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.

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Solution

$Given : s = t^{5} - 40 t^{3} + 30 t^{2} + 80 t - 250 \Rightarrow \frac{d s}{d t} = 5 t^{4} - 120 t^{2} + 60 t + 80 Acceleration, a = \frac{d^{2} s}{d t^{2}} = 20 t^{3} - 240 t + 60 \Rightarrow \frac{d a}{d t} = 60 t^{2} - 240 For maximum or minimum values of a, we must have \frac{d a}{d t} = 0 \Rightarrow 60 t^{2} - 240 = 0 \Rightarrow 60 t^{2} = 240 \Rightarrow t = 2 Now, \frac{d^{2} a}{d t^{2}} = 120 t \Rightarrow \frac{d^{2} a}{d t^{2}} = 240 > 0 So, acceleration is minimum at t = 2 . \Rightarrow a_{m i n =} 20 {(2)}^{3} - 240 (2) + 60 = 160 - 480 + 60 = - 260 ∴ At t = 2 : a = - 260$

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