Question

The length of a seconds hand in a watch is 1 cm. Magnitude of change in velocity of its tip in 15 sec is?

A: 0 cm/sec
B: `(pisqrt2)/60`

C: `pi/30`

D: `(pisqrt2)/30`

The answer is D for your reference

Answer 1

The seconds hand of the watch rotates `2pi` rad in `60` s and its length `r=1cm`

The angular velocity of its tip will be

`omega=(2pi)/60` rad/s

Let the initial position of the tip of the seconds hand be at 12 in the watch. At this position the direction of the velocity of its tip will be towards positive direction of X-axis (considering center of rotation as origin).After 15s the tip rotates `(2pi)/60xx15=pi/2` rad and the direction of velocity becomes towards negative direction of Y-axis.In both the positions the magnitude of the velocity will be `v=omegaxxr=(2pi)/60xx1=pi/30` cm/s

Initial velocity vector `vecv_i=vhati`

Final velocity vector `vecv_f=-vhatj`

So the change in velocity during 15s will be

`vec(Deltav)=vecv_f-vecv_i=-vhatj-vhati`

So magnitude of change in velocity will be

`abs(vec(Deltav))= abs(-vhatj-vhati)`

`=sqrt(v^2+v^2)`

`=sqrt2 v`

`=(sqrt2 pi)/30` cm/s