The motion of a body along a straight line is described by the equation $x = t^{3} + 4 t^{2} - 2 t + 5$ Where x is in meter and t is in second. Find e velocity from t = 0 to t = 4

m s^{- 1}

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m s^{- 1}

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m s^{- 1}

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m s^{- 1}

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Solution

The correct option is D 25 $m s^{- 1}$
$\begin{matrix} x = t^{2} + 4 t^{2} - 2 t + 5 a t t = 0 x = 5 a t t = 4 x = 64 + 64 - 8 + 5 = 125 V_{a v} = \frac{125}{5} = 25 m / s \end{matrix}$
Hence,
option $D$ is correct answer.

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The motion of a body along a straight line is described by the equation x=t3+4t2−2t+5 Where x is in meter and t is in second. Find e velocity from t = 0 to t = 4

The correct option is D 25 ms−1x=t2+4t2−2t+5att=0x=5att=4x=64+64−8+5=125Vav=1255=25m/sHence,option D is correct answer.

The motion of a body along a straight line is described by the equation $x = t^{3} + 4 t^{2} - 2 t + 5$ Where x is in meter and t is in second. Find e velocity from t = 0 to t = 4

The correct option is D 25 $m s^{- 1}$
$\begin{matrix} x = t^{2} + 4 t^{2} - 2 t + 5 a t t = 0 x = 5 a t t = 4 x = 64 + 64 - 8 + 5 = 125 V_{a v} = \frac{125}{5} = 25 m / s \end{matrix}$
Hence,
option $D$ is correct answer.