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Question

The acceleration of a particle which moves along the positive x-axis varies with its position as shown. The velocity of the particle is 0.8 m/s at x=0 and the acceleration becomes zero after x=1.4 m. The maximum velocity of the particle (in m/s) is

A
1.6 m/s
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B
1.2 m/s
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C
1.4 m/s
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D
1 m/s
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Solution

The correct option is B 1.2 m/s
As we know that the area under ax curve

(a=vdvdtvuvdv)

gives (v2u22) value
So, area of the given curve is
=0.4×0.2+0.4×0.2+0.4×0.2+12×0.4×0.2+0.6×0.2
Area =0.4
Now, v2u2=2 (Area)
And, as per given question u=0.8 m/s
v2=2×0.4+(0.8)2=1.44
v=1.2 m/s
Hence, the maximum velocity of the particle is 1.2 m/s

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