Question

ou are sitting in your car at rest at a traffic light with a bicyclist at rest next to you in the adjoining bicycle lane. As soon as the traffic light turns green, your car speeds up from rest to 49.0 mi/h with constant acceleration 7.00 mi/h/s and thereafter moves with a constant speed of 49.0 mi/h. At the same time, the cyclist speeds up from rest to 23.0 mi/h with constant acceleration 15.00 mi/h/s and thereafter moves with a constant speed of 23.0 mi/h.[HINT: this is a great problem to use some graphs to help you visualize what is going on. There are more than 2 important clock readings in this problem.] (a) For what time interval (in s) after the light turned green is the bicycle ahead of your car?

Answer 1

Till the time car is just adjacent to the bicycle we can say

distance moved by cycle = distance moved by car

Time taken by car to accelerate from rest

$t = \frac{v_f - v_i}{a}$

$t = \frac{49 - 0}{7} = 7 s$

Time taken by cycle to accelerate

$t = \frac{23 - 0}{15} = 1.53 s$

now the distance moved by cycle in time "t"

$d = \frac{23 + 0}{2}*1.53 + 23(t - 1.53)$

distance moved by car in same time

$d = \frac{7t + 0}{2}(t)$

now make them equal

$3.5t^2 = 17.595 - 35.19 + 23t$

$3.5 t^2 - 23t + 17.595 = 0$

$t = 5.68 s$

so cycle will move ahead of car for t = 5.68 s

Answer 2

Answer:

3.29s

Explanation:

See the sketch of  speed-time graph.

Since cyclists reach a constant speed of 23mi/h before car does to constant speed of 49 mi/h, we will calculate the time taken by car to reach 23mi/h to know the time interval for which after the light turned green the bicycle is ahead of your car

a=(vf-vi)/t

Let convert acceleration from 7mi/h/s to 7mi/h²

7mi/(1/3600)

25200mi/h²

25200=(23-0)/t

t= 0.000913h or 3.28s