Question

As soon as a traffic light turns green, a car speeds up from rest to its cruising speed 22.3 m/sec with constant acceleration 4.02 m/sec2 . In the adjoining bicycle lane, a cyclist speeds up from rest to its cruising speed 8.94 m/sec with constant acceleration 5.81 m/sec2 . Each vehicle maintains constant velocity after reaching its cruising speed. (a) For what time interval is the bicycle ahead of the car? (b) By what maximum distance does the bicycle lead the car?

Answer 1

Answer:  (a) The bicycle is ahead of the car for 4 s.

               (b) The bicycle leads the car by the maximum distance of 55 m.

Explanation:

(a)

Use the equation of the motion to calculate the time taken by the car.

$v=u+at$  

As it is given in the problem, a car speeds up from rest to its cruising speed 22.3 m/sec with constant acceleration 4.02 m/sec2 .

Put u=0, v=22.3 m/sec and a=4.02 m/sec^2.

$22.3=0+4.02t$

$t=\frac{22.3}{4.02}$

t= 5.5 s

Use the equation of the motion to calculate the time taken by the  bicycle.

$v=u+at_{1}$

As it is given in the problem, a cyclist speeds up from rest to its cruising speed 8.94 m/sec with constant acceleration 5.81 m/sec^2.

Put u=0, v=8.94 m/sec and a=5.81 m/sec^2.

$8.94=0+5.81t_{1}$

$t_{1}=\frac{8.94}{5.81}[tex]t_{1}=1.5 s$

Calculate the time interval for which the bicycle is ahead of the car.

$t-t_{1}= 5.5 s - 1.5s$

$t-t_{1}= 4s$

Therefore, the bicycle is ahead of the car for 4 s.

(b)

Use the equation motion to calculate the distance covered by the car.

$S=ut+\frac{1}{2}at^{2}$

As it is given in the problem, a car speeds up from rest to its cruising speed 22.3 m/sec with constant acceleration 4.02 m/sec^2 .

Put t= 5.5 s, u=0 s and a=4.02 m/sec^2.

$S=(0)t+\frac{1}{2}(4.02)(5.5)^{2}$

$S= 60.8 m$

Use the equation motion to calculate the distance covered by the bicycle.

$S_{1}=ut+\frac{1}{2}at^{2}$

As it is given in the problem, a cyclist speeds up from rest to its cruising speed 8.94 m/sec with constant acceleration 5.81 m/sec^2.

Put t= 1.5 s, u=0 s and a=5.81 m/sec^2.

$S_{1}=(0)t+\frac{1}{2}(5.18)(1.5)^{2}$

$S_{1}= 5.8 m$

Calculate the maximum distance covered by the bicycle to lead the car.

$S-S_{1}=60.8-5.8=55m$

Therefore, the bicycle leads the car by the maximum distance of 55 m.