Question

A police car is traveling at a velocity of 18.0 m/s due north, when a car zooms by at a constant velocity of 42.0 m/s due north. After a reaction time of 0.800 s the policeman begins to pursue the speeder with an acceleration of 5.00 m/s2. Including the reaction time, how long does it take for the police car to catch up with the speeder?

Answer 1

To find the time it takes for the police car to catch up with the speeder, we first calculate the distance the speeder travels during the police car's reaction time. Then, we use the equation for displacement to find the time it takes for the police car to catch up with the speeder. The police car takes approximately 1.2 seconds to catch up with the speeder.

Step-by-step explanation:

To determine how long it takes for the police car to catch up with the speeder, we need to find the time it takes for the police car to cover the same distance as the speeder.

First, let's calculate the distance the speeder travels during the reaction time of the police car:

Distance = Speed × Time = (42.0 m/s) × (0.800 s) = 33.6 m

Now, we can find the time it takes for the police car to catch up with the speeder, using the equation for displacement:

Displacement = Initial Velocity × Time + (1/2) × Acceleration × Time²

Since the initial velocity and the acceleration are both in the same direction (due north), we can consider them as positive values:

0 = (18.0 m/s) × t + (1/2) × (5.00 m/s²) × t²

Simplifying this equation, we get:

0 = 18 t + 2.5 t²

Using the quadratic equation (t = (-b ± √(b² - 4ac)) / (2a)), we can find the time it takes for the police car to catch up with the speeder:

t = (-18 ± √(18² - 4(2.5)(0))) / (2(2.5))

t ≈ -6 or 1.2

Since time cannot be negative, the police car takes approximately 1.2 seconds to catch up with the speeder.

Answer 2

t = 10.4 s

Step-by-step explanation:

given,

velocity of the police car = 18 m/s

reaction time of policeman = 0.8 s

acceleration of the policeman to catch up the speeder = 5 m/s²

time taken to catch the speeder be t₁

distance traveled by the police and the speeder will be same

`D_p = D_s`

`u t_1 + (1)/(2)at_1^2 = u' t_1`

`18* t_1+ 0.5 * 5* t_1^2= 42* t_1`

`2.5 t_1^2 - 24 t_1 = 0`

`(2.5 t_1 - 24)t_1 = 0`

`t_1 = (24)/(2.5)`

t₁ = 9.6 s

total time taken by the policeman

= reaction time + t₁

= 0.8 + 9.6

t = 10.4 s

Total time taken by the policeman to catch the speeder is 10.4 s