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 att=3s. [ (d (x ^(x)))/( dt ) = nx ^(n -1)]

The position of an object, moving in one dimension, is given by the fo

1.	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 att=3s. [ (d (x ^(x)))/( dt ) = nx ^(n -1)]
Answer» Solution :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)`