Chapter 3.4, Problem 38E

a.

To determine

To calculate:The time take by the falling ball and its average velocity for specific distance.

Answer to Problem 38E

First 160 cm is covered by falling balls in $\underset{\_}{0.571\sec }$ at an average velocity of $\underset{\_}{280\text{cm/sec}}$ .

Explanation of Solution

Given information:

1. Free fall equation for distance $s$ in cm and time $t$ in seconds is as follows.

  $s=490t^2$

2. Need to determine time take and average velocity for falling first $160\text{cm}$ .

Calculation:

Free fall equation governing motion of the falling body is as follows

  $s=490t^2$

Calculate time for falling first 160 cm by substituting $s$ as 160 in free fall equation

  $160=490t^2$

  $\frac{160}{490}=t^2$

  $t=\sqrt{\frac{16}{49}}$

  $=\frac{4}{7}=0.571\sec$

Calculate average velocity for first 160 cm.

  $Averagevelocity=\frac{Distancecovered}{timetaken}$

  $=\frac{160}{\left(\frac{4}{7}\right)}$

  $=\frac{160\times 7}{4}$

  $=280\text{cm/sec}$

Hence first 160 cm is covered by falling balls in $\underset{\_}{0.571\sec }$ at an average velocity of $\underset{\_}{280\text{cm/sec}}$ .

b.

To determine

To calculate:The instantaneous velocity and acceleration of the falling balls at specific distance.

Answer to Problem 38E

Velocity of falling balls at 160 cm mark is $560\text{cm/sec}$ and its acceleration is $980\text{cm}/{\sec }^2$ .

Explanation of Solution

Given information:

1. Free fall equation for distance $s$ in cm and time $t$ in seconds is as follows.

  $s=490t^2$

2. Need to determine the instantaneous velocity and acceleration for falling ball at $160\text{cm}$ .
3. Evaluated time at $160\text{cm}$ is $\frac{4}{7}\sec$

Calculation:

Free fall equation governing motion of the falling body is as follows

  $s=490t^2$

Instantaneous velocity at any time $t$ is given by first derivative of position function $s=490t^2$ .

  $v=\frac{ds}{dt}=490\frac{d}{dt}{t}^2$

  $=490\times 2t$

  $v=980t$

At 160 cm , calculated time is $\frac{4}{7}\sec$

So instantaneous velocity at 160 cm can by obtained by substituting $t=\frac{4}{7}\sec$ in expression of $v$ .

  $v=980t$

  $=980\times \frac{4}{7}$

  $=560\text{cm}/\sec$

As acceleration is change in velocity with respect to time that is $\frac{d}{dt}v$

  $\frac{d}{dt}v=980\frac{d}{dt}t$

  $=980\text{cm}/{\sec }^2$

Hence velocity of falling balls at 160 cm mark is $\underset{\_}{560\text{cm/sec}}$ and its acceleration is $\underset{\_}{980\text{cm}/{\sec }^2}$ .

c.

To determine

To calculate:The flashing speed of light .

Answer to Problem 38E

Required flashing speed is $29.75\text{flashes/sec}$ .

Explanation of Solution

Given information:

1. From given figure it can be observe that 17 pictures of the falling balls are capture till the 160 cm mark.
2. Calculated time to reach $160\text{cm}$ mark is $\frac{4}{7}\sec$ .

Calculation:

17 picture are captured till 160 cm mark means 17 flashes of light till 160 cm mark.

Time to reach $160\text{cm}$ mark is $\frac{4}{7}\sec$ .

Means in $\frac{4}{7}\sec$ , 17 flashes of light occurred

Speed of light flashing = light flashing in 1 sec = $\frac{17}{\left(\frac{4}{7}\right)}$

  = $\frac{119}{4}$

  = $29.75\text{flashes/sec}$

Hence required flashing speed is $29.75\text{flashes/sec}$ .