Chapter 5, Problem 15A

A toy car rolls off tables of various heights at different speeds as shown.

a. Rank them for the time in the air, from greatest to least.

b. Rank them for horizontal range, from greatest to least.

To determine

To Rank: The cars for the time in air.

Answer to Problem 15A

  $t_C>t_B>t_A$

Explanation of Solution

Given:

For car A:

Acceleration in vertical direction $=a_{Ay}=9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Time interval $=t_A=?$

Initial velocity along the vertical direction $=v_{Aoy}=0\frac{\text{m}}{\text{s}}$

Vertical displacement $=\Delta y_A=0.6\text{ m}$

For car B:

Acceleration in vertical direction $=a_{By}=9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Time interval $=t_B=?$

Initial velocity along the vertical direction $=v_{Boy}=0\frac{\text{m}}{\text{s}}$

Vertical displacement $=\Delta y_B=0.8\text{ m}$

For car C:

Acceleration in vertical direction $=a_{Cy}=9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Time interval $=t_C=?$

Initial velocity along the vertical direction $=v_{Coy}=0\frac{\text{m}}{\text{s}}$

Vertical displacement $=\Delta y_C=1\text{ m}$

Formula Used:

According to kinematics equation, displacement is given as

  $\Delta x=v_{o}t+(0.5)a{t}^2$

Where,

  $\Delta x$ is the displacement

  $v_o$ is the initial velocity

a is he acceleration

t is the time

Calculation:

For car A:

Using the kinematics equation for the vertical motion of car

  $\begin{array}{l}\Delta y_A=v_{Aoy}{t}_A+(0.5)a_{Ay}{t}_A{}^2\\ 0.6=(0)t_A+(0.5)(9.8)t_A{}^2\\ t_A=0.35\text{ s}\end{array}$

For car B:

Using the kinematics equation for the vertical motion of car

  $\begin{array}{l}\Delta y_B=v_{Boy}{t}_B+(0.5)a_{By}{t}_B{}^2\\ 0.8=(0)t_B+(0.5)(9.8)t_B{}^2\\ t_A=0.40\text{ s}\end{array}$

For car C:

Using the kinematics equation for the vertical motion of car

  $\begin{array}{l}\Delta y_C=v_{Coy}{t}_C+(0.5)a_{Cy}{t}_C{}^2\\ 1=(0)t_C+(0.5)(9.8)t_C{}^2\\ t_C=0.45\text{ s}\end{array}$

Conclusion:

Hence, the ranking comes out to be $t_C>t_B>t_A$

To determine

To Rank:The cars as per their horizontal range.

Answer to Problem 15A

  $\Delta x_A>\Delta x_C>\Delta x_B$

Explanation of Solution

Given:

For car A:

Acceleration in horizontal direction $=a_{Ax}=0\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Time interval $=t_A=0.35\text{ s}$

Initial velocity along the horizontal direction $=v_{Aox}=4\frac{\text{m}}{\text{s}}$

Horizontal displacement $=\Delta x_A=?$

For car B:

Acceleration in horizontal direction $=a_{Bx}=0\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Time interval $=t_B=0.40\text{ s}$

Initial velocity along the horizontal direction $=v_{Box}=3.2\frac{\text{m}}{\text{s}}$

Horizontal displacement $=\Delta x_B=?$

For car C:

Acceleration in horizontal direction $=a_{Cx}=0\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Time interval $=t_C=0.45\text{ s}$

Initial velocity along the horizontal direction $=v_{Cox}=3\frac{\text{m}}{\text{s}}$

Horizontal displacement $=\Delta x_C=?$

Formula Used:

According to kinematics equation, displacement is given as

  $\Delta x=v_{o}t+(0.5)a{t}^2$

Where

  $\begin{array}{l}\Delta x=\text{ displacement}\\ v_o=\text{ initial velocity}\\ a\text{ = acceleration}\\ t=\text{ time}\end{array}$

Calculation:

For car A:

Using the kinematics equation for the horizontal motion of car

  $\begin{array}{l}\Delta x_A=v_{Aox}{t}_A+(0.5)a_{Ax}{t}_A{}^2\\ \Delta x_A=(4)(0.35)+(0.5)(0){(}^{0.35}\\ \Delta x_A=1.4\text{ m}\end{array}$

For car B:

Using the kinematics equation for the horizontal motion of car,

  $\begin{array}{l}\Delta x_B=v_{Box}{t}_B+(0.5)a_{Bx}{t}_B{}^2\\ \Delta x_B=(3.2)(0.40)+(0.5)(0){(}^{0.40}\\ \Delta x_B=1.28\text{ m}\end{array}$

For car C:

Using the kinematics equation for the horizontal motion of car

  $\begin{array}{l}\Delta x_C=v_{Cox}{t}_C+(0.5)a_{Cx}{t}_C{}^2\\ \Delta x_C=(3)(0.45)+(0.5)(0){(}^{0.45}\\ \Delta x_C=1.35\text{ m}\end{array}$

Conclusion:

Hence, the ranking comes out to be $\Delta x_A>\Delta x_C>\Delta x_B$ .