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Standard XIII

Physics

Average velocity

The position ...

Question

The position $x$ of a moving particle at time $t$ is given by $x = t^{2} + 2 t + 3 m$ . The average velocity of the particle during the first second of its motion is

1 {ms}^{- 1}

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2 {ms}^{- 1}

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3 {ms}^{- 1}

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4 {ms}^{- 1}

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Solution

The correct option is C $3 {ms}^{- 1}$
Given, $x = t^{2} + 2 t + 3 m$
$V_{avg} = \frac{Displacement}{Time}$
Displacement = Change in position = $x_{1} - x_{0}$
At $t = 0 s$ , $x_{0} = 3 m$
and at $t = 1 s e c$ , $x_{1} = 1 + 2 + 3 = 6 m$
$V_{av} = \frac{6 - 3}{1} = 3 {ms}^{- 1}$

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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 first second of its motion is

The correct option is C 3 ms−1Given, x=t2+2t+3 m Vavg=DisplacementTime Displacement = Change in position = x1−x0 At t=0 s, x0=3 m and at t=1 sec , x1=1+2+3=6 m Vav=6−31=3 ms−1

The position $x$ of a moving particle at time $t$ is given by $x = t^{2} + 2 t + 3 m$ . The average velocity of the particle during the first second of its motion is

The correct option is C $3 {ms}^{- 1}$
Given, $x = t^{2} + 2 t + 3 m$
$V_{avg} = \frac{Displacement}{Time}$
Displacement = Change in position = $x_{1} - x_{0}$
At $t = 0 s$ , $x_{0} = 3 m$
and at $t = 1 s e c$ , $x_{1} = 1 + 2 + 3 = 6 m$
$V_{av} = \frac{6 - 3}{1} = 3 {ms}^{- 1}$