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
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
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.
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 ??
To calculate the braking distance for the given question, you need to understand the relationship between the braking distance and the initial speed of the car. From the table, it is mentioned that doubling the speed increases the distance by a factor of 4.
To find the equation that relates the braking distance (x) and the initial speed (u), you can use the following steps:
1. Look at the table and find two sets of data where one speed is double the other speed. In this case, 10m/s is double 5m/s and 40m/s is double 20m/s.
2. Calculate the ratio of the braking distances for these two sets of data. From the table, the braking distance for 10m/s is 8m and for 5m/s is 2m. So the ratio is 8/2 = 4.
3. Now, let's calculate the ratio for the other set of data. The braking distance for 40m/s is 128m and for 20m/s is 32m. So the ratio is 128/32 = 4.
4. Since both ratios are equal to 4, it indicates that the braking distance (x) is proportional to the square of the initial speed (u). Therefore, the equation can be written as x = k * u^2, where k is a constant.
5. To find the value of k, substitute the values from one set of data into the equation. Let's use the data for 10m/s and 8m: 8 = k * 10^2. Solving this equation, k = 0.08.
Now that we have the equation x = 0.08 * u^2, we can calculate the braking distance for a given initial speed of 30m/s.
1. Calculate the braking distance using the equation: x = 0.08 * 30^2 = 0.08 * 900 = 72m.
Note: The value in the mark scheme you mentioned (0.08 * u^2) is indeed correct for the equation of braking distance.
However, it is important to note that the question also mentions a reaction time of the driver, which needs to be considered while calculating the overall stopping distance. The reaction time is given as 0.60s and the thinking distance is calculated as 18m.
To find the total stopping distance, you need to add the thinking distance to the braking distance. Therefore, the total stopping distance would be 18m + 72m = 90m.
Additionally, "Xi" refers to the initial position of the car, which in this context would be the starting point or position of the car before it starts to brake.