Question

If initial velocity of particle is 2 m/s, the maximum velocity of particle from t=0 to t= 20 sec is :

(A) 20 m/s
(B) 18 m/s
(C) 22 m/s
(D) 24 m/s​

Answer 1

Explanation:

area under curve represents change in velocity.

so area will be maximum at t=0 and t=10 seconds and after 10seconds acceleration will be negative

as area =20

vf-vi=20 (vi=20)

vf= Vmax =22m/s

Answer 2

Concept:

• One dimensional motion
• Graphical representation of one-dimensional motion
• Kinematic equations

Given:

• Initial velocity of the particle u = 2 m/s
• Time period t = 0 to t = 20 s

Find:

• The maximum velocity of the particle in the given time period.

Solution:

In an acceleration - time graph, the area under the curve is the change in velocity.

From time t = 0 to t = 10 s, the graph is positive, so there is positive acceleration.

From time t = 10 s to t = 20 s, the graph is negative, so there is negative acceleration.

To find the maximum velocity of the particle, we have to consider positive acceleration only which is from t = 0 to t = 10 s

The change in velocity is the area under the graph.

Δv = 1/2 (2-0) (2-0) + (6-2) (2-0) + (8-6) (2-0) + 1/2 (8-6) (4-2) + 1/2 (10-8) (4-0)

Δv = 1/2 (2) (2) + (4)(2) + (2)(2) + 1/2 (2)(2) + 1/2 (2) (4)

Δv = 2 + 8 + 4 + 2 + 4

Δv = 20

The final velocity v = u + Δv = 2 + 20 = 22

The correct option is 22 m/s.

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