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प्रश्न
The following velocity-time graph represents a particle moving in the positive x-direction. Analyze its motion from 0 to 7 s. Calculate the displacement covered and distance travelled by the particle from 0 to 2 s.
बेरीज
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उत्तर
As per the graph,
(a) From 0 to 1.5 s the particle moving in the opposite direction.
- From 1.5 s to 2 s the particle is moving with increasing velocity.
- From 2 s to 5 s velocity of the particle is constant of magnitude 1 ms -1
- From 5 s to 6 s velocity of the particle is decreasing.
- From 6 s to 7 s the particle is at rest.
(b) Distance covered by the particle – Area covered under (v -t) graph
= `1/2 xx 2 xx 1.5 + 1/2 xx 1 xx 0.5` = 1.5 m + 0.25 m = 1.75 m
Displacement of the particle
= `-1/2 xx 2 xx 1.5 + 1/2 xx 1 xx 0.5`
= -1.5 m to 0.25 m = -1.25 m
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