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

12.974 m/s

Step-by-step 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=√(v^2-2\mu gs)\\\Rightarrow u=√(0^2-2* -(9.81* 0.66)* 13)\\\Rightarrow u=12.974\ m/s`

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