Question

the length of the second's hand in a watch is 1 cm. the magnitude of the channge in the velocity of its tip in 15 sec is

Answer 1

seconds hand of the watch rotates 2πrad in 60 s and its length r=1cm

The angular velocity of its tip will be

ω=2π60 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 2π60×15=π2 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=ω×r=2π60×1=π30 cm/s

Initial velocity vector →vi=vˆi

Final velocity vector →vf=−vˆj

So the change in velocity during 15s will be

−→Δv=→vf−→vi=−vˆj−vˆi

So magnitude of change in velocity will be

∣∣∣−→Δv∣∣∣=∣∣−vˆj−vˆi∣∣

=√v2+v2

=√2v

=√2π30 cm