Question

Ship A is moving with velocity 30m/s due east and ship B with velocity 40m/s due north. Initial separation between the ships is 10km
as shown in figure. After what time ships are closest to each other?
30m's
40m/s
h
-10km
(A) 80s
(B) 120s
(C) 160s
(D) 100s

Answer 1

Given :

▪ Velocity of ship A = 30m/s (east)

▪ Velocity of ship B = 40m/s (north)

▪ Distance b/w them at initial = 10km

To Find :

▪ The time after which the distance b/w them becomes shortest (closest to each other).

Diagram :

↗ Please, see the attachment for better understanding.

Solution :

It is clear from the diagram that the shortest distance b/w the ship A and B is PQ.

Here, sin53° = PQ/OQ

Therefore, PQ = 10×4/5 = 8km = 8000m

Also,

▪ Vab = √(Va² + Vb²) = √(30² + 40²)

▪ Vab = √(900+1600) = √2500

▪ Vab = 50m/s

So, time taken for them to reach shortest path,

☞ t = PQ/Vab

☞ t = 8000/50

☞ t = 160s

Option C is correct !

Answer 2

YOUR QUESTION :

Ship A is moving with velocity 30m/s due east and ship B with velocity 40m/s due north. Initial separation between the ships is 10km

as shown in figure. After what time ships are closest to each other?

30m's

40m/s

h

-10km

(A) 80s

(B) 120s

(C) 160s

(D) 100s

YOUR ANSWER :

Given :

• Velocity of ship A = 30m/s (East).
• Velocity of ship B = 40m/s (North).
• Distance b / w them at initial = 10km.

To find :

• The time after which the distance b/w them becoms shortest.

Solution :

Here, sin53° = PQ/OQ

Therefore, PQ = 10×4/5

= 8km (8000 metre)

Vab = √(Va² + Vb²) = √(30² + 40²)

=> Vab = √(900+1600) = √2500

=> Vab = 50m/s

So, time taken for them to reach shortest path,

t = PQ/Vab

=> t = 8000/50

=> t = 160s.

✔CORRECT OPTION : C