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Displacement of the bike is moving with respect to time as below -

\mathrm{x=\sqrt{t}+4}, the acceleration of the bike can be written as -

 

Option: 1

\mathrm{proportional \; to\; t^{3 / 2}}


Option: 2

\mathrm{Proportional\: to\: t^{-3 / 2}}


Option: 3

\mathrm{Proportional \: to \: t^2}


Option: 4

\mathrm{Proportional \: to\: t}


Answers (1)

best_answer

\mathrm{\begin{aligned} \text { Displacement }(x) & =\sqrt{t}+4 \\ (x) & =t^{1 / 2}+4 \\ \text { Velocity }(v)=\frac{d x}{d t} & =\frac{d}{d t}\left[t^{1 / 2}+4\right] \\ & =\frac{1}{2} t^{-1 / 2}+0=\frac{1}{2} t^{-1 / 2} \end{aligned}}

\mathrm{ \begin{aligned} \text { Acceleration }(a)=\frac{d^2 x}{d t^2} & =\frac{1}{2}\left(-\frac{1}{2}\right) t^{-3 / 2} \\ & =-\frac{1}{4} t^{-3 / 2} \end{aligned}}

\mathrm{Acceleration \propto t^{-3 / 2}}

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