Question

A point which has velocities represented by 7, 8
and 13 is at rest. Find the angle between the
direction of two smaller velocities.
(1) 30°
(2) 45°
(3) 60°
(4) 90°​

Answer 1

Hey Dear,

◆ Answer -

(3) θ = 60°

● Explaination -

Let x, y & z be velocity vectors of the point such that -

|x| = 7 units

|y| = 8 units

|z| = 13 units

For a particle to be at rest, velocity at that point must be balanced.

That is, resultant x+y must be equal in magnitude & opposite in direction to z.

|x + y| = |z|

√(|x|² + |y|² + 2|x||y|.cosθ) = |z|

Squaring both sides -

|x|² + |y|² + 2|x||y|.cosθ = |z|²

7² + 8² + 2×7×8×cosθ = 13²

49 + 64 + 112.cosθ = 169

cosθ = (169 - 49 - 64) / 112

cosθ = 56 / 112

cosθ = 1/2

Taking cos inverse,

θ = arccos(1/2)

θ = 60°

Therefore, angle between direction of two smaller velocities is 60°.

Thanks dear.