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
[tex]s=ut+\frac{1}{2}at^2[/tex]
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,
[tex]0\times t+\frac{1}{2}\times4t^2[/tex]= 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
