147k views
4 votes
A police car is traveling at a velocity of 19.0 m/s due north, when a car zooms by at a constant velocity of 41.0 m/s due north. After a reaction time 0.400 s the policeman begins to pursue the speeder with an acceleration of 6.00 m/s^2. Including the reaction time, how long does it take for the police car to catch up with the speeder?

User ITWitch
by
4.5k points

1 Answer

5 votes

Answer:

It takes the police car 8.42 s to catch the speeder.

Step-by-step explanation:

The equation for the position of the police car is as follows:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

For the speeder, the position will be:

x = x0 + v · t

Where "v" is the velocity.

To solve this problem, we need to know which is the position of the speeder, when the police car starts pursuing it.

So, using the equation for the position we can calculate the distance traveled by the speeder during the time it takes the police to react before starting the persecution. Let´s place the origin of the frame of reference at the point where the police car is located.

Then:

x = x0 + v · t

x = 0 m + 41.0 m/s · 0.400 s

x = 16.4 m

At the start of the persecution, the speeder is 16.4 m ahead of the police car.

When the police catches the speeder, the position for both cars is the same. Let´s place the origin of the frame of reference at the point where the police car is at the moment it starts the persecution. At that moment, the speeder is 16.4 m ahead of the police.

x police car = x speeder

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

0 m + 19.0 m/s · t +1/2 · 6.00 m/s² · t² = 16.4 m + 41.0 m/s · t

0 = 16.4 m + 41.0 m/s · t - 19.0 m/s · t - 3.00 m/s² · t²

0 = 16.4 m + 22.0 m/s · t - 3.00 m/s² · t²

Solving the quadratic equation.

t = 8.02 s

It takes the police car (8.02 s + 0.400 s) 8.42 s to catch the speeder

User Starrr
by
5.9k points