Question

39. | A car is traveling at a steady 80 km/h in a 50 km/h zone. A
police motorcycle takes off at the instant the car passes it, accel-
erating at a steady 8.0 m/s².
a. How much time elapses before the motorcycle is moving as
fast as the car?
b. How far is the motorcycle from the car when it reaches this
speed?

Answer 1

Answer:

a) 2.8 s b) 30.8 m

Explanation:

a) For a uniform acceleration a the equation for velocity v is given by:

$v = at + v_0$

For the police car the given values are:

v₀ = 0, v = 80 km/h = 22.2 m/s , a = 8 m/s²

Solving for time t:

$t = \frac{v - v_0}{a} = 2.8 s$

b) In this time t the distance x traveled by the car with uniform velocity v is given by:

$x = vt + x_o$

For the car the given values are:

v = 80 km/h = 22.2 m/s, x₀ = 0, t = 2.8 s

x = 61.7

The distance s traveled by the police car with uniform acceleration a is:

$s = \frac{1}{2} at^{2} + v_0t + x_0$

The given values are:

a = 8m/s², v₀ = 0, x₀ = 0, t = 2.8 s

s = 30.9 m

The difference between the distance x traveled by the car and the distance s traveled by the police is 30.8 m

Answer 2

Explanation:

Given that,

Acceleration = 8.0 m/s²

Initial velocity = 50 km/h =13.89 m/s

Final velocity = 80 km/h = 22.22 m/s

(a). We need to calculate the time

Using equation of motion

$v = u+at$

Put the value into the formula

$22.22=13.89+8.0t$

$t=\dfrac{22.22-13.89}{8.0}$

$t=1.04\ sec$

The time elapsed is 1.04 sec.

(b). We need to calculate the distance

Distance covered by the motorcycle

$d = v\times t$

$d=22.22\times1.04$

$d=23.11\ m$

Distance covered by the police

Using equation of motion

$s = ut+\dfrac{1}{2}at^2$

$s=13.89\times1.04+\dfrac{1}{2}\times8.0\times(1.04)^2$

$s=18.77\ m$

The distance of the motorcycle from the car

$d'=d-s$

$d'=23.11-18.77$

$d'=4.34\ m$

The distance of the motorcycle from the car is 4.34 m.

Hence, This is the required solution.