A police car is traveling at a velocity of 17.0 m/s due north, when a car zooms by at a constant velocity of 40.0 m/s due north. After a reaction time 0.300 s the policeman begins to pursue the speeder with an acceleration of 4.00 m/s^2. Including the reaction time, how long does it take for the police car to catch up with the speeder

8.13 seconds

Explanation:

Given that a police car is traveling at a velocity of 17.0 m/s due north, when a car zooms by at a constant velocity of 40.0 m/s due north. After a reaction time 0.300 s the policeman begins to pursue the speeder with an acceleration of 4.00 m/s^2. Including the reaction time, how long does it take for the police car to catch up with the speeder

First calculate the acceleration of the speeder

V = U - at

0 = 40 - a × 0.3

0.3a = 40

a = 40/0.3

a = 133.3 m/s^2

The speeder distance covered will be

V^2 = U^2 - 2as

0 = 40^2 - 2 × 133.3 × S

266.67S = 1600

S = 1600/266.7

S = 6 m

Let's calculate the time to reach the distance by using second equation of motion

S = Ut - 1/2at^2

6 = 17t - 0.5 × 4 t^2

6 = 17t - 2t^2

2t^2 - 17t + 6 = 0

t^2 - 17/2t = - 3

(t - 17/4)^2 = 18.06 - 3

(t - 17/4)^2 = 15.06

t - 17/4 = 3.88

t = 17/4 + 3.88

t = 8.13 seconds

Therefore, it take 8.13 seconds for the police car to catch up with the speeder

the temperature in kelvin of 2.0 dm3 of an ideal gas is doubled and its pressure is increased by a factor of four. what is the final volume of the gas