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kvasek [131]
3 years ago
15

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.
Physics
1 answer:
erastovalidia [21]3 years ago
3 0

Answer:

Car collides.

Explanation:

The (negative) acceleration experimented by the car will be given by Newton's 2nd Law a=\frac{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=\frac{-v_i^2}{2a}=\frac{-v_i^2}{2\mu g}

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

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

Which means the car crashes against the tree.

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