The figure shows a velocity-time graph of a particle moving along a straight line. If the particle starts from the position $x_{0} = 15 m$ , then its position at $t = 2 s$ will be:

25 m

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5 m

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10 m

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15 m

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Solution

The correct option is A $25 m$
Area under the velocity-time graph gives the displacement of the particle. In the first two seconds, displacement is given by:
$s = \frac{1}{2} \times 2 \times 10 = 10 m$
Given initial position is $x_{o} = 15$
Since displacement is change in position, $s = x_{f} - x_{o}$
$x_{f} - x_{o} = 10$
$x_{f} = 15 + 10 = 25 m$

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The figure shows a velocity-time graph of a particle moving along a straight line. If the particle starts from the position x0=15 m, then its position at t=2 s will be:

The correct option is A 25 mArea under the velocity-time graph gives the displacement of the particle. In the first two seconds, displacement is given by:s=12×2×10=10 mGiven initial position is xo=15Since displacement is change in position, s=xf−xoxf−xo=10xf=15+10=25 m

The figure shows a velocity-time graph of a particle moving along a straight line. If the particle starts from the position $x_{0} = 15 m$ , then its position at $t = 2 s$ will be: