The position x of a moving particle at time t is given by x=t2+2t+3 m. The average velocity of the particle during the t second of its motion is
A
(t+2)ms−1
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B
(t−2)ms−1
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C
t+2+3tms−1
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D
3ms−1
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Solution
The correct option is A(t+2)ms−1 Vavg=DisplacementTime Displacement = Change in position = xt−x0 At t=0s, x0=02+2×0+3=3m and at t=tsec , xt=(t2+2t+3)m Vavg=(t2+2t+3)−3t=t(t+2)t=(t+2)ms−1