The position of an object, moving in one dimension, is given by the formula `x (t) = (4.2 t ^(2) + 2.6)m.` Calculate its (i) average velocity in the time interval fromj t =0 to t = 3 s and (ii) Instantaneous velocity at ` t=3s.`
`[ (d (x ^(x)))/( dt ) = nx ^(n -1)]`

Explanation

The position of an object for `t =0,`
`x (0) = 4.2 (0) ^(2) + 2.6 = 2.6 m` (initial position)
`(because x (t) = 4.2 t ^(2) + 2.6)`
Similiarly, for `t = 3s, x (3) = 4.2(3) ^(2) +2.6 = 40.4m`
(final position)
Now, average velocity in the time interval from
`t =0` to `t =3s` is
`lt v gty = ("final position - initial position")/("time interval")`
`=(x (3) -x(0))/(3-0) = (40.4 -2.6)/(3)`
` therefore lt v gt = 12.6 ms ^(-1)`
To find the instantaneous velocity at 3s, first we have to differentiate the position equation,
`x (t) = (4.2 tg ^(2) + 2.6) m ` with respect to .t..
`therefore` Instantaneous velocity
`v = (dx)/(dt) = (d)/(dt) = (4.2 t ^(2) + 2.6)` `= 4.2 (d)/(dt) t ^(2) + (d)/(dt) (2.6)`
`=4.2 (2t) + 0 (because (d)/(dt) (2.6) =0)`
`therefore v = 8.4 ms ^(-1)`
Putting t = 3s in above equation, `v = 8.4(3)`
`therefore v = 25.2 ms ^(-1)`