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, `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)`
At `x=16m, t=(1.6//4)^(1//2) =2s y=10 xx 2 +2^(2)=24m`