A speeder traveling at a constant speed of 117 km/h races past a billboard. A patrol car pursues from rest with constant acceleration of (7.6 km/h)/s until it reaches its maximum speed of 175 km/h, which it maintains until it catches up with the speeder.(a) How long does it take the patrol car to catch the speeder if it starts moving just as the speeder passes? (b) How far does each car travel?
Speeder: Vo = 117,000m/3600s = 32.5 m/s. Ds = Vo*t = 32.5*t. = Distance traveled by speeder.
Patrol car: a = (7600m/3600s)/s = 2.11 m/s^2.
V = 175,000m/3600s = 48.6 m/s.
Vo = 0.
V^2 = Vo^2 + 2a*D, Dp = V^2/2a = (48.6)^2/4.22 = 560 m = Distance traveled by patrol car.
a. Ds = Dp, 32.5*t = 560, t =
17.2 s. To catch up.
b. Ds = Dp = 560 m.
To solve this problem, we need to use the kinematic equations of motion.
(a) To find the time it takes for the patrol car to catch the speeder, we first need to find the time it takes for the patrol car to reach its maximum speed.
Using the equation: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we have:
175 km/h = 0 km/h + (7.6 km/h)/s * t
Converting the velocities to meters per second:
175 km/h = 0 m/s + (7.6 km/h)/s * t
175 km/h = (7.6 * 1000 m)/(3600 s) * t
175 km/h = (7.6 * 1000)/(3600) * t
49 m/s = (7.6 * 1000)/(3600) * t
t = (49 * 3600) / (7.6 * 1000)
t ≈ 23.17 s
Now, we need to find the time it takes for the speeder to travel this amount of time, plus the additional time it takes for the patrol car to catch up.
Using the equation: t = d / v, where t is the time, d is the distance, and v is the velocity, we have:
t = d / 117 km/h
23.17 s = d / (117 * 1000 m)/(3600 s)
Simplifying:
23.17 s = d / (117 * 1000)/(3600)
23.17 s = d / 9.72
Solving for d:
d = 23.17 s * 9.72
d ≈ 225.25 m
Therefore, it takes approximately 23.17 seconds for the patrol car to catch the speeder.
(b) To find the distances traveled by each car, we can use the formula: d = ut + (1/2)at^2.
For the patrol car:
d1 = 0 m/s * 23.17 s + (1/2) * (7.6 km/h)/s * (23.17 s)^2
Converting the velocities to meters per second:
d1 = 0 * 23.17 + (1/2) * (7.6 * 1000)/(3600) * (23.17)^2
d1 = 0 + (1/2) * (7.6 * 1000)/(3600) * (23.17)^2
Simplifying:
d1 = (1/2) * (7.6 * 1000)/(3600) * (23.17)^2
d1 ≈ 99.52 m
For the speeder:
d2 = 117 km/h * 23.17 s
Converting the velocity to meters per second:
d2 = (117 * 1000)/(3600) * 23.17
Simplifying:
d2 = (117 * 1000)/(3600) * 23.17
d2 ≈ 685.69 m
Therefore, the patrol car travels approximately 99.52 meters and the speeder travels approximately 685.69 meters.
To solve this problem, we first need to find the time it takes for the patrol car to catch the speeder. Let's consider part (a) of the question.
(a) How long does it take the patrol car to catch the speeder?
Step 1: Convert the speeds to meters per second (m/s) for consistency.
Speeder's speed: 117 km/h = (117 × 1000) m / (60 × 60) s = 32.5 m/s
Patrol car's maximum speed: 175 km/h = (175 × 1000) m / (60 × 60) s = 48.6 m/s
Step 2: Find the time it takes for the patrol car to reach its maximum speed.
The patrol car starts from rest and has a constant acceleration of (7.6 km/h)/s.
To find the time taken to reach the maximum speed, use the formula:
final velocity = initial velocity + acceleration × time
48.6 = 0 + (7.6 × t)
48.6 = 7.6t
t = 48.6 / 7.6 ≈ 6.39 seconds
Step 3: Find the distance covered by the patrol car during acceleration.
To find the distance covered during acceleration, we use the formula:
distance = initial velocity × time + (1/2) × acceleration × time^2
distance = 0 × 6.39 + (1/2) × 7.6 × (6.39)^2
distance ≈ 122.66 meters
Step 4: Calculate the time it takes for the patrol car to catch the speeder.
To calculate the time, we can use the formula:
time = distance / relative velocity
The relative velocity is the difference between the speed of the patrol car and the speed of the speeder.
relative velocity = 48.6 - 32.5 = 16.1 m/s
time = 122.66 / 16.1 ≈ 7.62 seconds
Therefore, it takes approximately 7.62 seconds for the patrol car to catch the speeder.
(b) How far does each car travel?
For the speeder:
distance = speed × time
distance = 32.5 m/s × 7.62 s ≈ 247.65 meters
For the patrol car:
distance = distance covered during acceleration + distance covered at the maximum speed
distance = 122.66 m + (48.6 m/s × 7.62 s) ≈ 512.13 meters
Therefore, the speeder travels approximately 247.65 meters, and the patrol car travels approximately 512.13 meters.
Correction:
Speeder: Vo = 117,000m/3600s = 32.5 m/s. Ds = 32.5*t = Distance traveled by speeder.
Patrol car: a = (7600m/3600s)/s = 2.11 m/s^2.
a. 32.5t = 0.5*2.11*t^2, 1.05t = 32.5, t = 31 s.
b. Ds = Dp = 32.5*t = 32.5*31 = 1006 m.