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Instantaneous Velocity

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Question

The position of an object moving along x-axis is given by $x (t) = (4.2 t^{2} + 2.6) m$ , then find the velocity of particle at $t = 0 s$ and $t = 3 s$ , then find the average velocity of particle at $t = 0 s$ and $t = 3 s$ .

v (0) = 0 m s^{- 1}, v (3) = 35.2 m s^{- 1}

and

¯ ¯ ¯ v = 14.6 m s^{- 1}

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v (0) = 10 m s^{- 1}, v (3) = 25.2 m s^{- 1}

and

¯ ¯ ¯ v = 14.6 m s^{- 1}

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v (0) = 20 m s^{- 1}, v (3) = 25.2 m s^{- 1}

and

¯ ¯ ¯ v = 14.6 m s^{- 1}

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v (0) = 0 m s^{- 1}, v (3) = 25.2 m s^{- 1}

and

¯ ¯ ¯ v = 14.6 m s^{- 1}

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Solution

The correct option is C $v (0) = 0 m s^{- 1}, v (3) = 25.2 m s^{- 1}$ and $¯ ¯ ¯ v = 14.6 m s^{- 1}$
Given,
$x (t) = (4.2 t^{2} + 2.6) m$
Velocity, $v (t) = \frac{d x (t)}{d t}$
$v (t) = 8.4 t$
At $t = 0 s e c$ , $v (0) = 0 m / s$
At $t = 3 s e c$ , $v (3) = 8.4 \times 3 = 25.2 m / s$
The average velocity of the particle at $t = 0 s$ and $t = 3 s$
$¯ v = \frac{4.2 \times 3 \times 3 + 2.6}{3} = 14.6 m / s$
The correct option is D.

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The position of an object moving along x-axis is given by x(t)=(4.2t2+2.6)m, then find the velocity of particle at t=0s and t=3s, then find the average velocity of particle at t=0s and t=3s.

The correct option is C v(0)=0ms−1,v(3)=25.2ms−1 and ¯¯¯v=14.6ms−1Given,x(t)=(4.2t2+2.6)mVelocity, v(t)=dx(t)dtv(t)=8.4tAt t=0sec, v(0)=0m/sAt t=3sec, v(3)=8.4×3=25.2m/sThe average velocity of the particle at t=0s and t=3s¯v=4.2×3×3+2.63=14.6m/sThe correct option is D.

The position of an object moving along x-axis is given by $x (t) = (4.2 t^{2} + 2.6) m$ , then find the velocity of particle at $t = 0 s$ and $t = 3 s$ , then find the average velocity of particle at $t = 0 s$ and $t = 3 s$ .