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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 - Physics

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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

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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Distance and Displacement
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पाठ 2: Kinematics - Evaluation [पृष्ठ १०१]

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सामाचीर कलवी Physics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 2 Kinematics
Evaluation | Q IV. 8. | पृष्ठ १०१
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