Question

A 1000 kg car traveling is at 40.0 m/ s. To avoid hitting a second car the driver slams on his brakes.
The brakes provide a constant friction force of 8000 N. ( a) Find the minimum distance that the
brakes should be applied in order to avoid a collision with the vehicle in front? ( b) What would be
the speed of collision if the distance between the two vehicles is initially 45.0 m.

Answer 1

F = m a
-8000 = 1000 a
a = -8 m/s^2
v = 40 - 8 t
we want v = 0
8 t = 40
t = 5 seconds to stop (yikes!)

d = 40*5 - (1/2)(8)(25)
= 200 - 100
= 100 meters

for part b
how long to for 45 m?
45 = 40 t -4 t^2
solve quadratic for t
then
v = 40 - 8 t

Answer 2

a. F = M*a.
a = F/M = -8000/1000 = -8m/s^2.

V^2 = Vo^2 + 2a*d.
0 = 40^2 - 16d
d = 100m. to stop.

b. V^2 = Vo^2 + 2a*d.
V^2 = 40^2 - 16*45 =
V = ?.

Answer 3

It's helpful

Answer 4

Helpful