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?
s

(b) By what maximum distance does the bicycle lead the car?
ft

Answer 1

First convert the rates to ft/s instead of mi/hr, since we want to evaluate in terms of seconds
1 mi/hr = 22/15 ft/s
Also note that a(t) = d/dt v(t) and v(t) = d/dt x(t) , a is acceleration v is velocity, and x is position
Let t* represent time it takes to reach cruising speed

Bike:
a = 21.26 , v = 21.26t, x = 10.63t^2 ; when t <t*
v(t*) = 27.86 = 21.26t* --> t* = 1.31
It take the bicyclist 1.31 sec to reach cruising speed of 27.86 ft/s
x(t*) = 10.63(1.31^2) = 18.26
After 1.31 sec the cyclist has traveled 18.26 ft
x = 18.26 +27.86(t - 1.31) ; when t > t*

Car:
a = 14.66 , v = 14.66t, x = 7.33t^2 ; when t <t*
v(t*) = 85.07 = 14.66t* --> t* = 5.8
It take the car 5.8 sec to reach cruising speed of 85.07 ft/s
x(t*) = 7.33(5.8^2) = 246.8
After 5.8 sec the car will have traveled 246.8 ft
x = 246.8 +85.07(t - 5.8) ; when t > t*

Now when t < 1.31, we know that the bicycle is ahead of the car
--> 10.63t^2 > 7.33t^2
At t=1.31 the cyclist stops accelerating and then by setting their positions equal we can find the time t that the car catches up with the cyclist
--> 7.33t^2 = 18.26+27.86(t - 1.31)
solve the quadratic
7.33t^2 -27.86t +18.26 = 0
using quadratic formula
t = 0.84 or 2.96, however we know that t must be greater than 1.31
t = 2.96

part a) The cyclist is ahead of the car for first 2.96 seconds

For next part to find max distance, define distance d
d = x_cyclist - x_car
d = 18.26+27.86(t - 1.31) - 7.33t^2
d = -7.33t^2 +27.86t -18.26
Now find vertex of the parabola or find max by setting derivative equal to 0.
dd/dt = -14.66t +27.86 = 0
t = 1.9
--> max_d = 8.21 feet