Question

A car is being driven along a country road on a dark and rainy night at a speed of 20 m/s. The section of road is horizontal and straight. The driver sees that a tree has fallen and covered the road ahead. Panicking, the driver locks the brakes at a distance of 20 m from the tree. If the coefficient of friction between the wheels and road is 0.8, determine the outcome.

Answer 1

Car collides.

Step-by-step explanation:

The (negative) acceleration experimented by the car will be given by Newton's 2nd Law
`a=(f)/(m)`, where f is the friction force, which is the only force acting on the car on the horizontal direction. Its value is given by the formula
`f=\mu N=\mu mg` since the normal force and the weight of the car are balanced on the vertical direction. All this gives us the equation
`a=\mu g`

To calculate the distance traveled by the car, we will use the equation for accelerated motion
`v_f^2=v_i^2+2ad` on the form (since it comes to a rest
`v_f=0m/s`):

`d=(-v_i^2)/(2a)=(-v_i^2)/(2\mu g)`

Which for our values is (taking the direction of movement as positive, which makes a negative, so we add a -1 to account for this):

`d=(-(20m/s)^2)/(2(-1)(0.8)(9.8m/s^2))=25.5m`

Which means the car crashes against the tree.