Question

If x= 43-612-3t+5 then initial accleration is :-l





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Answer 1

Answer:

S=t

3

−6

2

+3t+4

v=

dt

ds

So =3t

2

−12+3=v

again differentiate ,a=

dt

dv

6t−12=0

t=2s

v=3(2)

2

−12×2+3=−9ms

−1

Answer 2

Explanation:

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