The position vectors of three particles are given by
`vecx_1 = (5hati +5hatj) m, vecx_2 = (5thati+5t hatj) m and vecx_3 = (5thati+10t^2 hatj)m` as a function of time t. Determine the velocity and acceleration for each, in SI units.

Explanation

`vecv_1 = d vecx_1//dt=0 " as " vecx_1` does not depend on time t.
Thus, the particle is at rest.
`vecv_2 = dvecx_2//dt = 5hati + 5hatj m//s. vecv_2 " does not change with time. " :. veca_2 = 0`
`vecv_2 = sqrt(5^2 +5^2) = 5sqrt2 m//s, tan theta = 5//5 = 1 " or " theta = 45^@`. Thus, the direction of `v_2`, makes an angle of `45^@` to the horizontal.
`vecv_3= d vecx_3 // dt=5hati +20t hatj`
`:. v_3 = sqrt(5^2+(20t)^2)m//s`. Its direction is along
`theta = tan^(-1)((20t)/5)` with the horizontal.
`veca_3 = (d vecv_3)/(dt) =20hatj m//s^2`
Thus, the particle 3 is getting accelerated along the y-axis at `20 m//s^2`.