A thief in a stolen car passes through a police check post at his top speed of `90 kmh^-1.` A motorcycle cop, reacting after 2 s, accelerates from rest at `5 ms^-2.` His top speed being `108 kmh^-1.` Find the maximum separation between policemen and thief.

Options

A. `112.5 m`
B. `115 m` (c) 116.5 m (1) None of these
C. `116.5 m`
D. None of these

Explanation

To solve the problem of finding the maximum separation between the thief and the police officer, we can follow these steps:

### Step 1: Convert Speeds from km/h to m/s
The speeds given in the problem are in kilometers per hour. We need to convert these speeds to meters per second for consistency with the acceleration units.

- Speed of the thief:
\[ 90 \text{ km/h} = 90 \times \frac{5}{18} = 25 \text{ m/s} \]

- Speed of the police officer:
\[ 108 \text{ km/h} = 108 \times \frac{5}{18} = 30 \text{ m/s} \]

### Step 2: Determine Time Until Maximum Separation
At maximum separation, the velocities of the thief and the police officer will be equal. The police officer starts accelerating after 2 seconds of delay.

- The police officer accelerates from rest at \(5 \text{ m/s}^2\). The formula for velocity under constant acceleration is:
\[ v = u + at \]
where \(u = 0\) (initial velocity), \(a = 5 \text{ m/s}^2\), and \(t\) is the time of acceleration.

- Setting the velocity of the police officer equal to the velocity of the thief:
\[ 30 = 0 + 5t \implies t = 6 \text{ seconds} \]

- However, since the police officer starts 2 seconds later, the total time the thief has been moving is:
\[ T = 2 + 6 = 8 \text{ seconds} \]

### Step 3: Calculate the Distance Traveled by the Thief
The thief travels at a constant speed of \(25 \text{ m/s}\) for \(8\) seconds:
\[ S_1 = \text{Speed} \times \text{Time} = 25 \text{ m/s} \times 8 \text{ s} = 200 \text{ m} \]

### Step 4: Calculate the Distance Traveled by the Police Officer
The police officer accelerates for \(6\) seconds (since he started 2 seconds later):
Using the formula for displacement under constant acceleration:
\[ S_2 = ut + \frac{1}{2} a t^2 \]
where \(u = 0\), \(a = 5 \text{ m/s}^2\), and \(t = 6 \text{ s}\):
\[ S_2 = 0 + \frac{1}{2} \times 5 \times (6)^2 = \frac{1}{2} \times 5 \times 36 = 90 \text{ m} \]

### Step 5: Calculate the Maximum Separation
The maximum separation between the thief and the police officer is the difference in the distances they have traveled:
\[ \text{Maximum Separation} = S_1 - S_2 = 200 \text{ m} - 90 \text{ m} = 110 \text{ m} \]

### Final Answer
The maximum separation between the policeman and the thief is **110 meters**.

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