If acceleration of an object is given as $a = - 6 s^{- 3}$ , where's is the displacement. Motion starts with infinite displacement from rest, then what will be the velocity of the object at 6 m?

0.167 m/s

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2.3 m/s

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0.408 m/s

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0.68 m/s

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Solution

The correct option is C 0.408 m/s
$v \frac{d v}{d s} = a$

$- 6 s^{- 3} = v \frac{d v}{d s}$

$\int_{\infty}^{6} - 6 s^{- 3} d s = \int_{0}^{v} v d v$

${[- \frac{6}{- 2} s^{- 2}]}_{\infty}^{- 2} = \frac{v^{2}}{2}$

$⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} a = \frac{d v}{d t}, d t = \frac{d S}{v} a = \frac{d v}{(\frac{d S}{v})} = v (\frac{d v}{d S}) \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

${[\frac{3}{s^{2}}]}_{\infty}^{6} = \frac{v^{2}}{2}$

$\Rightarrow v^{2} = \frac{6}{6 \times 6} = \frac{1}{6}$

$v = 0.408 m / s$

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If acceleration of an object is given as a=−6s−3, where's is the displacement. Motion starts with infinite displacement from rest, then what will be the velocity of the object at 6 m?

The correct option is C 0.408 m/svdvds=a −6s−3=vdvds ∫6∞−6s−3ds=∫v0vdv [−6−2s−2]6∞=v22 ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣a=dvdt,dt=dSva=dv(dSv)=v(dvdS)⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦ [3s2]6∞=v22 ⇒v2=66×6=16 v=0.408 m/s

If acceleration of an object is given as $a = - 6 s^{- 3}$ , where's is the displacement. Motion starts with infinite displacement from rest, then what will be the velocity of the object at 6 m?