Question

4. Robbers in a car travelling at 20 m/s pass
a policeman on a motorcycle at rest. The
policeman immediately starts chasing the
robbers. The policeman acceleraltes at 3 meter per second for 12s and thereafter travels at a constant velocity. calculate the distance covered by the policeman before he overtakes the car.​

Answer 1

Given:

Initial elocity of Robbers car (u) = 20 m/s

Initial velocity of Police car (u) = 0 m/s

The policeman acceleraltes at 3 meter per second for 12s.

Acceleration = 3 m/s²

Time = 12 sec

Distance covered by Policeman in 12 sec

s = ut + 1/2 at²

Substitute the known values

→ s = (0)(12) + 1/2 (3)(12)²

→ s = 0 + 1/2 × 3 × 144

→ s = 3 × 72

→ s = 216 m

v = u + at

Substitute the known values

→ v = (0) + 3(12)

→ v = 36 m/s

Now,

Distance covered by Robber in (12 + t)sec

Distance = Speed × Time

Substitute the known values

→ Distance = 20 × (12 + t)

→ Distance = (240 + 20t)m

Now, the difference in distance covered by Policeman and Robber in 12 sec

Difference = Distance covered by Robber - Distance covered by Policeman

→ Difference = (240 - 216) m

→ Difference = 24 m

Let us assume that in (t) time the bike has covered (x+24) m distance and the car has covered (x) m distance.

• Distance covered by bike

→ 36t = x + 24

• Distance covered by car

→ 20t = x

Substitute value of x in (36t = x + 24)

→ 36t = 20t + 24

→ 36t - 20t = 24

→ 16t = 24

→ t = 24/16

→ t = 1.5 sec

During this time, the Police man bike was in constant motion.

So, Distance = 216 + 36t

→ Distance = 216 + 36(1.5)

→ Distance = 216 + 54

→ Disatnce = 270 m

Answer 2

|| ✰✰ ANSWER ✰✰ ||

||✪✪ Given That ✪✪|| :-

→ initial of Policeman = u = 0 m/s.

→ Acceleration = 3m/s.

→ Time = 12 seconds.

So, The Total distance covered by the Policeman in First 12 seconds is = s = ut + (1/2)at²

→ s = 0 * 12 + (1/2) * 3 * (12)²

→ s = 0 + 3 * 6 * 12

→ s = 216m.

___________________

Now, Final Velocity After 12 seconds will be = v = u + at .

→ v = 0 + 3 * 12

→ v = 36m/s.

______________

Now Lets Assume That, The Policeman Chase The robbers for x seconds .

So,

→ Before Chasing Robbers cover distance in 12 seconds = Speed * Time .

→ D = 20*12

→ D = 240m.

And, After (12+x) Seconds he overtake , in which he Travels = s + (v * x)

→ 216 + (36 * x)

→ (216 + 36x)m. -------- Equation (1)

And , in This Time Robbers covered = 20(12 + x) = (240+20x) m . -------- Equation (2)

______________

Now, we can say That, when the Policeman overtake The car , The distance will be same in both case .

So,

Comparing Equation (1) & (2),

→ (216 + 36x) = (240 + 20x)

→ 36x - 20x = 240 - 216

→ 16x = 24

→ x = (24/16)

→ x = (3/2) seconds.

______________

Putting value of x in Equation (1) now, we get :-

→ Distance Covered by The Policeman = (216 + 36x) = [ 216 + 36*(3/2) ] = (216 + 18*3) = (216 + 54) = 270m.

Hence , The distance covered by the policeman before he overtakes the car is 270m.