posted by cylinder .
The table below shows how the braking distance x for a car depends on its initial speed u
u / ms-1 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= 30ms-1.
The mark scheme says:
Thinking distance= 30 X 0.6 =18m which I understand fully.
BRAKING DISTANCE =0.08 * U^2= 0.08 * 30^2= 72m IS THE BIT I DON’T UNDERSTAND. CAN SOMEONE PLEASE EXPALIN THE BRAKING DISTANCE STAGE AND WHY AND HOW YOU DO THIS BIT AS WELL AS ALL EQUATIONS INVOLVED!
Stopping distance= 18 +72= 90m which I understand fully.
Explain the working out of the braking distance for this question.
As physics - cylinder Thursday, November 9, 2017 at 7:37am
I wrote what I do know and the caps lock is what I dont know, and that is the working out for braking distance shown above is confusing to me.
As physics - Damon Thursday, November 9, 2017 at 8:39am
your table makes no sense to me
v = Vi + a t where t is AFTER 0.6 s
v = 30 + a t
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
remember total stopping time = t + .6
As physics - cylinder Thursday, November 9, 2017 at 9:27am
I am in as and we dont use those symbols so im confused
Sorry for the repost i didnt understand and i just wanted to show how the working and answer went so someone can better help me
I had this question too and didnt understand it either when i went through thisquestionin a level we dont usae Xi or any of those please write in ful lsentencs ad clearly explai
What i think
someone used the equation v=u+at and s=ut+1/2at^2
if t is zero then i think v is 30 after putting the numbers in. then.. oim confused using suvat and where the numbers came from please help
Please help i hvae been confused on this for ages
You have not posted all the information needed.
9.81 is the acceleration of gravity which has nothing to do with this. Why is there a 4 above 128?
Please post the complete problem you were given as given.
if acceleration and force is constant
(1/2) v^2 = F d
or initial kinetic energy = work done by F
Q11.The table below shows how the braking distance x for a car depends on its initial speed u.
| u / ms-1 | 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 30ms-1.
Thinking distance= 18m
and i didntknow 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:
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 wa for the table
5 is above 2
10 is above 8
the space i leftr arent posted/shown