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प्रश्न
The space s described in time t by a particle moving in a straight line is given by S = \[t5 - 40 t^3 + 30 t^2 + 80t - 250 .\] Find the minimum value of acceleration.
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उत्तर
\[\text { Given: } \hspace{0.167em} s = t^5 - 40 t^3 + 30 t^2 + 80t - 250\]
\[ \Rightarrow \frac{ds}{dt} = 5 t^4 - 120 t^2 + 60t + 80\]
\[\text { Acceleration, } a = \frac{d^2 s}{d t^2} = 20 t^3 - 240t + 60\]
\[ \Rightarrow \frac{da}{dt} = 60 t^2 - 240\]
\[\text { For maximum or minimum values of a, we must have }\]
\[\frac{da}{dt} = 0\]
\[ \Rightarrow 60 t^2 - 240 = 0\]
\[ \Rightarrow 60 t^2 = 240\]
\[ \Rightarrow t = 2\]
\[\text { Now, }\]
\[\frac{d^2 a}{d t^2} = 120t\]
\[ \Rightarrow \frac{d^2 a}{d t^2} = 240 > 0\]
\[\text { So, acceleration is minimum at t } = 2 . \]
\[ \Rightarrow a_{min =} 20 \left( 2 \right)^3 - 240\left( 2 \right) + 60 = 160 - 480 + 60 = - 260\]
\[ \therefore \text { At t }= 2: \]
\[a = - 260\]
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