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A particle starts from the origin at t=O...

A particle starts from the origin at `t=Os` with a velocity of `10.0 hatj m//s` and moves in the `xy`-plane with a constant acceleration of `(8hati+2hatj)m//s^(-2)`. What time is the `x`-coordinate of the particle `16m`?

Text Solution

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Here, `vecu=10.0 hatj ms^(-1)" at t = 0". Veca=(dvecv)/(dt)=(8.0hati+2.0 hatj)ms^(-2)" So " dvecv=(8.0hati+2.0hatj)jdt`
Integrating it within the limits of motion i.e. as time changes from 0 to t, velocity changes from u to v, we have `vecv-vecu=(8+0hati+2.0hatj) t or vecv=vecu+8.0t hati+2.0t hatj`
As `vecv=(dvecr)/(dt) or dvecr=vecv dt So,`
`dvecr=(vecu+8.0 thati+2.0 t hatj)dt`
Integrating it within the conditions of motion i.e. as time changes from 0 to t, displacement changes from 0 to r, we have:
`vecr=vecut+1/2 xx8.0 t^(2) hati+1/2 xx 2.0 t^(2) j or x hati+y hatj+10 hatjt+4.0 t^(2)hati+ t^(2) hatj=4.0 t^(2)hati+(1.0 t+t^(2))hatj`
Here, we have `x=4.0 t^(2) and y=10t+t^(2) therefore t=(x//4)^(3//2)`
Velocity of the particle at time t is `vecv=10hatj+8.0thati+2.0t hatj`
When t=2, s then, `vecv=10hatj+8.0 xx 2hati+2.0 xx 2hatj=16hati+14hatj`
Speed `=|vecv|=sqrt(16^(2)+14^(2))=21.26ms^(-1)`.
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