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: `pi/60sqrt2`
C: `pi/30`
D: `pisqrt2/30`
The answer is D for your reference

Answer 1

`Deltavecv=`Deltavecv= romega hat x + rwhaty# #|Deltav|=sqrt(2)romega#

Explanation:

The velocity of the tip is tangent to the circular arc traced out by the seconds hand. Therefore In 15 seconds, the tangential velocity of the tip has changed direction 90 degrees or `pi/2`, but the speed of the tip remains the same.

`r = 1 cm`
`omega = (2pi)/60s = pi/(30s) larr` angular speed of the seconds hand
`v=romega`

At t=0, `vecv_0 = romega hatx`
i.e.., @ 12 o'clock, tip velocity is in horizontal direction facing East

15 seconds later, `vec v = romega haty`
i.e., @ 3 o'clock, tip velocty is in vertical direction facing East

`rArr Deltavecv= vec v-vec v_0= romegahaty - rwhatx`

`|Deltav|=sqrt(2) romega = sqrt(2) (1cm)((pi)/(30s))`

`|Deltav| = (sqrt(2)pi)/(30) (cm)/s`