Chapter 5, Problem 91GP

To determine

To find : The acceleration of the tip of second hand.

Answer to Problem 91GP

  $1.64\times {10}^{-4}{\text{ m/s}}^2$

Explanation of Solution

Given:

  $T=60$ sec

  $l=1.5$ cm

Formula Used :

The acceleration and time are related as,

  $a=\frac{4{\pi }^2}{T^2}l$

Calculation :

Substituting the value

  $a=\frac{4{\pi }^2}{{\left(60\right)}^2}\times \frac{3}{2}\times {10}^{-2}$

That is $a=1.64\times {10}^{-4}{\text{ m/s}}^2$

Conclusion :

Therefore the acceleration is $a=1.64\times {10}^{-4}{\text{ m/s}}^2$