A burglar's car in operation alone townA has a start with an acceration of 2m/s^2. A police van came after 5sec and continue to chase the burglar's car wit uniform velocity of 20m/s.find the time take in which the police patroj van will overtake the burglar's car.

solve for t when the two distances are equal:

(1/2)(2)t^2 = (20)(t-5)

You will find that 20 m/s is the minimum speed at which the patrol car can catch the burglar.

To find the time it takes for the police van to overtake the burglar's car, we need to consider the relative motion between the two cars.

Let's calculate the distance covered by the burglar's car in the first 5 seconds using the kinematic equation:

Distance = Initial Velocity * Time + (1/2) * Acceleration * Time^2

Given:
Initial Velocity of the burglar's car (u1) = 0 (since it starts from rest)
Acceleration of the burglar's car (a1) = 2 m/s^2
Time (t) = 5 seconds

Using the equation:
Distance1 = 0 * 5 + (1/2) * 2 * 5^2
Distance1 = 0 + 1/2 * 2 * 25
Distance1 = 25 meters

So, after 5 seconds, the burglar's car will have covered a distance of 25 meters.

Now, let's determine the relative velocity between the two cars once the police van starts chasing the burglar's car.

Relative Velocity = Velocity of the police van (v2) - Velocity of the burglar's car (v1)

Given:
Velocity of the police van (v2) = 20 m/s
Velocity of the burglar's car (v1) = 2 m/s^2 * 5 seconds = 10 m/s

Relative Velocity = 20 m/s - 10 m/s
Relative Velocity = 10 m/s

Now, we can calculate the time it takes for the police van to overtake the burglar's car by using the relative velocity and the distance covered by the burglar's car after 5 seconds.

Time taken = Distance covered by the burglar's car / Relative velocity

Time taken = 25 meters / 10 m/s
Time taken = 2.5 seconds

Therefore, it will take the police van 2.5 seconds to overtake the burglar's car.