Question

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

Answer 1

Answer:

sqrt2×(pi/30)

Explanation:

Since v(velocity) = r(radius)×w(angular veocity)

r= 1cm, w= (2pi/60) radian/sec, so v= (pi/30)cm/sec.

Now the sec hand has rotated by 90* and since the magnitude of the initial and final velocity of tip of sec hand is same. The change in magnitude can be calculated as V= sqrt(v sq +v sq)

So chnge in magnitude = sqrt2×(pi/30)