As soon as a traffic light turns green, a car speeds up from rest to 50.0 mi/h with a constant acceleration of 8.10 mi/h·s. In the adjoining bike lane, a cyclist speeds up from rest to 20.0 mi/h with a constant acceleration of 13.0 mi/h·s. Each vehicle maintains a constant velocity after reaching its cruising speed.
(a) For how long is the bicycle ahead of the car?
(b) By what maximum distance does the bicycle lead the car?
a) Write equations for distance from the light vs time for bike and car. Call them X1(t) and X2(t)
Set X1 = X2 and solve for t. These are piecewise continous functions, so you cannot solve single equations.
If t is in seconds and X in feet,
X1(car) = 5.94 t^2
Car acceleration lasts for 6.17 seconds
X2(bike) = 9.533 t^2
Bike acceleration lasts for 1.538 s.
After 1.538s, the bike is ahead by X1-X2 = 22.55 - 14.05 = 8.5 feet. The car is traveling at 18.2 ft/s, which is slower than the bike's 29.3 ft/s but the car still accelerating. Let t' be the time measured after 1.538 seconds, when the bike reaches constant velocity.
X1 = 14.05 + 18.2 t' + 5.94 t'^2
X2 = 22.55 + 29.3 t'
You can now set up a quadratic equation for when X1 = X2
You can also differentiate X2 - X1 and set the derivative equal to zero to get the maximum separation time and distance.
To solve this problem, we need to find the time at which both the bicycle and the car reach their cruising speeds and then calculate the relative positions between them.
To find the time it takes for each vehicle to reach its cruising speed, we can use the equation for final velocity in terms of initial velocity, acceleration, and time:
(vf)² = (vi)² + 2aΔx
Where vf is the final velocity, vi is the initial velocity, a is the acceleration, and Δx is the displacement. We can assume that the initial velocity is 0 for both the car and the bicycle.
For the car:
(vf car)² = 0 + 2(8.10 mi/h·s)(Δx car)
(50.0 mi/h)² = 16.20 mi/h·s (Δx car)
Δx car = (50.0 mi/h)² / (16.20 mi/h·s)
Δx car = 154.32 mi
For the bicycle:
(vf bicycle)² = 0 + 2(13.0 mi/h·s)(Δx bicycle)
(20.0 mi/h)² = 26.0 mi/h·s (Δx bicycle)
Δx bicycle = (20.0 mi/h)² / (26.0 mi/h·s)
Δx bicycle = 15.38 mi
(a) To find how long the bicycle is ahead of the car, we need to calculate the time it takes for the car and the bicycle to reach their respective displacements:
For the car:
Δx car = (1/2)at²
154.32 mi = (1/2)(8.10 mi/h·s)t²
t² = (2*154.32 mi) / (8.10 mi/h·s)
t² = 37.94 mi²·h² / mi²·h·s
t = √37.94 h
t ≈ 6.16 h
For the bicycle:
Δx bicycle = (1/2)at²
15.38 mi = (1/2)(13.0 mi/h·s)t²
t² = (2*15.38 mi) / (13.0 mi/h·s)
t² = 2.37 mi²·h² / mi²·h·s
t = √2.37 h
t ≈ 1.54 h
The bicycle is ahead of the car for approximately 1.54 hours (or 1 hour and 32 minutes).
(b) To find the maximum distance by which the bicycle leads the car, we can multiply the time it takes for the bicycle to reach its cruising speed by the velocity of the car:
d = vt
d = (50.0 mi/h)(1.54 h)
d ≈ 77 mi
The bicycle leads the car by a maximum distance of approximately 77 miles.
To solve this problem, we can use the kinematic equations of motion. Let's start working on part (a) of the question.
(a) For how long is the bicycle ahead of the car?
To find the time when the bicycle is ahead of the car, we'll need to find when their positions are equal.
Let's assume that the time taken by both the car and the bicycle to reach their cruising speed (constant velocity) is t seconds.
For the car:
Initial velocity (u1) = 0 (rest)
Final velocity (v1) = 50.0 mi/h
Acceleration (a1) = 8.10 mi/h·s
For the bicycle:
Initial velocity (u2) = 0 (rest)
Final velocity (v2) = 20.0 mi/h
Acceleration (a2) = 13.0 mi/h·s
The position (s) of an object can be calculated using the formula:
s = ut + (1/2)at^2
For the car:
s1 = (1/2)a1t^2
= (1/2)(8.10 mi/h·s)(t^2)
For the bicycle:
s2 = (1/2)a2t^2
= (1/2)(13.0 mi/h·s)(t^2)
Setting the two positions equal, we get:
(1/2)(8.10 mi/h·s)(t^2) = (1/2)(13.0 mi/h·s)(t^2)
Simplifying, we have:
8.10 mi/h·s t^2 = 13.0 mi/h·s t^2
Dividing both sides by t^2, we get:
8.10 mi/h·s = 13.0 mi/h·s
Since the accelerations are not equal, it implies that the positions are never equal. Therefore, the bicycle is never ahead of the car, and the answer to part (a) is that the bicycle is never ahead of the car.
(b) By what maximum distance does the bicycle lead the car?
Since the bicycle is never ahead of the car, the maximum distance that the bicycle leads the car is 0 mi. Therefore, the answer to part (b) is 0 mi.