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 accleration of ` ( 8.0 hat I + 2. 0 hat j ) ms^(-2)` . (a) At wht 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

Here `vecr(0)=0, veca(t)=(8.0hati+2.0hatj) ms^(-2)`
(a) `x-x_(0)=u_(x)t+(1)/(2)a_(x)=8 ms^(-2), x=16 m `
`16-0=0+(1)/(2)xx8xxt^(2)`
`rArr t=2 s`
`u_(y)=10 m//s , a_(y)=2 m//s^(2), t=2 s`
`therefore y-y_(0) =u_(y)t+(1)/(2) a_(y)t^(2)`
`y=y_(0)+u_(y)t+(1)/(2)a_(y)t^(2)`
`=0+10xx2+(1)/(2)xx2xx2^(2)`
`=20+4=24m`
(b) At `t=2s`
`v_(x)=u_(x)+a_(x)t`
`=0+8xx2=16m//s`
`u_(y)=u_(y)+a_(y)t`
`=10+2xx2=14m//s`
`therefore` Speed of particle at `t=2` s is
`v=sqrt(v_(x)^(2)+v_(y)^(2))`
`=sqrt((16)^(2)+(14)^(2))=21.26 m//s`.