Chapter 4, Problem 21A

Below we see before and after snapshots of a car’s velocity. The time interval between before and after for each is the same.

a. Rank the cars in terms of the change in velocity, from most positive to most negative. (Negative numbers rank lower than positive ones, and remember, tie scores can be part of your ranking.)

b. Rank them in terms of acceleration, from greatest to least.

To determine

The rank of the cars in terms of the change in velocity, from most positive to most negative.

Answer to Problem 21A

The rank of the cars in terms of the change in velocity, from most positive to most negative is $\Delta v_{\text{A}}=\Delta v_{\text{C}}>\Delta v_{\text{D}}>\Delta v_{\text{B}}$ .

Explanation of Solution

Formula used:

The expression for the change in velocity as follows:

  $\Delta v=v_{\text{f}}-v_{\text{i}}$

Here, $v_{\text{f}}$ is the final velocity and $v_{\text{i}}$ is the initial velocity.

Calculation:

The change in velocity for case A as follows:

  $\begin{array}{l}\Delta v=v_{\text{f}}-v_{\text{i}}\\ \Delta v_{\text{A}}=\left(20\text{m}/\text{s}\right)-\left(10\text{m}/\text{s}\right)\\ \Delta v_{\text{A}}=10\text{m}/\text{s}\end{array}$

The change in velocity for case B as follows:

  $\begin{array}{l}\Delta v=v_{\text{f}}-v_{\text{i}}\\ \Delta v_{\text{B}}=\left(15\text{m}/\text{s}\right)-\left(20\text{m}/\text{s}\right)\\ \Delta v_{\text{B}}=-5\text{m}/\text{s}\end{array}$

The change in velocity for case C as follows:

  $\begin{array}{l}\Delta v=v_{\text{f}}-v_{\text{i}}\\ \Delta v_{\text{C}}=\left(10\text{m}/\text{s}\right)-\left(0\text{m}/\text{s}\right)\\ \Delta v_{\text{C}}=10\text{m}/\text{s}\end{array}$

The change in velocity for case D as follows:

  $\begin{array}{l}\Delta v=v_{\text{f}}-v_{\text{i}}\\ \Delta v_{\text{D}}=\left(15\text{m}/\text{s}\right)-\left(15\text{m}/\text{s}\right)\\ \Delta v_{\text{D}}=0\text{m}/\text{s}\end{array}$

By comparing the values, the change in velocity is $\Delta v_{\text{A}}=\Delta v_{\text{C}}>\Delta v_{\text{D}}>\Delta v_{\text{B}}$ .

Conclusion:

Thus, the rank of the cars in terms of the change in velocity, from most positive to most negative is $\Delta v_{\text{A}}=\Delta v_{\text{C}}>\Delta v_{\text{D}}>\Delta v_{\text{B}}$ .

To determine

The rank of the cars in terms of acceleration from greatest to least.

Answer to Problem 21A

The rank of the cars in terms of acceleration from greatest to least is $a_{\text{A}}=a_{\text{C}}>a_{\text{D}}>a_{\text{B}}$ .

Explanation of Solution

Formula used:

The expression for the change in velocity as follows:

  $\Delta v=v_{\text{f}}-v_{\text{i}}$

Here, $v_{\text{f}}$ is the final velocity and $v_{\text{i}}$ is the initial velocity.

Calculation:

Acceleration is the variation in velocity during a given period of time. That is acceleration is directly proportional to the change in velocity.

Refer part (a).

The change in velocity for case A is

  $\Delta v_{\text{A}}=10\text{m}/\text{s}$ .

The change in velocity for case B is $\Delta v_{\text{B}}=-5\text{m}/\text{s}$ .

The change in velocity for case C is

  $\Delta v_{\text{C}}=10\text{m}/\text{s}$ .

The change in velocity for case D is

  $\Delta v_{\text{D}}=0\text{m}/\text{s}$ .

By comparing the values, the change in velocity is $\Delta v_{\text{A}}=\Delta v_{\text{C}}>\Delta v_{\text{D}}>\Delta v_{\text{B}}$ .

Hence, the rank of the cars in terms of acceleration from greatest to least is $a_{\text{A}}=a_{\text{C}}>a_{\text{D}}>a_{\text{B}}$ .

Conclusion:

Thus, the rank of the cars in terms of acceleration from greatest to least is $a_{\text{A}}=a_{\text{C}}>a_{\text{D}}>a_{\text{B}}$ .