Question

A table clock has a minute hand 5.0 cm long. find the average Velocity of the tip of the minute hand between 6:00 am and 6:15 am?

Answer 1

Dear student,
(a) When the needle is at 6.00 A.M., then the tip of the minute hand will be at mark 12. Similarly at 6.15 AM. then it is at 90^o away from the before position. Therefore, the distance in between the initial and final positions of a tip of the needle will be equal to the hypotenuses of the right angle triangle. Thus, the displacement will be,
$disp=\sqrt{5^2+5^2}=7.07cm=0.071m$

The total time taken from 6.00 A.M. to 6.15 A.M. will be 15 minutes that is 900 s. Therefore, the average velocity will become,
$av.vel.=\frac{disp}{time}=\frac{0.071m}{900s}=7.85\times {10}^{-5}m/s$
Regards