A speeding car is traveling at a constant 30.0 m/s when it passes a stationary police car. If the police car delays for 1.00 s before starting, what must be the magnitude of the constant acceleration of the police car to catch the speeding car after the police car travels a distance of 300 m? A) 6.00 m/s² B) 3.00 m/s² C) 7.41 m/s² D) 1.45 m/s² E) 3.70 m/s²

Question

Quizplus · Accepted Answer

First, we can find the time it takes for the police car to catch up to the speeding car. Let t be the time it takes for the police car to catch up, then:
distance traveled by speeding car = distance traveled by police car
30.0t = 300 + (30.0)(t + 1.00)
Solving for t, we get t = 9.48 s.
Next, we can use the equation of motion: 
distance = initial velocity × time + 0.5 × acceleration × time^2
for the police car. Since the initial velocity is 0, the equation simplifies to:
distance = 0.5 × acceleration × time^2
We know that the distance the police car needs to travel is 300 m and the time it takes is 9.48 s. Plugging in these values, we get:
300 = 0.5 × acceleration × (9.48)^2
Solving for acceleration, we get:
acceleration = 7.41 m/s^2 (rounded to two significant figures)
Therefore, the answer is (C) 7.41 m/s^2.

A Speeding Car Is Traveling at a Constant 30