A particle starts from the origin at ` t=0` with a velocity of ` 10.0 hat j m//s` and moves in the x-y plane with a constant acceleration of ` ( 8.0 hat i + 2. 0 hat j ) ms^(-2)` . (a) At what time is the x-coordinate of the particle ` 16 m` ? What is the y-coordinate of the particle at that time ? (b) What is the speed of the particle at that time ?

Explanation

It is given that `vecr_(t=0s)=0, vecv_(0)=10.0hat(j)m//s` and `veca(t) = (8.0hat(i)+ 2.0hat(j))ms^(-2)`
a) It means `x_(0) = 0, u_(x)=0, a_(x)=8.0ms^(-2)` and x=16m
Using relation `s=x-x_(0)= u_(x)t+1/2a_(x)t^(2)`, we have
`16-0 = 0+1/2xx 8.0 xx t^(2) rArr t=2s`
`y=y^(0)+u_(y)t+1/2a_(y)t^(2)=0+10.0+2+1/2xx 2.0 xx (2)^(2)`.
`20+4=24m`.
b) velocity of particles at t=2s along x-axis.
`v_(x)= u_(x) + a_(x)t=0+8.0xx2=16.0m//s`
and along y-axis `v_(y)= u_(y)+a_(y)t= 10.0+2.0 xx 2= 14.0m//s`.
`therefore` Speed of particle at t=2s.
`v=sqrt(v_(x)^(2)+v_(y)^(2))` =`sqrt((16.0)^(2)+(14.0)^(2))`= `21.26 ms^(-1)`.