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

The bicycle is ahead of the car for a time interval of 5.52 seconds and the maximum distance the bicycle leads the car is 19.06 meters.

Step-by-step explanation:

To solve this problem, we can use the equations of motion. Let's first calculate the time it takes for the car and the bicycle to reach their cruising speeds:

The car:

Final velocity (v) = 22.3 m/s

Acceleration (a) = 4.02 m/s²

Using the equation v = u + at, where u is the initial velocity and t is the time:

22.3 = 0 + 4.02t

t = 22.3 / 4.02 = 5.52 seconds

The bicycle:

Final velocity (v) = 8.94 m/s

Acceleration (a) = 5.81 m/s²

Using the same equation:

8.94 = 0 + 5.81t

t = 8.94 / 5.81 = 1.54 seconds

(a) To find the time interval when the bicycle is ahead of the car:

Since the bicycle reaches its cruising speed faster, it will be ahead of the car for the time it takes the car to reach its cruising speed:

Time interval = 5.52 seconds.

(b) To find the maximum distance the bicycle leads the car:

We can use the equation of motion s = ut + (1/2)at², where s is the distance, u is the initial velocity, a is the acceleration, and t is the time:

For the bicycle:

s = 0 + (1/2)(5.81)(1.54)² = 7.03 m

For the car:

s = 0 + (1/2)(4.02)(5.52)² = 26.09 m

The maximum distance the bicycle leads the car is 26.09 - 7.03 = 19.06 m.