Question

A ship A is moving Westwards with a speed of 10 km $h^{ - 1}$ and a ship B 100 km South of A, is moving Northwards with a speed of 10 km $ h^{ - 1}$. The time after which the distance between them becomes shortest, is

• A. 5 $ \sqrt 2 $ h
• B. 10 $ \sqrt 2 $ h
• C. 0 h
• D. 5 h

Answer 1

The Correct Option is D

To solve this problem, let's consider the movement of the two ships, A and B, in a coordinate system. We need to determine the point in time when the distance between them is the shortest.

1. Initial Position:
   • Ship A starts at point (0, 0) and moves Westward (negative x-direction) at a speed of 10 km/h.
   • Ship B is initially 100 km South of Ship A, so it starts at point (0, -100) and moves Northward (positive y-direction) at a speed of 10 km/h.
2. Position Functions:
   • The position of Ship A after time t hours is (-10t, 0).
   • The position of Ship B after time t hours is (0, -100 + 10t).
3. Distance Between the Ships:
   • The distance d(t) between Ship A and Ship B at time t is given by the distance formula:
   d(t) = \sqrt{((-10t) - 0)^2 + (0 - (-100 + 10t))^2}
   • Simplifying further,
   d(t) = \sqrt{(100t^2) + ((100 - 10t)^2)}
   d(t) = \sqrt{100t^2 + (100 - 10t)^2} = \sqrt{100t^2 + (10000 - 2000t + 100t^2)} = \sqrt{200t^2 - 2000t + 10000}
4. Minimize the Distance:
   • To find when the distance is minimized, take the derivative of d(t)^2 with respect to t and set it to zero:
   d(t)^2 = 200t^2 - 2000t + 10000
   • Derivative: \frac{d}{dt}(d(t)^2) = 400t - 2000
   • Set the derivative to zero:
   400t - 2000 = 0 \\ t = \frac{2000}{400} \\ t = 5
5. Conclusion:
   • The distance between the two ships becomes shortest after 5 hours.

Hence, the correct option is 5 hours.