The intantaneous coordinates of a particle are ` x= (8t-1)m` and `y=(4t^(2))m`. Calculate (i) the average velocity of the particle during time interval from `t =2s ` to `t=4s` (ii) the instantaneous velocity of the particle at `t = 2s`.

Explanation

(i) Given that `x(t) = (8t-1)` m and `y=(4t^(2))m`
`r(t) = (8t- 1)hati + (4 t^(2) )hatj" ".....(i)`
At `t=2s," "vecr_(1) = 15 hati + 16 hatj`
At `t= 4s," "vecr_(2) = 31 hati + 64 hatj`
`because` Displacement, `vecr_(2) - vecr_(1) = (31 hati + 64 hatj)- (15 hati + 16hatj)`
`" "=16 hati + 48 hatj`
`therefore` Average velocity `vecv_(av) =(vecr_(2) -vecr_(1))/(vect_(2) + vect_(1))`
`" "(16hati + 48hatj)/(4-2)`
`" "= (8hati + 24hatj) ms^(-1)`
(ii) Differentianting Eqn. (i) w.r.t. time,
`" "vecv= (dvecr)/(dt)= 8hati + 8thatj`
At `t=2 2s, vecv = 8 hati + 16 hatj`