As soon as a traffic light turns green, a car speeds up from rest to 50.0 mi/h with constant acceleration 9.00 mi/h/s. In the adjoining bicycle lane, a cyclist speeds up from rest to 20.0 mi/h with constant acceleration 13.0 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? (1mil=1.6 km)
here hope this helps
Ok - first of all this is a interesting problem because it is pretty challenging.
1st) Convert the miles/hr to feet/sec and acceleration to ft/sec^2. Here are examples of how to do both.
50 miles/hr x 1hr/3600 sec x 5280 ft/mile = 73 ft/sec
To solve this problem, we will use the equations of motion and kinematic equations.
First, we need to convert the given speeds from miles per hour (mi/h) to meters per second (m/s):
Car's final speed = 50.0 mi/h * (1.6 km/mi * 1000 m/km) / (1 h/3600 s) ≈ 22.35 m/s
Cyclist's final speed = 20.0 mi/h * (1.6 km/mi * 1000 m/km) / (1 h/3600 s) ≈ 8.94 m/s
Now we can calculate the time it takes for each vehicle to reach its cruising speed.
Car's acceleration = 9.00 mi/h/s * (1.6 km/mi * 1000 m/km) / (1 h/3600 s) ≈ 4.02 m/s^2
Cyclist's acceleration = 13.0 mi/h/s * (1.6 km/mi * 1000 m/km) / (1 h/3600 s) ≈ 5.81 m/s^2
Let's denote the time it takes for the car and cyclist to reach their cruising speeds as tc and tb, respectively.
Using the equation of motion vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time, we can solve for tc and tb:
Car's final speed = Car's initial speed + Car's acceleration * tc
22.35 m/s ≈ 0 + 4.02 m/s^2 * tc
tc ≈ 5.56 s
Cyclist's final speed = Cyclist's initial speed + Cyclist's acceleration * tb
8.94 m/s ≈ 0 + 5.81 m/s^2 * tb
tb ≈ 1.54 s
Now we can move on to answering the questions:
(a) For what time interval is the bicycle ahead of the car?
To determine this, we need to compare the positions of the car and the bicycle at different times.
Let's assume t is the time in seconds after the traffic light turns green.
The position of the car at time t is given by the equation:
Car's position = Car's initial position + Car's initial velocity * t + (1/2) * Car's acceleration * t^2
Since the car starts from rest, its initial position and velocity are zero. Thus, the equation simplifies to:
Car's position = (1/2) * Car's acceleration * t^2
The position of the bicycle at time t is given by the equation:
Cyclist's position = Cyclist's initial position + Cyclist's initial velocity * t + (1/2) * Cyclist's acceleration * t^2
Again, since the cyclist starts from rest, its initial position and velocity are zero:
Cyclist's position = (1/2) * Cyclist's acceleration * t^2
To find the time interval when the bicycle is ahead of the car, we need to equate the positions of the car and the bicycle and solve for t:
(1/2) * Car's acceleration * t^2 = (1/2) * Cyclist's acceleration * t^2
Since the accelerations are different, t will cancel out, and we can solve for t:
Car's acceleration = 4.02 m/s^2
Cyclist's acceleration = 5.81 m/s^2
4.02 * t^2 = 5.81 * t^2
t^2 (5.81 - 4.02) = 0
t^2 = 0
Since t^2 = 0, this implies that t = 0. Therefore, the bicycle and the car are never at the same position simultaneously. Hence, the bicycle is always ahead of the car.
(b) By what maximum distance does the bicycle lead the car?
Since the bicycle is always ahead of the car, we want to find the maximum distance between them.
To find this, we need to determine the position of each vehicle when they reach their cruising speeds.
Car's final position = 0 + (1/2) * Car's acceleration * tc^2
Cyclist's final position = 0 + (1/2) * Cyclist's acceleration * tb^2
Calculating the values:
Car's final position = 0 + (1/2) * 4.02 m/s^2 * (5.56 s)^2 ≈ 31.3 m
Cyclist's final position = 0 + (1/2) * 5.81 m/s^2 * (1.54 s)^2 ≈ 7.48 m
The maximum distance the bicycle leads the car is the difference between these two positions:
Maximum distance = Car's final position - Cyclist's final position
≈ 31.3 m - 7.48 m ≈ 23.82 m
Therefore, the bicycle leads the car by a maximum distance of approximately 23.82 meters.