The position of a particle is given by ` vec r= 3.0 t hat i - 2.0 t^2 hat j + 4.0 hat k m`, where (t) in seconds and the coefficients have the proper units for `vec r` to be in metres. (a) Find the ` vec v` and ` vec a` of the particle ? (b) What is the magnitude and direction of velocity of the particle at ` t= 2 s`?

Explanation

`varr(t) = (3.0that(i)-2.0t^(2)hat(j)+4.0hatk))`m
a) `therefore` `vecv(t) = vec(dr)/(dt)= (3.0hat(i)-4.0hat(j))m//s`
and `veca(t) = (vec(dv)/(dt)= (-4.0hatj)m//s^(2)`
b) Magnitude of velocity at t=2.0 s,
`v_(t=2s) = sqrt((3.0)^(2)+(-4.0xx2)^(2))` = `sqrt(9+64)`= `sqrt(73)`.
`=8.54ms^(-1)`
This velocity will subtend an angle `beta` from x-axis, where `tanbeta= (-4.0xx2)/(3.0) = -2.667 = -2.6667`
`therefore beta=tan^(-1)(-2.6667)= -69.44^(@)= 69.44^(@)` from negative x-axis.