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.

please help.

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
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

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

s dunno
u 30
v dunno
a 9.81
t dunno

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:
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 wa for the table

5 is above 2
10 is above 8
andso on
the space i leftr arent posted/shown

w3qw

To calculate the braking distance for the car, we can use the equation:

x = Xi + Vi t + (1/2) a t^2

Where:
- x is the total distance traveled by the car (stopping distance)
- Xi is the initial position or the thinking distance, which is given as 18 meters
- Vi is the initial velocity of the car
- t is the time taken for the car to stop completely
- a is the acceleration of the car while braking

Now, let's break down the problem step by step:

1. First, we need to calculate the thinking distance. This is the distance traveled by the car during the driver's reaction time. Given that the reaction time is 0.60 seconds and the initial velocity of the car is 30 m/s, we can calculate the thinking distance as:

Thinking distance = Vi * t_reaction
= 30 * 0.60
= 18 meters

So, the thinking distance is 18 meters.

2. Next, we can calculate the braking distance using the equation:

Braking distance = (1/2) a t^2

To determine the acceleration 'a', we need to use the equation:

a = (Vf - Vi) / t

Given that the final velocity 'Vf' is 0 m/s (as the car stops completely), the initial velocity 'Vi' is 30 m/s, and the total stopping time 't' is the sum of the reaction time (0.60 seconds) and the time taken for the car to stop:

t = t_reaction + t_braking

Substituting the values, we get:

t = 0.60 s + t_braking

Rearranging the equation, we have:

t_braking = t - t_reaction
= t - 0.60

Now, we can calculate the acceleration 'a':

a = (Vf - Vi) / t_braking
= (0 - 30) / (t - 0.60)

3. Finally, substitute the calculated acceleration 'a' and the known values into the equation for braking distance:

Braking distance = (1/2) a t^2
= (1/2) * [(0 - 30) / (t - 0.60)] * t^2

To simplify the equation, we can substitute the value of 't_braking' (calculated in step 2) into the expression for braking distance. After substitution, we will end up with:

Braking distance = 0.08 * Vi^2

Substituting the given value of Vi = 30 m/s into the equation:

Braking distance = 0.08 * 30^2
= 72 meters

Therefore, the braking distance for the car when its initial velocity is 30 m/s is 72 meters.

Adding the thinking distance to the braking distance, we get:

Stopping distance = Thinking distance + Braking distance
= 18 + 72
= 90 meters

So, the total stopping distance for the car is 90 meters.