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Question

# The acceleration of a particle which moves along the positive x-axis varies with its position as shown in the figure. If the velocity of the particle is 0.8 m/s at x=0, the velocity of the particle at x=1.4 is (in m/s)

A
1.6
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B
1.2
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C
1.4
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D
none
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Solution

## The correct option is B 1.2We know that, a=vdvdx ⇒adx=vdv On integrating the above equation using given limits, we get ∫x=1.4x=0adx=∫vuvdv ∫x=1.4x=0adx=v2−u22...(1) The left hand side ∫adx represents area under the curve on ′x′ axis. Area under curve from x=0 to x=1.8 will be, ∫x=1.4x=0adx=0.4×0.4+12(0.8−0.4)(0.4−0.2)+(1.4−0.4)(0.2−0) =0.16+0.0.04+0.2 =0.40 Substituting it in equation (1), we get 0.40=v2−u22 Given, u=0.8 m/s ⇒0.80=v2−(0.8)2 ⇒v2=0.80+0.64 ⇒v2=1.44 ⇒v=√1.44 ∴v=1.2 m/sec Hence, option (b) is the correct answer.

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