Question

A 1100 kg car is traveling at 24 m/s.
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?
(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)
seconds.
(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 and road is reduced to μk = 0.2.
Now what is the minimum distance required to stop?

Answer 1

(a) 4.54 m/s^2

Answer 2

(b) 63.36 m

Answer 3

(c) 5.28 seconds

Answer 4

well, i'll leave the rest to you, i have to go to sleep

Answer 5

aw man i was waiting for d

Answer 6

Thank you very much.

Answer 7

Have a good night

Answer 8

The force of friction is defined as\[F_f = F_N \mu ~~\text {where} ~ \mu ~ \text{is the static coefficient of friction and}~F_N~\text{ is the normal force}\]Assuming during the braking process the tires never slip, the Force of friction must equal the force of braking. Therefore, \[F_f = 5000 [N] = 1100 [kg] ~9.81 [m/s^2]~\mu\]Solve for mu.

Answer 9

"Assuming during the braking process the tires never slip, the Force of friction must equal the force of braking"
thats the part i did not understand.. thanks