`` A particle moves on a horizontal line with acceleration a=16-12t where x is measured in metres and time in seconds. Initially the particle is at rest at the origin.

Find its maximum positive displacement and

Find the average spped of partile during time interval from t=0 and t=4 (which is again at the origin)

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Given acceleration 16-12t ,let x is displacement at any time ' t'

But we know

`(d^2x)/(dt^2)=16-12t`

Integrating w.r.t.' t ' once ,we have

`dx/dt=16t-6t^2+c` ,where c is integrating constant determined by initial conditions,

when t=0 ,initial velocity `dx/dt=0` ,thus we get c=0

`dx/dt=16t-6t^2`

again ,integrating w. r.t ' t ' ,we have

`x=8t^2-2t^3+d`

where d is integrating constant.

when t=0 ,x=0 ,so we have d=0

`x=8t^2-2t^3`

`t=0 , x=0`

`t=4 , x=0`

velocity is zero.

average velocity =16.4-6.4^2=64-96=-32m/sec

max x if `16t-6t^2=0`

`t=0 ,t=8/3 sec`

`maxx=8(8/3)^2-2(8/3)^3=512/27`

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