A particle moves along a straight line such that its displacement is given by s=t?-6t?+3t+4m. find the velocity when the acceleration is zero.

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Dear Student,
$A s g i v e n, s = t^{3} - 6 t^{2} + 3 t + 4, N o w, v e l o c i t y, v = \frac{d s}{d t} = 3 t^{2} - 12 t + 3 a n d a c c e l e r a t i o n, a = \frac{d^{2} s}{d t^{2}} = 6 t - 12 N o w i f, a = 0 \Rightarrow 6 t - 12 = 0 \Rightarrow t = 2 s e c . S o, v e l o c i t y a t, t = 2 s e c, v = 3 \times 2 \times 2 - 12 \times 2 + 3 = - 9 m / s .$

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A particle moves along a straight line such that its displacement is given by s=t?-6t?+3t+4m. find the velocity when the acceleration is zero.

Dear Student, As given , s =t3-6t2+3t+4, Now, velocity, v =dsdt=3t2-12t+3 and acceleration, a =d2sdt2 =6t-12 Now if, a =0 ⇒6t-12=0 ⇒t = 2 sec.So, velocity at, t= 2sec, v = 3×2×2-12×2+3 =-9 m/s. Regards