where the other 26m comes from
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3 years ago
#10
If you calculate the magnitude the deceleration from the values in the table, (using SUVAT) it'll show you a constant value of deceleration at -6.25 (negative as it's decelerating), assuming those values are all for the same vehicle.

S=? U=30 V=0 A= -6.25 T=X s=0^2-30^2/2*(-6.25) = 72
72+18 = 90m
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3 years ago
#11
(Original post by dreamerz11)
I got 64m from doing:

Thinking = 30 x 0.60 = 18m (C1)
Braking =
u= 30
v= 0
a= 9.81
So, v^2 = u^2 + 2as (rearranged for “s”) = (u^2-v^2)/2xa. (0^2-30^2)/2x9.81 = -45.871....
45.871.....+18 = (rounded to 2sf) 64m
9.81 is only ever used in the vertical plane, the vehicle is not falling towards Earth so you don't assume g = 9.81
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#12
(Original post by Invective)
9.81 is only ever used in the vertical plane, the vehicle is not falling towards Earth so you don't assume g = 9.81
How are you getting -6.25 for acceleration? Sorry if I’m being stupid, I really do appreciate the help! 😊
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3 years ago
#13
(Original post by dreamerz11)
How are you getting -6.25 for acceleration? Sorry if I’m being stupid, I really do appreciate the help! 😊
Lol, getting something wrong doesn't make you stupid otherwise I'd be a complete idiot.

S=(any value from the table, 128 for example) U=(because I just chose 128, u= 40) V=0, A= ? T= X
v^2-u^2/2s = a = 0^2-40^2/2*128 = -6.25 which make sense as the vehicle is decelerating.
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#14
(Original post by Invective)
Lol, getting something wrong doesn't make you stupid otherwise I'd be a complete idiot.

S=(any value from the table, 128 for example) U=(because I just chose 128, u= 40) V=0, A= ? T= X
v^2-u^2/2s = a = 0^2-40^2/2*128 = -6.25 which make sense as the vehicle is decelerating.
Thank you! I understand this now! That’s makes loads more sense now! Thank you soooo much 😊
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