Question

A car moving at some speed hits the brakes and skids to a stop after 13 m on a level road. If the coefficient of friction for the road conditions of dry concrete is 0.66, what was the car's original speed (in m/s) before braking?

Answer 1

Answer:

12.974 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s²

$\mu$ = Coefficient of friction =0.66

a = Acceleration = $\mu g$

$v^2-u^2=2as\\\Rightarrow -u^2=2\mu gs-v^2\\\Rightarrow u=\sqrt{v^2-2\mu gs}\\\Rightarrow u=\sqrt{0^2-2\times -(9.81\times 0.66)\times 13}\\\Rightarrow u=12.974\ m/s$

Car's original speed before braking was 12.974 m/s