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.
Time taken by the inspector to catch the thief is:
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.
Time taken by the inspector to catch the thief is:
Time taken by the inspector to catch the thief is:
Show Hint
Split the thief's velocity into East and North parts, then find the time at which the straight-line distance from the inspector's start point equals the distance the inspector himself covers.
Updated On: Jul 10, 2026
- 12 minutes
- 15 minutes
- 18 minutes
- 20 minutes
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Set up coordinates.
Place the inspector at the origin, with East as the positive \(x\)-direction and North as the positive \(y\)-direction. The thief starts \(7\) km to the west of the inspector, so the thief's starting point is \((-7,0)\).
The thief rides at \(9\sqrt{2}\) km/hr at \(45^{\circ}\) to the road towards the North-East, so his velocity splits equally into an East part and a North part, each of size \(9\sqrt{2}\cos 45^{\circ} = 9\sqrt{2}\times \dfrac{1}{\sqrt{2}} = 9\) km/hr.
So after time \(t\) hours, the thief is at \((-7+9t,\ 9t)\).
Step 2: Write the condition for the inspector to catch the thief.
The inspector rides in a straight line at a constant \(15\) km/hr and reaches the thief exactly at time \(T\), so the straight-line distance from the origin (the inspector's start) to the thief's position at time \(T\) must equal the distance the inspector himself covers, \(15T\).
\[ \sqrt{(9T-7)^2+(9T)^2} = 15T \]
Step 3: Solve the equation for \(T\).
Squaring both sides:
\[ (9T-7)^2+(9T)^2 = 225T^2 \]
\[ 81T^2-126T+49+81T^2=225T^2 \]
\[ 162T^2-126T+49=225T^2 \]
\[ 63T^2+126T-49=0 \]
Dividing through by \(7\):
\[ 9T^2+18T-7=0 \]
By the quadratic formula, \(T=\dfrac{-18\pm\sqrt{18^2+4\times9\times7}}{2\times9}=\dfrac{-18\pm\sqrt{324+252}}{18}=\dfrac{-18\pm24}{18}\). Taking the positive root, \(T=\dfrac{6}{18}=\dfrac{1}{3}\) hour.
Step 4: Convert to minutes.
\(\dfrac{1}{3}\) hour equals \(20\) minutes. Options A, B, C and E do not match this value. \[ \boxed{20 \text{ minutes}} \]
Place the inspector at the origin, with East as the positive \(x\)-direction and North as the positive \(y\)-direction. The thief starts \(7\) km to the west of the inspector, so the thief's starting point is \((-7,0)\).
The thief rides at \(9\sqrt{2}\) km/hr at \(45^{\circ}\) to the road towards the North-East, so his velocity splits equally into an East part and a North part, each of size \(9\sqrt{2}\cos 45^{\circ} = 9\sqrt{2}\times \dfrac{1}{\sqrt{2}} = 9\) km/hr.
So after time \(t\) hours, the thief is at \((-7+9t,\ 9t)\).
Step 2: Write the condition for the inspector to catch the thief.
The inspector rides in a straight line at a constant \(15\) km/hr and reaches the thief exactly at time \(T\), so the straight-line distance from the origin (the inspector's start) to the thief's position at time \(T\) must equal the distance the inspector himself covers, \(15T\).
\[ \sqrt{(9T-7)^2+(9T)^2} = 15T \]
Step 3: Solve the equation for \(T\).
Squaring both sides:
\[ (9T-7)^2+(9T)^2 = 225T^2 \]
\[ 81T^2-126T+49+81T^2=225T^2 \]
\[ 162T^2-126T+49=225T^2 \]
\[ 63T^2+126T-49=0 \]
Dividing through by \(7\):
\[ 9T^2+18T-7=0 \]
By the quadratic formula, \(T=\dfrac{-18\pm\sqrt{18^2+4\times9\times7}}{2\times9}=\dfrac{-18\pm\sqrt{324+252}}{18}=\dfrac{-18\pm24}{18}\). Taking the positive root, \(T=\dfrac{6}{18}=\dfrac{1}{3}\) hour.
Step 4: Convert to minutes.
\(\dfrac{1}{3}\) hour equals \(20\) minutes. Options A, B, C and E do not match this value. \[ \boxed{20 \text{ minutes}} \]
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