Chapter 2, Problem 102P

(a)

To determine

To plot: The velocity-time graph and acceleration-time graph for given situation on a single graph.

Explanation of Solution

Given:

The free fall of the professor is for $8.0\text{s}$ .

Initial velocity of the professor is $0\text{m}/{\text{s}}^{\text{2}}$ .

The acceleration opposite to gravity applied after free fall is $15\text{m}/{\text{s}}^{\text{2}}$ .

The speed which is maintained until she reaches the ground is $5.0\text{m}/\text{s}$ .

Formula used:

Write expression for the final velocity of the professor during free fall.

  $v_f=v_i+g\cdot t_1$..........(1)

Here, $v_f$ is the final velocity of professor during free fall, $v_i$ is the initial velocity during free fall, $g$ is the acceleration due to gravity and $t_1$ is the duration of free fall.

The initial speed when she starts slowing her rate of descent will be same as the final velocity of the free fall.

Write expression for final speed of professor when she reaches $5\text{m}/\text{s}$ velocity.

  ${v^{\prime }}_f={v^{\prime }}_i+a\cdot t_2$

Rearrange above expression for $t_2$ .

  $t_2=\frac{{v^{\prime }}_f-{v^{\prime }}_i}{a}$..........(2)

Here $t_2$ is the time taken to reach the velocity of $5.0\text{m}/\text{s}$ , ${v^{\prime }}_f$ is the final velocity, ${v^{\prime }}_i$ is the initial velocity of when she starts slowing her rate of descent and $a$ is the rate of descent.

Calculation:

Substitute $0\text{m}/\text{s}$ for $v_i$ , $-9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $8.0\text{s}$ for $t_1$ in equation (1).

  $\begin{array}{c}{v}_f=0+\left(-9.81\right)\left(8.0\right)\\ =-78.5\text{m}/\text{s}\end{array}$

Substitute $-78.5\text{m}/\text{s}$ for ${v^{\prime }}_i$ , $-5\text{m}/\text{s}$ for ${v^{\prime }}_f$ and $15\text{m}/{\text{s}}^{\text{2}}$ for $a$ in equation (2).

  $\begin{array}{c}{t}_2=\frac{\left(-5\right)-\left(-78.5\right)}{15}\\ =\frac{73.5}{15}\\ =4.9\text{s}\end{array}$

The graph for the velocity and acceleration of the professor is shown in figure 1.

  

(b)

To determine

To find:The speed of professor at the end of first $8.0\text{s}$ .

Answer to Problem 102P

The speed of professor at the end of $8\text{s}$ is $78.5\text{m}/\text{s}$ .

Explanation of Solution

Given:

The initial speed of the professor is $0\text{m}/\text{s}$ .

The time of fall of the professor is $8\text{s}$ .

The acceleration due to gravity on the surface of earth is $9.81\text{m}/{\text{s}}^{\text{2}}$ .

Formula used:

The professor jumps out of the helicopter and falls freely under the action of acceleration due to gravity, the motion of the professor for these first $8\text{s}$ is accelerated motion where the speed continues to increase as she comes down.

The first equation of motion relates the initial and final velocity of a body falling freely under the gravity. Therefore, the velocity of the professor can be obtained from this expression at the instant of $8\text{s}$ .

Write expression for the final velocity of the professor during free fall.

  $v_f=v_i+g\cdot t_1$..........(1)

Calculation:

Substitute $0\text{m}/\text{s}$ for $v_i$ , $-9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $8.0\text{s}$ for $t_1$ in equation (1).

  $\begin{array}{c}{v}_f=0+\left(-9.81\right)\left(8.0\right)\\ =-78.5\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the speed of professor at the end of $8\text{s}$ is $78.5\text{m}/\text{s}$ .

(c)

To determine

To find: The duration for which the professor descends her speed.

Answer to Problem 102P

  $4.9\text{s}$

Explanation of Solution

Given:

The initial speed of the professor during descend is $78.5\text{m}/\text{s}$ .

The final speed of the professor at which time is measured is $5\text{m}/\text{s}$ .

The deceleration of the professor is $15\text{m}/{\text{s}}^{\text{2}}$ .

Formula used:

The time taken by the professor during her slow rate of descent can be obtained from Newton’s first equation of motion.

Write expression for final speed of professor.

  ${v^{\prime }}_f={v^{\prime }}_i+a\cdot t_2$

Rearrange above expression for $t_2$ .

  $t_2=\frac{{v^{\prime }}_f-{v^{\prime }}_i}{a}$..........(2)

Calculation:

Substitute $-78.5\text{m}/\text{s}$ for ${v^{\prime }}_i$ , $-5\text{m}/\text{s}$ for ${v^{\prime }}_f$ and $15\text{m}/{\text{s}}^{\text{2}}$ for $a$ in equation (2).

