he police officer starts rom rest ust as he speeder passes it, and accelerates a

ID: 1876868 • Letter: H

Question

he police officer starts rom rest ust as he speeder passes it, and accelerates at d constant rate of 2.1m s2 Ari automobile is driving excessive y as a constant speed of4.2m An officer notices he speeding ve 49.8m/s. whe the office's vahiela raaches maximum spad, it moves at a cnnstant sp a, How long does the officer take to catch the speeder? b. What distance has each vehicle traveled? c. It the police car starts 1.is after the speeder passes it, how long does it take to catch the speeder? he maximum speed o he police car is e

Explanation / Answer

let the officer catch the speeder after time "t"

for the officer's motion :

vi = initial velocity = 0 m/s

vf = final maximum velocity = 49.8 m/s

a = acceleration = 2.1 m/s2

ta = time during which the officer's vehicle accelerate = ?

Using the equation

vf = vi + a ta

49.8 = 0 + (2.1) ta

ta = 23.71 sec

tc = time during which the officer's vehicle move at constant velocity = t - ta = t - 23.71

Da = distance travelled by officer's vehicle during acceleration

using the equation

Da = vi ta + (0.5) a t2a

Da = (0) (23.71)+ (0.5) (2.1) (23.71)2= 590.3 m

Dc = distance travelled by the officer's vehicle at constant velocity = 49.8 tc = 49.8 (t - 23.71)

Dofficer = total distance travelled by officer = Da + Dc

Dofficer = 590.3 + 49.8 (t - 23.71) eq-1

for the speeder :

Vs = speed of speeder = 34.2 m/s

t = time of travel

Ds = distance travelled by speeder = Vs t

Ds = 34.2t eq-2

for the officer to catch the speeder

Dofficer = Ds

590.3 + 49.8 (t - 23.71) = 34.2 t

t = 37.85 sec

using eq-2

Ds = 34.2t

Ds = 34.2(37.85)

Ds = 1294.5 m

hence distance travelled by each is 1294.5 m