A particle moves aling x-axis with an initial speed `v_(0)=5 m s^(-1)`. If its acceleration varies with with time asshown in `a-t` graph in .
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a. Find the time when the particle starts moving along - x direction

Explanation

The velocity of the particle at `t=4 s` can given as
`vec v4=vec v_(0)+Delta vec v` …(i)
where ` Delta vec v-=A`
(=area under `a-t` graaph during four seconds )
Referring to `a-t` graph , we have
`A=A_(1) +A_(2)-A_(3)-A_(4)` ....(ii)
where `A_(1)=5xx1=5,A_(2) =(1)/(2) xx x xx5`,
`A_(3)=(1)/(2)xx(-1-x)xx10`, and `A_(4)=(1)/(2)xx2xx10=10`
We can find `x` as following:
Using properties of similar triangles, we have `(x)/(5)=(1-x)/(10)`
This yields `x=(1)/(2)`.
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Substituting `x=(1)/(2) in A_(2) =(1)/(2)xx x xx5`
and `A_(3)=(1)/(2)(1-x)xx10`,
we have `A_(2) =-(5)/(6)` and `A_(3)=(10)/(3)`.
Then substitution `A_(1),A_(2),A_(3),` and `A_(4)` in (ii), we have `A=-7.5.`
Negative area tells us that change in velcity is along `-x` direction
`Delta vec v =-7.5 m s^(-1)`
Hence, substituting (i),
`vec v_(0)=5 m s^(-1)` and `Delta vec v=-7.5m s^(-1)`,
we have
`vec v _(4)=vec v_(0)+ Delta vec v=5-7.5=-2.5 m s^(-1)`.