Question

The minimum stopping distance for a car from an initial 100 km/h is 60 m on level ground. What is the stopping distance when it moves (a) down a 10 degree incline; (b) up a 10 degree incline ? Assume that the initial speed and the surface are unchanged.

Answer 1

Given : initial velocity = 100 km/hr = 27.78 m/s, minimum distance travelled = 60 m.

SOLUTION:

We use equation v² - u² = 2.a.s ( where , v is final velocity, u is initial velocity, a is acceleration, s is distance )

Put in values gives acceleration,

  v² - u² = 2 x a x s

0² - 27.78² = 2 x a x 60

   a_levelled = - 6.431 m/s² [ negative sign means deceleration ,on level ground ]

part (a) car moves down a 10 degree incline plane (as shown in figure)

we will have additional acceleration component due to gravity along the plane ( that is g .cos 80),

so total  acceleration will be:

a _down =    a_levelled + a gravity ( considering g as 9.8 m/s²)

  a _down = - 6.431 m/s² + 9.8 x cos 80 m/s²

a _down = - 4.729 m/s²

{ Important point : as coefficient of friction is not given in question we ignore friction force and acceleration due to friction along the incline is not considered)

Put in values in     v² - u² = 2 x a x s  gives distance,

     v² - u² = 2 x a x s

0² - 27.78² = 2 x -4.729 x s  

s = 81.590 m

ANSWER : For part (a) is car moves 81.590 m down a 10 degree incline plane

Similarly

for part (b) car moves up a 10 degree incline plane (as shown in figure)

total  acceleration will be:

a _up =    a_levelled + a gravity

  a _up = - 6.431 m/s² - 9.8 x sin 10 m/s²

a _up= - 8.132 m/s²

Put in values in     v² - u² = 2 x a x s  gives distance,

     v² - u² = 2 x a x s

0² - 27.78² = 2 x -8.132 x s  

s = 47.45 m

ANSWER : For part (a) is car moves 47.45 m down a 10 degree incline plane