  $\begin{array}{c}{t}_2=\frac{\left(-5\right)-\left(-78.5\right)}{15}\\ =\frac{73.5}{15}\\ =4.9\text{s}\end{array}$

Conclusion:

Thus, the duration for which she descends her speed is $4.9\text{s}$ .

(d)

To determine

To find:Distance covered by the professor when she descends her speed.

Answer to Problem 102P

  $204.5\text{m}$

Explanation of Solution

Given:

The initial speed of the professor during descend is $78.5\text{m}/\text{s}$ .

The deceleration of the professor is $15\text{m}/{\text{s}}^{\text{2}}$ .

Formula used:

Write expression for final speed of the professor during the descent.

  ${v^{\prime }}_f^2={v^{\prime }}_i^2+2as$

Rearrange above expression for $\text{s}$ .

  $s=\frac{{v^{\prime }}_f^2-{v^{\prime }}_i^2}{2a}$ ........ (3)

Calculation:

Substitute $-78.5\text{m}/\text{s}$ for ${v^{\prime }}_i$ , $-5\text{m}/\text{s}$ for ${v^{\prime }}_f$ and $15\text{m}/{\text{s}}^{\text{2}}$ for $a$ in equation (3).

  $\begin{array}{c}s=\frac{{\left(-5\right)}^2-{\left(-78.5\right)}^2}{2\left(15\right)}\\ =-204.5\text{m}\end{array}$

The negative sign shows the direction of motion.

Conclusion:

Thus, the distance covered by the professor during the descent is $204.5\text{m}$ .

(e)

To determine

To find:Time required by the professor to reach the ground.

Answer to Problem 102P

  $24.2\text{s}$

Explanation of Solution

Given:

The altitude of helicopter is $575\text{m}$ from the ground.

Formula used:

Write expression for distance travelled during free fall of the professor.

  $s^{\prime }=v_i\cdot t_1^2-\frac{1}{2}g\cdot t_1^2$......... (4)

Write expression for the distance covered when she maintains constant speed of $5\text{m}/\text{s}$ .

  $S=575\text{m-}\left(s+s^{\prime }\right)\text{m}$......... (5)

Here, S is the distance covered during constant speed motion.

Write expression for time taken to when professor maintained constant speed.

  $t_3=\frac{S}{{v^{\prime }}_f}$......... (6)

Write expression for total time of journey.

  $T=t_1+t_2+t_3$..........(7)

Calculation:

Substitute $0\text{m}/\text{s}$ for $v_i$ , $\text{8}\text{s}$ for $t_1$ and $9.81\text{m}/{\text{s}}^2$ for $g$ in equation (4).

  $\begin{array}{c}{s}^{\prime }=0+\frac{1}{2}\left(-9.81\right)\left(8^2\right)\\ =-314\text{m}\end{array}$

Here, negative sign shows the direction of motion.

Substitute $314\text{m}$ for $s^{\prime }$ and $204.5\text{m}$ for $s$ in equation (5).

  $\begin{array}{c}S=575-\left(314+204.5\right)\\ =56.5\text{m}\end{array}$

Substitute $56.5\text{m}$ for $S$ and $5\text{m}/\text{s}$ for ${v^{\prime }}_f$ in equation (6).

  $\begin{array}{c}{t}_3=\frac{56.5}{5}\text{s}\\ \text{=11}\text{.3}\text{s}\end{array}$

Substitute $8\text{s}$ for $t_1$ , $4.9\text{s}$ for $t_2$ and $11.3\text{s}$ for $t_3$ in equation (7).

  $\begin{array}{c}T=8+4.9+11.3\\ =24.2\text{s}\end{array}$

Conclusion:

Thus, the total time taken by the professor for complete journey is $24.2\text{s}$ .

(f)

To determine

To find:The average velocity of professor for entire trip.

Answer to Problem 102P

  $-24\text{m}/\text{s}$

Explanation of Solution

Given:

The distance through which the professor falls is $575\text{m}$ .

The time for which the professor falls is $24.2\text{s}$ .

Formula used:

Average velocity is defined as the ratio of total distance covered to the total time taken during the journey.

Write expression for average velocity of the professor.

  $v_{av}=\frac{D}{T}$......... (8)

Here $v_{av}$ is the average velocity and $D$ is the total distance covered by the professor.

Calculation:

Substitute $-575\text{m}$ for $D$ and $24.2\text{s}$ in equation (8).

  $\begin{array}{c}{v}_{av}=\frac{\left(-575\right)}{24.2}\\ =-23.7\text{m}/\text{s}\\ \approx -24\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the average velocity of the professor is $-24\text{m}/\text{s}$ .