Question

A Chevrolet Corvette convertible can brake to a stop from a speed of 60.0 mi/h in a distance of 123 ft on a level roadway. What is its stopping distance on a roadway sloping downward at an angle of 26.0°?

Answer in ft

Answer 1

using the formulas

F = m a = u mg so u = a/g

vf^2=v0^2 + 2ad

here u is the coefficient of friction

converting ,

v0=60mi/hr = 26.9m/s
a is the acceleration
d=distance = 123ft = 37.5m

now , substituting the values

0=26.9^2 + 2*a*37.5

a = - 9.65m/s^2

now findiing the coefficient

coeff of friction = a/g = 0.98  

since forces are acting down the plane

mg sin(theta) - u mg cos(theta) = ma

a = 9.8(sin 26- 0.98 cos 26) = -4.335m/s^2

now using the formula 'vf^2=v0^2 + 2ad

vf=0
v0=26.9m/s
a=-4.335m/s^2
here d is the distance
0=26.9^2 - 2*(4.335m/s/s)*distance

d= 83.46 m

d = 273.8189 ft

therefore stopping distance = 273.8189 ft