A burglar's car had a start with an acceleration of `2ms^(-1)` . A police van came after 5 second and continued to chase the burglar's car with a uniform velocity of `20 ms ^(-1)`. Find the time in which the police van overtakes the burglar's car .

Answer

To solve the problem, we need to determine the time it takes for the police van to overtake the burglar's car after it starts moving. Let's break down the solution step by step.

### Step 1: Determine the distance traveled by the burglar's car in the first 5 seconds.
The burglar's car starts from rest and accelerates at `2 m/s²`. We can use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( u = 0 \) (initial velocity),
- \( a = 2 \, m/s^2 \) (acceleration),
- \( t = 5 \, s \) (time).

Substituting the values:
\[ s = 0 + \frac{1}{2} \times 2 \times (5)^2 \]
\[ s = \frac{1}{2} \times 2 \times 25 \]
\[ s = 25 \, m \]

### Step 2: Determine the velocity of the burglar's car after 5 seconds.
Using the formula:
\[ v = u + at \]
Substituting the values:
\[ v = 0 + 2 \times 5 \]
\[ v = 10 \, m/s \]

### Step 3: Write the expression for the distance traveled by the burglar's car after the police van starts chasing.
Let \( t \) be the time in seconds after the police van starts chasing. The total time for the burglar's car when the police van starts is \( t + 5 \). The distance traveled by the burglar's car after the police van starts can be calculated using:
\[ s_{\text{burglar}} = s_{\text{initial}} + vt + \frac{1}{2} a t^2 \]
Where:
- \( s_{\text{initial}} = 25 \, m \) (distance traveled in the first 5 seconds),
- \( v = 10 \, m/s \) (velocity after 5 seconds),
- \( a = 2 \, m/s^2 \) (acceleration),
- \( t \) is the time after the police van starts.

So,
\[ s_{\text{burglar}} = 25 + 10t + \frac{1}{2} \times 2 \times t^2 \]
\[ s_{\text{burglar}} = 25 + 10t + t^2 \]

### Step 4: Write the expression for the distance traveled by the police van.
The police van travels at a constant speed of \( 20 \, m/s \). The distance traveled by the police van after \( t \) seconds is:
\[ s_{\text{police}} = 20t \]

### Step 5: Set the distances equal to find when the police van overtakes the burglar's car.
To find the time when the police van overtakes the burglar's car, we set the distances equal:
\[ 25 + 10t + t^2 = 20t \]
Rearranging gives:
\[ t^2 - 10t + 25 = 0 \]

### Step 6: Solve the quadratic equation.
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -10, c = 25 \).
\[ t = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 25}}{2 \cdot 1} \]
\[ t = \frac{10 \pm \sqrt{100 - 100}}{2} \]
\[ t = \frac{10 \pm 0}{2} \]
\[ t = 5 \, s \]

### Step 7: Find the total time from the start.
Since the police van started chasing after 5 seconds, the total time from the start is:
\[ t_{\text{total}} = t + 5 = 5 + 5 = 10 \, s \]

### Final Answer:
The police van overtakes the burglar's car after **10 seconds** from the start.