Question

a body moves of the total of nine seconds starting fro rest with uniform velocity retardation which is twice the value of acceleration and then stops the duration of acceleration is
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Answer 1

$\huge\star\underline\mathfrak\red{Answer:-}$

Consider the acceleration of the is α and retardation is β

and according to question retardation is twice that acceleration then

β=2a

Body moves with acceleration α for t1 sec

v=u+at

v=0+αt1 .....(1)

Body retards with 2α for time t2 sec and then body becomes rest .

v=u+at

Here, 

u=αt1v=0a=−2α

0=αt1−2αt2t1=2t2

As

t1+t2=9sec

Then,

t1=3sec

t2=6sec

Therefore, 3sec body moves with uniform acceleration.

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Answer 2

Answer:

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