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.
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.
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
I am in as and we dont use those symbols so im confused
To understand the working out of the braking distance in this question, let's start by explaining the equation involved:
The equation used to calculate the braking distance is given by:
Braking Distance = 0.08 * u^2
In this equation, "u" represents the initial speed of the car in meters per second (ms^-1). The equation states that the braking distance is equal to 0.08 times the square of the initial speed.
To calculate the braking distance when u = 30 ms^-1, we can plug this value into the equation:
Braking Distance = 0.08 * (30^2)
= 0.08 * 900
= 72 meters
Therefore, the braking distance of the car when its initial speed is 30 ms^-1 is 72 meters.
It's important to note that the equation used in this calculation is based on an approximation and may not be entirely accurate in all situations. However, it is a common formula used to estimate braking distances in certain scenarios.