Question

A suspicious looking car (A) travels along a road at a constant velocity of 13.3m/s. As it passes a parked, stationary police car (P), the police car accelerates from rest at 4m/s 2 in pursuit. The police car continues to accelerate until it reaches a velocity of 17.3m/s. (a)How long does it take, from the moment the police car starts moving, for the police car to pull alongside the pursued car? (b)How far has the police car travelled at this instant?

Answer 1

a) 9.4 sec

b) 125.02 m

Explanation:

a)

For the police car :

v₀ = initial velocity of the police car = 0 m/s

a = acceleration of the car = 4 m/s²

v' = final constant velocity gained by police car = 17.3 m/s

t' = time taken to attain the final constant velocity

using the equation

v' = v₀ + a t'

17.3 = 0 + (4) t'

t' = 4.325 sec

d' = distance travelled by the police car while accelerating

distance travelled by the police car while accelerating is given as

d' = v₀ t' + (0.5) a t'²

d' = (0) (4.325) + (0.5) (4) (4.325)²

d' = 37.4 m

Consider the motion the suspicious looking car for the time when the police car accelerate:

v = velocity of suspicious looking car = 13.3 m/s

t' = time taken to accelerate = 4.325 sec

d = distance travelled by suspicious looking car = v t' = (13.3) (4.325) = 57.5 m

t = time after which the suspicious car is caught

for the police car to catch the suspicious car

d' + v' (t - t') = v t

37.4 + (17.3) (t - 4.325) = 13.3 t

t = 9.4 sec

b)

D = distance travelled by police car = distance travelled by suspicious car = v t

D = 13.3 x 9.4

D = 125.02 m

Answer 2

a) 9.4 sec

b) 125.02 m

Step-by-step explanation:

a)

For the police car :

v₀ = initial velocity of the police car = 0 m/s

a = acceleration of the car = 4 m/s²

v' = final constant velocity gained by police car = 17.3 m/s

t' = time taken to attain the final constant velocity

using the equation

v' = v₀ + a t'

17.3 = 0 + (4) t'

t' = 4.325 sec

d' = distance traveled by the police car while accelerating

distance traveled by the police car while accelerating is given as

d' = v₀ t' + (0.5) a t'²

d' = (0) (4.325) + (0.5) (4) (4.325)²

d' = 37.4 m

Consider the motion the suspicious looking car for the time when the police car accelerate:

v = velocity of suspicious looking car = 13.3 m/s

t' = time taken to accelerate = 4.325 sec

d = distance traveled by suspicious looking car = v t' = (13.3) (4.325) = 57.5 m

t = time after which the suspicious car is caught

for the police car to catch the suspicious car

d' + v' (t - t') = v t

37.4 + (17.3) (t - 4.325) = 13.3 t

t = 9.4 sec

b)

D = distance traveled by police car = distance traveled by suspicious car = v t

D = 13.3 x 9.4

D = 125.02 m