A particle starts from the origin at t = 0s with a velocity of `10.0 hat(j) m//s` nad 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 16m ? What is the y-coordinate of the particle at that time ?
(b) what is the speed of the particle at the time ?

Explanation

Here, `vec(u) = 10.0 hat(j) ms^(-1)` at t = 0
`vec(a) = (vec(dv))/(dt) = (8.0 hat(i) + 2.0 hat(j)) ms^(-2)`
So d `vec(v) = (8.0 hat(i) + 2.0 hat(j)) dt`
Integrating it with in the limits of motion i.e as time changes from 0 to t, velocity changes is from u to v, we have
`vec(v) - vec(u) = (8.0 hat(i) + 2.0 hat(j))t`
or `vec(v) = vec(u) + 8.0 hat(i) + 2.0 t hat(j)`
As `vec(v) = (vec(d r))/(d t)` or `d vec(r) = vec(v) d t`
So, `d vec(r) = (vec(u) + 8.0 t hat(i) + 2.0 t hat(j)) dt`
Integrating it within the conditions of motion i.e as time changes from 0 to t, displacement is from 0 to r, we have
`r = vec(u) t + (1)/(2) xx 8.0 t hat(i) + (1)/(2) xx 2.0 t^(2)hat(j)`
or `x hat(i) + y hat(j) = 10 hat(j) + 4.0 t^(2) hat(i) + t^(2) hat(j)`
`= 4.0 t^(2) hat(i) + (10 t - t^(2)) hat(j)`
Here, we have `x = 4.0 t^(2)` and `y = 10t + t^(2)`
`t = (x//4)^(1//2)`
(a) At x = 16m, `t = (16//4)^(1//2) = 2s`
`y = 10 xx 2 + 2^(2) = 24m`.
(b) Velocity of the particle at time t is
`vec(v) = 10 hat(j) + 8.0 t hat(i) + 2.0 t hat(j)`
when t = 2s, then `vec(v) = 10 hat(j) + 8.0 xx 2hat(i) + 2.02 hat(j)`
`= 16 hat(i) + 14 hat(j)`
Speed `= |vec(v)| = sqrt(16^(2) + 14^(2)) = 21.26 ms^(-1)`