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A particle moves towards east with velocity 5 m/s. After 10 seconds its direction changes towards north with same velocity. The average acceleration of the particle is
  • a)
    zero
  • b)
    1/√2 m/s2N-W
  • c)
    1/√2 m/s2N-E
  • d)
    1/√2 m/s2S-W
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
A particle moves towards east with velocity 5 m/s. After 10 seconds it...
The average acceleration of the particle is not zero since its velocity changed direction.

Using the formula for average acceleration,

average acceleration = (change in velocity) / (time taken)

The change in velocity can be calculated as the final velocity minus the initial velocity.

Initial velocity towards east = +5 m/s
Final velocity towards north = +5 m/s

To calculate the final velocity vector, we can use Pythagoras' theorem.

Magnitude of velocity = 5 m/s
Angle between velocity and north direction = 90 degrees

Using trigonometry,

Magnitude of northward component of velocity = 5 sin(90) = 5
Magnitude of eastward component of velocity = 5 cos(90) = 0

Therefore, the final velocity vector is +5 m/s towards north.

The change in velocity is then:

(change in velocity) = (+5 m/s towards north) - (+5 m/s towards east)
(change in velocity) = +5 m/s towards north

The time taken for the change in velocity to occur is 10 seconds.

Therefore,

average acceleration = (change in velocity) / (time taken)
average acceleration = (+5 m/s towards north) / (10 s)
average acceleration = +0.5 m/s^2 towards north

The answer is (b) 0.5 m/s^2 towards north.
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A particle moves towards east with velocity 5 m/s. After 10 seconds it...
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A particle moves towards east with velocity 5 m/s. After 10 seconds its direction changes towards north with same velocity. The average acceleration of the particle isa)zerob)1/√2 m/s2N-Wc)1/√2 m/s2N-Ed)1/√2 m/s2S-WCorrect answer is option 'C'. Can you explain this answer?
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