PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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The Space S Described in Time T by a Particle Moving in a Straight Line is Given by S = T 5 − 40 T 3 + 30 T 2 + 80 T − 250 . Find the Minimum Value of Acceleration. - PUC Karnataka Science Class 12 - Mathematics

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Question

The space s described in time 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.

Solution

\[\text { Given: } \hspace{0.167em} s = t^5 - 40 t^3 + 30 t^2 + 80t - 250\]

\[ \Rightarrow \frac{ds}{dt} = 5 t^4 - 120 t^2 + 60t + 80\]

\[\text { Acceleration, } a = \frac{d^2 s}{d t^2} = 20 t^3 - 240t + 60\]

\[ \Rightarrow \frac{da}{dt} = 60 t^2 - 240\]

\[\text { For maximum or minimum values of a, we must have }\]

\[\frac{da}{dt} = 0\]

\[ \Rightarrow 60 t^2 - 240 = 0\]

\[ \Rightarrow 60 t^2 = 240\]

\[ \Rightarrow t = 2\]

\[\text { Now, }\]

\[\frac{d^2 a}{d t^2} = 120t\]

\[ \Rightarrow \frac{d^2 a}{d t^2} = 240 > 0\]

\[\text { So, acceleration is minimum at t } = 2 . \]

\[ \Rightarrow a_{min =} 20 \left( 2 \right)^3 - 240\left( 2 \right) + 60 = 160 - 480 + 60 = - 260\]

\[ \therefore \text { At t }= 2: \]

\[a = - 260\]

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Solution The Space S Described in Time T by a Particle Moving in a Straight Line is Given by S = T 5 − 40 T 3 + 30 T 2 + 80 T − 250 . Find the Minimum Value of Acceleration. Concept: Graph of Maxima and Minima.
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