Question

a burglar's car had a start with an acceleration of 2m/s . A police vigilant party cane after 5seconds and continued to chase the burglar's car with a uniform velocity of 20m/s . find the time taken in which the police van will overtake the burglar's car​

Answer 1

Step-by-step explanation:

Let the distance covered after which the police party catches the burglar is ‘s ’.

Let ‘t ’ be the time and given that the police came after 5sec, therefore the time taken by police to catch the burglar is ‘t−5 ’.

The burglar’s car will start moving from rest, therefore initial velocity is, u=0

Given that the acceleration is 2m/s

Applying the equation of motion,

S1=ut+12at2

⇒S1=0×t+12×2×t2

⇒S1=0×t+12×2×t2

⇒S1=t2 -----------(i)

Since the police’s car is moving at uniform velocity, therefore acceleration will be =0, Also u=20m/s.

Again applying the equation of motion to the police car,

⇒S2=ut+12at2

On substituting the corresponding values, we get

⇒S2=20(t−5)+12×0×t2

⇒S2=20t−100 -------(ii)

Since the time at which the police car catches the burglar’s car is to be calculated, therefore

S1=S2

⇒t2=2t−100

⇒t2−20t+100=0

we can rewrite the above equation as,

⇒(t−10)2=0

⇒(t−10)(t−10)=0

On solving,

⇒t=10

The time taken by the police is t−5=10−5=5sec.

hope this helps you.