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Flura [38]
3 years ago
15

A bike travels at 7.5 m/s along a straight road, whereas a car travels at 10.0 m/s along the same road and in the same direction

. When the car is 5.0 m ahead of the bike, it starts slowing down at a constant rate of 0.75 m/s^2 . a) How long it will take the bike to reach the car, and at what location (relative to its position when the car began slowing down) is the bike then? b) Plot the positions of the car and the bike as functions of time.

Physics
1 answer:
Ilya [14]3 years ago
3 0

Answer:

t = 8.3 s

Explanation:

As we know that

velocity of bike = 7.5 m/s

velocity of car is 10 m/s

deceleration of car is 0.75 m/s^2

part a)

velocity of bike with respect to car is given as

v_r = 7.5 - 10 = -2.5 m/s

acceleration of bike with respect to car is given as

a_r = 0 - (-0.75) = 0.75 m/s^2

now the distance of the bike with respect to car is given as

d = v_i t + \frac{1}{2}at^2

5 = (-2.5) t + \frac{1}{2}(0.75)t^2

t = 8.3 s

Part b)

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Hello! In this problem we have a uniformly varied rectilinear movement.

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