Question

A ship A is travelling due north with speed of 24 m/s. Another ship B is travelling due east with a speed of 10 m/s. The
displacement of B as seen from A in 10 sec is : (direction of displacement is approx.)
(1) 260 m north-west (2) 260 m north-east (3) 260 m south-west (4) 260 m south-east

Answer 1

Answer:

Let the speed of ship be x,

Thus x×cos60=10 (since the ship is always north of the ship due east)

thus 2x=10

Therefore, speed of the second ship, x=20km/h

Answer 2

The displacement of B as seen from A in 10 seconds is (2)260 m north-east.

Given,

Speed of the ship-A=24 m/s due north

Speed of the ship-B=10 m/s due east.

To find,

the displacement of B as seen from A in 10 seconds.

Solution:

• The concept that will be used here is that of relative motion in one dimension.

Displacement made by the ship-A in 10 seconds,

$Displacement=VelocityXTime\\Displacement=24 X10\\Displacement=240 m(due north).$

Displacement made by the ship-B in 10 seconds,

$Displacement=VelocityXTime\\Displacement=10 X10\\Displacement=100 m(due east).$

Taking north direction as positive y-axis and the east direction as positive x-axis, and the angle between them will be 90°.

Displacement made by the ship-A in 10 seconds=240 i m

Displacement made by the ship-B in 10 seconds=100 j m.

Their resultant vector will give the resultant displacement.

Displacement of B as seen from A, Sba will be:

$Sba=\sqrt{Sb^{2}+Sa^{2} } \\Sba=\sqrt{(240)^{2+(100)^{2} } } \\Sba=\sqrt{57600+10000} \\Sba=\sqrt{67600} m\\Sba=260m(due north-east).$

Hence, the displacement is 260 m due north-east.

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