If the distance s covered in time t by a part | vedclass

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(B) Given that,$s = t^5 - 40t^3 + 30t^2 + 80t - 250$. Velocity $v = \frac{ds}{dt} = 5t^4 - 120t^2 + 60t + 80$. Acceleration $a = \frac{dv}{dt} = 20t^3

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$-260$

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(B) Given that,$s = t^5 - 40t^3 + 30t^2 + 80t - 250$. Velocity $v = \frac{ds}{dt} = 5t^4 - 120t^2 + 60t + 80$. Acceleration $a = \frac{dv}{dt} = 20t^3 - 240t + 60$. Let $f(t) = 20t^3 - 240t + 60$. To find the minimum acceleration,we find the critical points by setting $f'(t) = 0$: $f'(t) = 60t^2 - 240 = 0 \Rightarrow t^2 = 4 \Rightarrow t = \pm 2$. Now,check the second derivative $f''(t) = 120t$. At $t = 2$,$f''(2) = 120(2) = 240 > 0$,so $f(t)$ has a minimum at $t = 2$. Minimum acceleration $a_{\min} = f(2) = 20(2)^3 - 240(2) + 60 = 160 - 480 + 60 = -260$.

If the distance $s$ covered in time $t$ by a particle moving on a straight line is given by $s = t^5 - 40t^3 + 30t^2 + 80t - 250$,then its minimum acceleration is

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