Question

A police inspector spots a thief standing 7 km away from him on a straight road that runs East-West. The inspector is standing on the eastern side while the thief is on the western side of the road. On spotting the inspector, the thief takes his bicycle and tries to cut across the field next to the road, riding away at a uniform speed of \(9\sqrt{2}\) km/hour in a direction making an angle of \(45^{\circ}\) with the road towards North-East. The inspector starts on his scooter at the same instant, moving at a uniform speed of \(15\) km/hour, and catches the thief.

The distance the inspector has to travel is:

• A. 3 km
• B. 3.75 km
• C. 5 km
• D. 6 km

Hint:

Once the time taken to catch the thief is known, the distance is simply the inspector's speed multiplied by that time.

Answer 1

The Correct Option is C

Step 1: Set up coordinates.
Place the inspector at the origin, East as the positive \(x\)-direction and North as the positive \(y\)-direction. The thief starts at \((-7,0)\), \(7\) km west of the inspector.
The thief's speed of \(9\sqrt{2}\) km/hr at \(45^{\circ}\) to the road splits into an East part and a North part, each \(9\sqrt{2}\cos45^{\circ}=9\) km/hr, so at time \(t\) the thief is at \((-7+9t,\ 9t)\).

Step 2: Write the interception condition.
The inspector rides in a straight line at \(15\) km/hr and meets the thief at time \(T\), so
\[ \sqrt{(9T-7)^2+(9T)^2} = 15T \]

Step 3: Solve for \(T\).
Squaring and simplifying (as in the time calculation):
\[ 81T^2-126T+49+81T^2=225T^2 \ \Rightarrow\ 63T^2+126T-49=0 \ \Rightarrow\ 9T^2+18T-7=0 \]
\[ T=\dfrac{-18+\sqrt{324+252}}{18}=\dfrac{-18+24}{18}=\dfrac{1}{3} \text{ hour} \]

Step 4: Find the distance travelled by the inspector.
Distance \( = \) speed \(\times\) time \( = 15 \times \dfrac{1}{3} = 5\) km. Options A, B, D and E do not match this value. \[ \boxed{5 \text{ km}} \]