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  • Displacement equation is given by- <img alt="\mathrm{s=\alpha \sqrt{t}+\beta t^2}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bs%3D%5Calpha%20%5Csqrt%7Bt%7D&plus;%5C

Displacement equation is given by-
\mathrm{s=\alpha \sqrt{t}+\beta t^2} where \mathrm{\alpha \, \, and\, \, \, \beta } are 
constants. The acceleration at  \mathrm{t=1 \mathrm{sec}} will be -

Option: 1

 Zero 


Option: 2

1


Option: 3

constant Acceleration


Option: 4

None of these


Answers (1)

best_answer

Given,
\mathrm{ \begin{aligned} & s=\alpha \sqrt{t}+\beta t^2 \\ & s=\alpha t^{1 / 2}+\beta t^2 \end{aligned}}
 To find Acceleration firstly find Velocity - 
\mathrm{v=\frac{d s}{d t}=\frac{d}{d t}\left(\alpha t^{1 / 2}+\beta t^2\right)}
\mathrm{v=\frac{d s}{d t}=\frac{\alpha}{2} \alpha^{-1 / 2}+2 \beta t \}
\mathrm{Acceleration\, \, \, \, \, \frac{d v}{d t} \, \, or \, \, \, \frac{d^2 s}{d t^2}=\frac{d}{d t}\left[\frac{\alpha}{2} \alpha^{-1 / 2}+2 \beta t\right]}
\mathrm{\begin{aligned} \frac{d v}{d t} & =-\frac{4}{4} \alpha^{-3 / 2}+2 \beta \\ \frac{d v}{d t} & =-\frac{1}{4 \alpha^{3 / 2}}+2 \beta \\ \frac{d v}{d t} & =\frac{1}{2 \beta-\frac{1}{4 \alpha^{3 / 2}}} \rightarrow \text { Constant Acceleration } \end{aligned}}

Posted by

himanshu.meshram

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