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# Solution for 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. - CBSE (Science) Class 12 - Mathematics

#### 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 for question: 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. For the courses CBSE (Science), CBSE (Commerce), PUC Karnataka Science, CBSE (Arts)
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