Question

Ship A is moving westwards with 10km/h and Ship B is at 100km in the south and moving with 10km/h towards north. The time taken by the ship B to cover the minimum distance is
a) 5h. b)5√2h.
c)10h. d)0h​

Answer 1

Answer:

The time taken by the ship B to cover the minimum distance is 5h.

Explanation:

To find the time taken by the ship B to cover the minimum the distance = ?

Velocity of ship A, $\sf V_A$ = 10 km/h towards west.

Velocity of ship B, $\sf VB$ = 10 km/h towards north.

• OS = 100 km
• OP is he shortest distance.

Now,

Relative velocity of A and B is :-

$: \implies \sf V_{AB} = \sqrt{V_A^2 + V_B^2}$

$: \implies \sf 10 \sqrt{2} km/h$

$: \implies \sf cos 45^{\circ} = \dfrac{OP}{OS} ; \dfrac{1}{\sqrt{2}} = \dfrac{OP}{100}$

$: \implies \sf OP = \dfrac{100}{\sqrt{2}}$

$: \implies \sf OP = \dfrac{100 \sqrt{2}}{2}$

$: \implies \sf 50 \sqrt{2} km$

Now,

$\sf Time, t = \dfrac{OP}{V_{AB}}$

$: \implies \sf \dfrac{50 \sqrt{2}}{10 \sqrt{2}}$

$: \implies \sf 5 hours$

Therefore,

The time taken by the ship B to cover the minimum distance is 5 hours.