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`` A particle moves on a horizontal line with acceleration a=16-12t where x is measured...

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xdsmith | (Level 2) eNoter

Posted March 4, 2013 at 11:05 AM via web

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`` 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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pramodpandey | College Teacher | (Level 3) Valedictorian

Posted March 17, 2013 at 10:34 AM (Answer #1)

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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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