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A particle moving with displacement  \frac{\alpha }{2}t+\frac{\beta }{2} t^2 what will be the acceleration with respect to time-

Option: 1

\alpha t+\beta


Option: 2

\beta \text { (constant) }


Option: 3

t+\beta


Option: 4

\frac{\beta}{2}+t


Answers (1)

best_answer

Displacement  = \mathrm{\frac{\alpha}{2} t+\frac{\beta }{2} t^2}
Now, Acceleration with respect to time, firstly calculate
\mathrm{\text { velocity } \frac{d s_0(t)}{d t}=\frac{d}{d t}\left[\frac{\alpha}{2} t+\frac{\beta}{2} t^2\right] }
\mathrm{=\frac{\alpha}{2}+\beta t }
\mathrm{ \text { Acceleration }(a)=\frac{d V_c(t)}{d t}=\frac{d}{d t}\left[\frac{\alpha}{2}+\beta t\right] }
\mathrm{=0+\beta=\beta \rightarrow \text { constant. } }

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Rishabh

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