The figure shows a velocity-time graph of a p | vedclass

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Question

(A) The displacement of a particle in a given time interval is equal to the area under the velocity-time graph.For the interval $t = 0 \, s$ to $t = 2

Vedclass · Accepted Answer

$-5$

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(A) The displacement of a particle in a given time interval is equal to the area under the velocity-time graph.For the interval $t = 0 \, s$ to $t = 2 \, s$,the area is a triangle with base $b = 2 \, s$ and height $h = 10 \, m/s$.Displacement $\Delta x = 	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 2 	imes 10 = 10 \, m$.We know that displacement $\Delta x = x_f - x_0$,where $x_f$ is the final position and $x_0$ is the initial position.Given $x_0 = -15 \, m$,we have $10 = x_f - (-15)$.$10 = x_f + 15$.$x_f = 10 - 15 = -5 \, m$.Therefore,the position at $t = 2 \, s$ is $-5 \, m$.

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 ........ $m$.

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