Question

2.10 A car is going 35 m/s when the driver applies the brakes. a) If the braking acceleration is −5.7 m/s2 , how far will the car go as its velocity drops from 35 m/s to 22 m/s? b) How much time will the change from 35 m/s to 22 m/s take?

2.11 As hockey puck slides to a stop, it goes 40 m across the ice in 3.7 s. a) What was the initial velocity of the puck? b) What was the acceleration of the puck?

Answer 1

2.10)
(a)
a = - 5.7 m/s^2
vi = 35 m/s
vf = 22 m/s

We know,
vf^2 = vi^2 + 2*a*s
22^2 = 35^2 - 2*5.7*s
s = 65 m

Distance moved by car, s = 65 m

(b)
vf = vi + a*t
22 = 35 - 5.7 * t
t = 2.28 s

2.11)
s = 40 m
t = 3.7 s

(a)
vf = vi + a*t
0 = vi - a*3.7
vi = a*3.7 ------1

s = vi*t + 1/2*a*t^2
40 = vi*3.7 - 1/2*a*3.7^2 --------2

Solving 1 & 2,
vi = 21.6 m/s
a = - 5.84 m/s^2

Initial Velocity of the puck, vi = 21.6 m/s

(b)
Acceleration of the puck, a = - 5.84 m/s^2

-ve sign means it is de accelerating