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Displacement equation as below of the particle s=a t^{3 / 2}, where a is constant.
Acceleration of the particle will be proportional to-
 

Option: 1

\quad t^{1 / 2}


Option: 2

t^{1/3}


Option: 3

t^{3 / 2}


Option: 4

t^{-1 / 2}


Answers (1)

best_answer

Given, 

\quad s=a t^{3 / 2}
 v=\frac{d s}{d t}=\frac{3}{2} a t^{1 / 2} \\
a=\frac{d v}{d t}=\frac{3}{4} a t^{-1 / 2}
a \alpha t^{-1 / 2}

Posted by

Ritika Kankaria

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