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.

Respuesta :

AMB000

Answer:

Car collides.

Explanation:

The (negative) acceleration experimented by the car will be given by Newton's 2nd Law [tex]a=\frac{f}{m}[/tex], 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 [tex]f=\mu N=\mu mg[/tex] since the normal force and the weight of the car are balanced on the vertical direction. All this gives us the equation [tex]a=\mu g[/tex]

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

[tex]d=\frac{-v_i^2}{2a}=\frac{-v_i^2}{2\mu g}[/tex]

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):

[tex]d=\frac{-(20m/s)^2}{2(-1)(0.8)(9.8m/s^2)}=25.5m[/tex]

Which means the car crashes against the tree.