Damon/BobPursley/Physicstutor/whichoneisright
posted by cylinder .
Q11.The table below shows how the braking distance x for a car depends on its initial speed u.
 u / ms1  5  10  20  40 
 X / M  2  8  32  128 
The relationship between x and u is: doubling speed increases distance by a factor of 4.
i)The reaction time of a driver is 0.60s. Calculate the stopping distance of the car when u is 30ms1.
Thinking distance= 18m
and i didn't know how to work out braking distance so i looked at the mark scheme to get this number
so braking distance=72m
then i used this number to get the stoppign distance and got:
stopping idstamce=
18m + 72m = 90m stopping distance.
Please show mehow to work out braking distance for this questions, show where you got numbersfrom, what equations used and full wqriting not jut symbols please/
by the way for the table
5 is above 2
10 is above 8
and so on
the space i left aren't posted/shown
I reposted like you asked Damon and i included the full question and my working out and what i need help with so pplease help Damon or anyone else
Physics  bobpursley today at 6:26am
distance=constant*v^2
32=constant*10^2
constant=.32
now try it for 20
distance=.32*20^2=128, so you have the formula. So at 30m/s,
distance=.32*30^2=288
now you have to add the distance traveled before breaking, 30*.6=18m
total distance=18+288 m.
The72 m is bogus.
Physics  cylinder today at 7:14am
Thank you, finally someone who answered, I have been waiting ages for this questions answer. So you saying that your 100 percent sure that is the answer and the mark scheme which the examiner made is wrong??
This is what the mark scheme said for braking distance:
BRAKING DISTANCE =0.08 * U^2= 0.08 * 30^2= 72m which I went over many times but did not understand at all.
Just so you know their working out
Physics  cylinder today at 8:00am
Please tell me what you think of the mark scheme answer what they did and if now looking at this mark schee answer you stillthink that it is wrong and your answer is right? just as i want to make sure my notes are right.
Physics  cylinder today at 8:07am
Damon said:
v = Vi + a t where t is AFTER 0.6 s
v = 30 + a t
so
x = Xi + Vi t + (1/2) a t^2
a will be negative of course
Xi is 18 when t = 0
x = 18 + 30 t + (1/2) a t^2
i dont know what xi means
if this wiill result in yuor answer or the mark scheme answer but just sent it for you to confirm which working out is right and what answer is rightand what the mark scheme is saying as i dontget where the 0.08 came from
Physics  cylinder today at 8:09am
i also dont get how to work out damons last line if his is right
Which working outand answer is right please explain and where did the 0.08 come from

Damon/BobPursley/Physicstutor/whichoneisright 
bobpursley
I see the table is formatted wrong
10m/s >8m
20m/s >32m
40m/s >128m
so the question is: what is the stopping distance at 32m/s if it takes .6sec to hit the brakes?
distance=constant*v^2 From the table:
8=constant*10^2
constant=.08
now try it for 20
distance=.08*20^2=128, so you have the formula. So at 30 m/s,
distance=.08*30^2=72m
now you have to add the distance traveled before breaking, 30*.6=18m
total distance=18+72 m.
This entire thread is a history maker, one that I shall not soon forget. 
Damon/BobPursley/Physicstutor/whichoneisright 
Cylinder
This is the clearest explanation given for this thread thank you.
But may I ask what equation you used at the beginning as I can’t seem to find it when I searched it online but at first I thought it was d=vt but b=velocity isn’t squared in this equation.
Also, what do you mean by constant? I know what the word means: a number that stays the same but I just don’t understand why this number is the constant and what is this the speed time ??
Respond to this Question
Similar Questions

AP Physics
To stop a car, you require first a certain reaction time to begin braking. Then the car slows under the constant braking deceleration. Suppose that the total distance moved by your car during these two phases is 56.7 m when its initial … 
college math
Braking distance The braking distance y in feetthat it takes for a car to stop on wet ,level pavement can be estimated by y = 1/9 x^2, where the x is the speed of the car in miles per hour Find the speed associated with with each braking … 
Physics
The braking distance of a car is 29m. If the speed of the car is increased by 45%, what is the car's new braking distance? 
Maths
The braking distance of a car is directly proportional to te square of it's speed. When the speed is p metres per second, the braking distance is 6m. When the speed is increased by 300%, find (a) an expression for speed of the car … 
As physics
The table below shows how the braking distance x for a car depends on its initial speed u u / ms1 5.0 10 20 4 x / m 2.0 8.0 32 128 the reaction time of a driver is 0.60s. Calculate the sopping distance of the car when u= 30ms1. The … 
As physics
The table below shows how the braking distance x for a car depends on its initial speed u u / ms1 5.0 10 20 4 x / m 2.0 8.0 32 128 the reaction time of a driver is 0.60s. Calculate the sopping distance of the car when u= 30ms1. The … 
As physics
The table below shows how the braking distance x for a car depends on its initial speed u u / ms1 5.0 10 20 4 x / m 2.0 8.0 32 128 the reaction time of a driver is 0.60s. Calculate the sopping distance of the car when u= 30ms1. The … 
physics As
The table below shows how the braking distance x for a car depends on its initial speed u u / ms1 5.0 10 20 4 x / m 2.0 8.0 32 128 the reaction time of a driver is 0.60s. Calculate the sopping distance of the car when u= 30ms1. The … 
As physics
Q11.The table below shows how the braking distance x for a car depends on its initial speed u.  u / ms1  5  10  20  40   X / M  2  8  32  128  The relationship between x and u is: doubling speed increases distance by a … 
Physics
Q11.The table below shows how the braking distance x for a car depends on its initial speed u.  u / ms1  5  10  20  40   X / M  2  8  32  128  The relationship between x and u is: doubling speed increases distance by a …