A particle starts from origin t = 0 with velocity `10.0 hatj ms^(-1)` and moves in the `x-y` plane with a constant acceleration of `8.0 hati+2.0hatjms^(-2)`
(a) At what time the x - coordinate of the particle is 16 m?
(b) What is the y - coordinate of the particle at the time?
(c) What is the speed of the particle at that time?

Explanation

Given `" "vecv_(y)=10.0hatj and v_(y)=10ms^(-1)`
`veca=(8.0hatt+2.0hatj)ms^(-2)`
`"i.e., "|veca_(x)|=8.0ms^(-2) and |veca_(y)|=2.0 ms^(-2)`
`"but "vecv=vecv_(0)+vecat`
`"i.e., "vecv=10hatj+(8.0hati+2.0hatj)t`
`therefore" "vecv=8.0hatit+(10+2.0t)hatj`
`vecs=vecv_(y)t+(1)/(2)veca t^(2)=10hatj t+(1)/(2)(8hati+2hatj)t^(2)=10hatjt+(4hati+hatj)t^(2)`
`=(4t^(2))hati+(10t+t^(2))hatj`
`vecs=s_(x)hatt+s_(y)hatj`
Where`" "s_(x)=4t^(2) and s_(y)=10t+t^(2)`
`t^(2)=(s_(x))/(4)=(16)/(4)" "s_(y)=10(2)+2^(2)`
`t=2s" "=24m`
`v_(x)=8xx2=16ms^(-1) and v_(y) =2xx2+10=14ms^(-1)`
Where `v=sqrt(v_(x)^(2)+v_(y)^(2))=sqrt(16^(2)+14^(2))=21.26ms^(-1)`