a point which has velocity is represented by 78 and 13 is at rest find the angle between the direction of two smaller velocities
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Answered by
10
Hey Dear,
◆ Answer -
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..
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5
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