Question

Suppose that a 1100 kg car is traveling at 24 m/s (≈52.8 mph).
Its brakes can apply a maximum force of 5000 N while skidding to a stop.
(a) What is the maximum acceleration that the car can manage with its brakes?
(slowing down is also acceleration. Give your answer as a positive number.)
(b) What is the minimum distance required for the car to stop?
(c) How much time does this take? (from first application of the brakes until the car is stopped)
(d) What is the maximum coefficient of kinetic friction between tires and road?
(e) It starts to rain and the coefficient of friction between tires

Answer 1

(a) \[F = ma \rightarrow a = \frac{F}{m} = -4,5m/s^2\]
(c)
\[v_f = v_i + at \rightarrow t = \frac{v_f-v_i}{a} = \frac{0-24}{-4,5} = 5,3seconds.\]
Knowing the time we can find the distance.
\[x_f = x_i + v_i t+\frac{1}{2}at^2\]
Where x_i = 0 , v_i = 24m/s, t = 5,3 seconds, a = -4,5m/s.

Answer 2

d.

\[\sum F_x = breaks + friction = ma. \rightarrow 5000+\mu m g = ma \rightarrow \mu = \frac{ma - 5000}{mg}\]

friction = 0.46