A thief starts his car from a point with an acceleration of `2m//s^(2)`. The police chasing the thief arives at the point 10s later and continues to chase the thief with a uniform velocity of 40m/s I their car. In what time the police's car will overtake the thief car?

Explanation

To solve the problem of the thief and the police, we can break it down into steps:

### Step 1: Understand the Initial Conditions
- The thief starts from rest (initial velocity \( u = 0 \)) and accelerates at \( a = 2 \, \text{m/s}^2 \).
- The police arrive at the starting point of the thief 10 seconds later and travel with a constant speed of \( v = 40 \, \text{m/s} \).

### Step 2: Calculate the Distance Covered by the Thief in 10 Seconds
Using the equation of motion:
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