Chapter 4, Problem 43A

a. If a freely falling rock were equipped with a speedometer, by how much would its speed readings increase with each second of fall?

b. Suppose the freely falling rock were dropped near the surface of a planet where $g=20m/s^2$. By how much would its speed readings change each second?

To determine

Increase in speed reading each second.

Answer to Problem 43A

Increase in speed reading is, $9.8\text{ m/s}$

Explanation of Solution

Given :

A freely falling rock were equipped with a speedometer.

Time is given per second, t =1 s

Formula used :

According to first equation of motion.

  $v=u+at$

Where, v is final velocity, u is initial velocity, a is acceleration and t is time.

Calculation :

As there is free fall therefore initial velocity $u=0\text{ m/s}$

Acceleration due to gravity, $g=9.8{\text{ m/s}}^2$

Now, substituting $a=g$ , u and t and solve.

  $\begin{array}{l}v=0+9.8\times 1\\ =9.8\text{ m/s}\end{array}$

Conclusion :

Therefore, increase in speed reading per second is 9.8 m/s

To determine

Change in speed reading each second.

Answer to Problem 43A

Increase in speed reading is, $20\text{ m/s}$

Explanation of Solution

Given :

A freely falling rock were equipped with a speedometer.

Time is given per second, t =1 s

Acceleration due to gravity near the surface of planet is, $g=20{\text{ m/s}}^2$

Formula used :

According to first equation of motion.

  $v=u+at$

Where, v is final velocity, u is initial velocity, a is acceleration and t is time.

Calculation :

As there is free fall therefore initial velocity $u=0\text{ m/s}$

Now, substituting $a=g$ , u and t and solve.

  $\begin{array}{l}v=0+20\times 1\\ =20\text{ m/s}\end{array}$

Conclusion :

Therefore, increase in speed reading per second is 20 m/s