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 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
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.
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
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.
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
i also dont get how to work out damons last line if his is right
plz answer
To calculate the braking distance for the car at a given initial speed, we need to use the relationship mentioned in the table, which states that doubling the speed increases the distance by a factor of 4.
Let's take a look at how the braking distance changed as the initial speed increased:
| u / ms-1 | 5 | 10 | 20 | 40 |
| X / m | 2 | 8 | 32 | 128 |
From the table, we can observe that as the initial speed doubles, the braking distance increases by a factor of 4. For example, at an initial speed of 20 m/s (which is double the initial speed of 10 m/s), the braking distance is 32 m (which is 4 times the braking distance at 10 m/s, which is 8 m).
Now, let's calculate the braking distance for an initial speed of 30 m/s.
We know that the braking distance doubles when the initial speed doubles. Therefore, if the initial speed was 20 m/s and the braking distance was 32 m, then when the initial speed is 30 m/s, the braking distance will be double that amount.
Braking distance at 30 m/s = 2 * 32 m = 64 m.
So, the braking distance for an initial speed of 30 m/s is 64 m.
To find the stopping distance, we need to consider both the thinking distance and the braking distance.
Thinking distance is given as 18 m.
Stopping distance = Thinking distance + Braking distance = 18 m + 64 m = 82 m.
Therefore, the stopping distance for a car with an initial speed of 30 m/s is 82 m.