Question

A policeman, parked at arn intersection, is passed by a speeder traveling at 120 km/hr. Two seconds later, the policeman starts and uniformly accelerates at 4.0 m/s^2. How far will the policeman travel before he catches the speeder?

Answer 1

Answer:

682.32 m

Explanation:

Speed of the passing speeder = 120 km/hr = 120 × 0.2777 = 33.33 m/s

time after which police man starts = 2 seconds

Acceleration of the policeman = 4.0 m/s²

let the time taken to catch be 't' seconds

Now,

The total distance to be covered by the policeman

= Distance covered by the speeder in the 2 seconds + Distance further traveled by the speeder in time t

thus,

From Newton's equation of motion

$s=ut+\frac{1}{2}at^2$

where,  

s is the total distance traveled by the police man

u is the initial speed  = 0

a is the acceleration

t is the time

thus,

$0\times t+\frac{1}{2}\times4t^2$= 33.33 × 2 + 33.33 × t

or

2t² = 66.66 + 33.33t

or

2t² - 33.33t - 66.66 = 0

on solving the above equation, we get

t = 18.47 seconds (negative value is ignored as time cannot be negative)

therefore,

the total distance covered = 33.33 × 2 + 33.33 × 18.47 = 682.32 m