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Question: A burglar’s car had a start with an acceleration of \(2m/{s^2}\). A police vigilant party came after...

A burglar’s car had a start with an acceleration of 2m/s22m/{s^2}. A police vigilant party came after 5sec⁡5\sec and continued to chase the burglar’s car with a uniform velocity of 20m/s20m/s. Find the time in which the police van overtakes the burglar’s car.

Explanation

Solution

Displacement of an object moving is directly proportional to the velocity of the object and the square of the time taken. When the displacement of a particle is to be calculated using velocity and time, then the acceleration of the body should be kept constant.

Complete step by step answer:
Let the distance covered after which the police party catches the burglar is ‘ss ’.
Let ‘tt ’ be the time and given that the police came after 5sec⁡5\sec , therefore the time taken by police to catch the burglar is ‘t−5t - 5 ’.
The burglar’s car will start moving from rest, therefore initial velocity is, u=0u = 0
Given that the acceleration is 2m/s2m/s
Applying the equation of motion,
S1=ut+12at2{S_1} = ut + \dfrac{1}{2}a{t^2}
⇒S1=0×t+12×2×t2\Rightarrow {S_1} = 0 \times t + \dfrac{1}{2} \times 2 \times {t^2}
⇒S1=t2\Rightarrow {S_1} = {t^2} -----------(i)
Since the police’s car is moving at uniform velocity, therefore acceleration will be =0 = 0, Also u=20m/su = 20m/s.
Again applying the equation of motion to the police car,
⇒S2=ut+12at2\Rightarrow {S_2} = ut + \dfrac{1}{2}a{t^2}
On substituting the corresponding values, we get
⇒S2=20(t−5)+12×0×t2\Rightarrow {S_2} = 20(t - 5) + \dfrac{1}{2} \times 0 \times {t^2}
⇒S2=20t−100\Rightarrow {S_2} = 20t - 100 -------(ii)
Since the time at which the police car catches the burglar’s car is to be calculated, therefore
S1=S2{S_1} = {S_2}
⇒t2=2t−100\Rightarrow {t^2} = 2t - 100
⇒t2−20t+100=0\Rightarrow {t^2} - 20t + 100 = 0
we can rewrite the above equation as,
⇒(t−10)2=0\Rightarrow {(t - 10)^2} = 0
⇒(t−10)(t−10)=0\Rightarrow (t - 10)(t - 10) = 0
On solving,
⇒t=10\Rightarrow t = 10

The time taken by the police is t−5=10−5=5sec⁡t - 5= 10 - 5 = 5\sec

Note:
It is to be noted that constant velocity is not to be mixed with constant acceleration. This is because velocity is the speed in the given direction. But acceleration is the change of speed with time. Therefore, if the velocity is constant, this means that the speed is constant. Since the speed is not changing, therefore there will be no acceleration.