Question

A man is driving his car with speed 55.0 mi/h on a horizontal stretch of road. (a) When the road is wet, the coefficient of static friction between the road and the tires is 0.102. Find the minimum stopping distance (in m).

Answer 1

302.076411995 m

Step-by-step explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

`\mu` = Coefficient of friciton = 0.102

`a=-(f)/(m)\\\Rightarrow a=-(\mu mg)/(m)\\\Rightarrow a=-\mu g`

`v^2-u^2=2as\\\Rightarrow s=(v^2-u^2)/(2(-\mu g))\\\Rightarrow s=(0^2-((55* 1609.34)/(3600))^2)/(2* 0.102* -9.81)\\\Rightarrow s=302.076411995\ m`

The stopping distance is 302.076411995 m.