As soon as a traffic light turns green, a car speeds up from rest to 46.0 mi/h with constant acceleration 8.50 mi/h-s. In the adjoining bike lane, a cyclist speeds up from rest to 24.0 mi/h with constant acceleration 12.00 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?
To solve this problem, we need to first find the time it takes for each vehicle to reach its cruising speed. Then we can compare their positions at that time.
Given:
For the car:
Initial velocity (u1) = 0 (since it starts from rest)
Final velocity (v1) = 46.0 mi/h
Acceleration (a1) = 8.50 mi/h-s
For the bicycle:
Initial velocity (u2) = 0 (since it starts from rest)
Final velocity (v2) = 24.0 mi/h
Acceleration (a2) = 12.00 mi/h-s
(a) To find the time interval when the bicycle is ahead of the car, we can equate their positions at the cruising speed:
Distance covered by the car (s1) = Distance covered by the bicycle (s2)
For the car:
s1 = (u1 + v1) * t1 / 2
For the bicycle:
s2 = (u2 + v2) * t2 / 2
Since they both start from rest, their initial distance is zero for both vehicle and bicycle.
Therefore,
(u1 + v1) * t1 / 2 = (u2 + v2) * t2 / 2
Replacing the values given:
(0 + 46.0) * t1 / 2 = (0 + 24.0) * t2 / 2
23.0 * t1 = 12.0 * t2
Simplifying the equation:
23.0 t1 = 12.0 t2 ------------- (Equation 1)
Now, to find the time taken for each vehicle to reach its cruising speed, we can use the following formula:
For the car:
v1 = u1 + a1 * t1
46.0 = 0 + 8.50 * t1
For the bicycle:
v2 = u2 + a2 * t2
24.0 = 0 + 12.00 * t2
Simplifying these equations:
8.50 * t1 = 46.0
12.00 * t2 = 24.0
Now solve for t1 and t2:
t1 = 46.0 / 8.50 ≈ 5.41 seconds
t2 = 24.0 / 12.00 = 2.00 seconds
(b) To find the maximum distance by which the bicycle leads the car, we need to find the distance covered by each vehicle during the time interval t2.
For the car:
Distance covered by car = (u1 * t2) + (0.5 * a1 * t2^2)
Distance covered by car = (0 * 2) + (0.5 * 8.50 * 2^2)
Distance covered by car = 17.0 feet
For the bicycle:
Distance covered by bicycle = (u2 * t2) + (0.5 * a2 * t2^2)
Distance covered by bicycle = (0 * 2) + (0.5 * 12.00 * 2^2)
Distance covered by bicycle = 24.0 feet
Therefore, the bicycle leads the car by a maximum distance of 24.0 - 17.0 = 7.0 feet.
To solve this problem, we need to analyze the motion of the car and the bicycle separately and then compare their positions.
Given:
For the car:
Initial velocity (u1) = 0 mph
Final velocity (v1) = 46.0 mph
Acceleration (a1) = 8.50 mph/s
For the bicycle:
Initial velocity (u2) = 0 mph
Final velocity (v2) = 24.0 mph
Acceleration (a2) = 12.00 mph/s
(a) For what time interval is the bicycle ahead of the car?
To find the time interval when the bicycle is ahead of the car, we need to find the time taken by both the car and the bicycle to reach their respective final velocities.
Using the formula: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
For the car: v1 = u1 + a1t1
46 mph = 0 mph + (8.50 mph/s)t1
t1 = 5.41 s (approx.)
For the bicycle: v2 = u2 + a2t2
24 mph = 0 mph + (12.00 mph/s)t2
t2 = 2.00 s
The bicycle is ahead of the car for the time interval t2, which is 2.00 seconds.
(b) By what maximum distance does the bicycle lead the car?
To find the maximum distance by which the bicycle leads the car, we need to calculate the distance traveled by each vehicle in the time interval t2.
Using the formula: s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time taken.
For the car: s1 = u1t2 + (1/2)a1t2^2
s1 = 0 mph * 2.00 s + (1/2)(8.50 mph/s)(2.00 s)^2
s1 = 8.50 mph * 2.00 s
s1 = 17.00 mi
For the bicycle: s2 = u2t2 + (1/2)a2t2^2
s2 = 0 mph * 2.00 s + (1/2)(12.00 mph/s)(2.00 s)^2
s2 = 12.00 mph * 2.00 s
s2 = 24.00 mi
Therefore, the bicycle leads the car by a maximum distance of 24.00 miles.
Note: Please ensure that all units (mph, s) are consistent in the calculations to get accurate results.