Question

Describe the motion shown by the following velocity-time graphs.

Accepted Answer

For graph $(a)$

During
interval $A B$ Velocity is $+ v e$ , so the particle is moving in $+ v e$ direction, but it is slowing down as acceleration (slope of $v - t$ curve) is negative.

During
interval $B C$ Particle remains at rest as the velocity is zero. Acceleration is also zero.

During
interval $C D$ Velocity is $- v e$ , so the particle is moving in $- v e$
direction, and is speeding up as acceleration is also negative.

For graph $(b)$

During
interval $A B$ Velocity is $+ v e$ , so the particle is moving in $+ v e$ direction, but the acceleration (slope of $v - t$ curve) is zero. Hence, it is moving with constant speed.

During
interval $B C$ Velocity is $+ v e$ , so the particle is moving in $+ v e$ direction and the acceleration is negative. Hence, particle is slowing down.

During
interval $C D$ Velocity is $- v e$ , so the particle is moving in $- v e$
direction, and is speeding up as acceleration is also negative.

B M Sharma Solutions for Chapter: Kinematics I, Exercise 59: ILLUSTRATION 4.55

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics I, Exercise 59: ILLUSTRATION 4.55

Questions from B M Sharma Solutions for Chapter: Kinematics I, Exercise 59: ILLUSTRATION 4.55 with Hints & Solutions

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