Question

A car of 1000 kg with good tires on a dry road can decelerate (slow down) at a steady rate of about 5.0 m/s2 when braking. If a car is initially traveling at 20 m/s (45 mi/h),

(a) How much time does it take the car to stop?

(b) What is its stopping distance?

(c) What is the deacceleration?

(d) How big is the net force to be applied to stop this car?

(e) Calculate the work done by this force (Work = Force * distance).

(f) During the stopping process, what happens to the car's kinetic energy?

Answer 1

a)

a = acceleration = - 5 m/s²

v_o = initial velocity = 20 m/s

v = final velocity = 0 m/s

t = time taken

Using the equation

v = v_o + at

0 = 20 + (- 5) t

t = 4 s

b)

d = stopping distance

Using the equation

v² = v_o² + 2 a d

0² = (20)² + 2 (-5) d

d = 40 m

c)

a = - 5 m/s²

d)

net force is given as

F_net = ma

F_net = 1000 x (- 5) = - 5000 N

e)

work done is given as

W = F d

W = - 5000 x 40 = - 2 x 10⁵ J

f)

as the car slow down, the speed of car decrease hence the kinetic energy of the car decrease to zero when the car stops