Question

please....5th.... question....answer

Answer 1

x = 2t^3 - 3t^2 + 1

velocity is zero

v = dx/dt = 0

d(2t^3 -3t^2 + 1) / dt = 0

6t^2 - 6t = 0

6t(t - 1) = 0

t = 1s

Hence at t= 1s its velocity would be zero

Answer 2

v=dx/dt=6t^2 - 6t
v=0
6t(t-1)=0
t=0,t=1
At origin x=0
2t^3-3t^2 +1=0
(t-1)^2 *(2t+1)=0
t=1,t=-1/2
t=-1/2 is not possible
Therefore at origin velocity =0
Hope it helps
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