Question

As soon as a traffic light turns green, a car speeds up from rest to 58.0 mi/h with constant acceleration 10.00 mi/h/s. In the adjoining bike lane, a cyclist speeds up from rest to 19.0 mi/h with constant acceleration 14.50 mi/h/s. 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 following are the answers to the questions presented:

The bicycle is ahead of the car for a time interval of 2.96 seconds.

The bicycle leads the car by a maximum distance of 8.21 feet.

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Answer 2

Answer:

Part a)

$t = 2.96 s$

Part b)

$d = 1.55 \times 10^{-3} miles$

Explanation:

Part a)

cruising velocity of bicycle = 19 mi/h

acceleration = 14.50 mi/h/s

For the motion of car

cruising velocity of car = 58 mi/h

acceleration = 10 mi/h/s

Both start from rest

Now we have

time taken by bicycle to reach its cruising speed is given as

$t = \frac{\Delta v}{a}$

$t = \frac{19 - 0}{14.50} = 1.31 s$

Time taken by car to reach its cruising speed

$t_2 = \frac{58- 0}{10} = 5.8 s$

now the distance moved by the car = distance moved by bicycle

till the time cycle moving in front of car

distance moved by the car = $\frac{1}{2}at^2$

distance moved by bicycle = $\frac{1}{2}at_1^2 + v(t - t_1)$

$\frac{1}{2}(10)(t^2) = \frac{1}{2}(14.50)(1.31^2) + 19(t - 1.31)$

$5t^2 = 12.44 + 19t - 24.89$

$t = 2.96 s$

Part b)

Bicycle will lead the car till the speed of car and speed of bicycle is same

so we have

$V_{cycle} = V_{car}$

$19.0 mi/h = 0 + 10(t)$

$t = 1.9 s$

so the distance between them is given as

$d = (\frac{1}{2}(14.50)(1.31^2) + 19(1.9 - 1.31)) - (\frac{1}{2}(10)(1.9^2))$

$d = (6.56 \times 10^{-3}) - (5.01 \times 10^{-3})$

$d = 1.55 \times 10^{-3} miles$