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Question

# The acceleration-displacement graph of a partice moving in a straight line is shown in the figure. Initial velocity of the particle is zero. Find the velocity of the particle when displacement of the particle is 12 m.

A
32 m/s
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B
25 m/s
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C
4 m/s
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D
43 m/s
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Solution

## The correct option is D 4√3 m/sThe area under a−S graph gives v2−u22, Thus, ∫adS=v2−u22...(1) Given, Initial velocity, u=0 From the given figure, ∫adS=Area of ΔOAE+Area of rectangle ABFE+ Area of trapezium FGCB + Area of ΔCGD Now, substituting the values from graph ∫adS=(12×2×2)+[2×(8−2)]+12(10−8)(2+4)+12(12−10)4 ⇒∫adS=2+12+6+4=24...(2) From equation (1) and (2), we get v2−022=24 ⇒v22=24 ⇒v=√48 ∴v=4√3 m/s Hence, option (d) is the correct answer.

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