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 need to use the equation of motion for the speeding car to find out how far it travels during the police car delay.

distance = speed × time 
distance = 30.0 m/s × 1.00 s 
distance = 30.0 m

This means that the police car must travel 300 m - 30 m = 270 m to catch up with the speeding car.

Next, we can use the equation of motion for constant acceleration to find out the acceleration needed for the police car to travel this distance in the given time.

distance = (initial velocity × time) + (0.5 × acceleration × time²)

Since the police car starts from rest, its initial velocity is 0 m/s. We can rearrange the equation to solve for acceleration:

acceleration = (2 × distance) ÷ (time²) 
acceleration = (2 × 270 m) ÷ (1.00 s)² 
acceleration = 540 m/s²

However, the answer choices are in units of m/s², so we must convert.

acceleration = 540 m/s² ÷ 9.81 m/s² 
acceleration ≈ 55.09 m/s²

Rounding to 2 significant figures, we get an acceleration of 55 m/s², which is equivalent to 7.41 m/s² (Choice C).

A Speeding Car Is Traveling at a Constant 30