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  • Displacement relation of any particle is given by - <img alt="\mathrm{s=\alpha t+\beta t^3+5 }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bs%3D%5Calpha%20t&plus;%5C

Displacement relation of any particle is given by -
\mathrm{s=\alpha t+\beta t^3+5 }
The Acceleration is proportional to -

Option: 1

\mathrm{t^{1 / 2}}


Option: 2

\mathrm{t}


Option: 3

\mathrm{t^{3 / 2}}


Option: 4

\mathrm{t^2}


Answers (1)

best_answer

Given,
\mathrm{s=\alpha t+\beta t^3+5 }
velocity is given by.
\mathrm{ \frac{d s}{d t}=\alpha+3 \beta t^2+0 }


\mathrm{\frac{d s}{d t}=v=\alpha+3 \beta t^2 }

Now,

Acceleration 

\mathrm{ \frac{d v}{d t}=a=0+6 \beta t }


\mathrm{ \frac{d v}{d t}=a=6 \beta t \\ }


\mathrm{\frac{d V}{d t} \alpha t \quad }.

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mansi

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