2.7: Determining how to solve a motion equation problem
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Calculate any of the following from a relevant word problem about the motion of an object: distance, displacement, average velocity, or average acceleration.
We have considered several terms to describe the motion of an object. In the next two sections we will look at relationships between three of those terms: displacement, velocity, and acceleration. Each of these terms is a vector quantity related to change in position. For displacement, this is a complete description of the term. For velocity and acceleration we are also considering time. Velocity is the change in displacement over a period of time. Acceleration is the change in velocity over a period of time. Additional relationships that involve multiple values can be derived. For example, there is a specific equation that relates acceleration and displacement. These equations are covered in more detail in a textbook for a full year course sequence about physics. For the purpose of this textbook, we will confine our interests to those equations already developed.
The rest of this section will consist of example problems pulled from content covered in the earlier sections of this chapter. As you consider these example problems, think about which of the movement concepts are most applicable to the problem before devising a solution method.
The next several examples consider the motion of the subway train shown in Figure \(\PageIndex{6}\). In (a) the shuttle moves to the right, and in (b) it moves to the left. The examples are designed to further illustrate aspects of motion and to illustrate some of the reasoning that goes into solving problems.
What are the magnitude and sign of displacements for the motions of the subway train shown in parts (a) and (b) of Figure \(\PageIndex{6}\)?
Strategy
A drawing with a coordinate system is already provided, so we do not need to make a sketch, but we should analyze it to make sure we understand what it is showing. Pay particular attention to the coordinate system. To find displacement, we use the equation \(\Delta x=x_{\mathrm{f}}-x_{0}\). This is straightforward since the initial and final positions are given.
Solution
1. Identify the knowns. In the figure we see that \(x_{\mathrm{f}}=6.70 \mathrm{~km}\) and \(x_{0}=4.70 \mathrm{~km}\) for part (a), and \(x_{\mathrm{f}}^{\prime}=3.75 \mathrm{~km}\) and \(x_{0}^{\prime}=5.25 \mathrm{~km}\) for part (b).
2. Solve for displacement in part (a).
\[\Delta x=x_{\mathrm{f}}-x_{0}=6.70 \mathrm{~km}-4.70 \mathrm{~km}=+2.00 \mathrm{~km} \nonumber\]
3. Solve for displacement in part (b).
\[\Delta x^{\prime}=x_{\mathrm{f}}^{\prime}-x_{0}^{\prime}=3.75 \mathrm{~km}-5.25 \mathrm{~km}=-1.50 \mathrm{~km} \nonumber\]
Discussion
The direction of the motion in (a) is to the right and therefore its displacement has a positive sign, whereas motion in (b) is to the left and thus has a negative sign.
What are the distances traveled for the motions shown in parts (a) and (b) of the subway train in Figure \(\PageIndex{6}\)?
Strategy
To answer this question, think about the definitions of distance and distance traveled, and how they are related to displacement. Distance between two positions is defined to be the magnitude of displacement, which was found in Example \(\PageIndex{2}\). Distance traveled is the total length of the path traveled between the two positions. In the case of the subway train shown in Figure \(\PageIndex{6}\), the distance traveled is the same as the distance between the initial and final positions of the train.
Solution
1. The displacement for part (a) was +2.00 km. Therefore, the distance between the initial and final positions was 2.00 km, and the distance traveled was 2.00 km.
2. The displacement for part (b) was −1.5 km. Therefore, the distance between the initial and final positions was 1.50 km, and the distance traveled was 1.50 km.
Discussion
Distance is a scalar. It has magnitude but no sign to indicate direction.
Suppose the train in Figure \(\PageIndex{6}\) (a) accelerates from rest to 30.0 km/h in the first 20.0 s of its motion. What is its average acceleration during that time interval?
Strategy
It is worth it at this point to make a simple sketch:
This problem involves three steps. First we must determine the change in velocity, then we must determine the change in time, and finally we use these values to calculate the acceleration.
Solution
1. Identify the knowns. \(v_{0}=0\) (the trains starts at rest), \(v_{\mathrm{f}}=30.0 \mathrm{~km} / \mathrm{h}\), and \(\Delta t=20.0 \mathrm{~s}\).
2. Calculate \(\Delta v\). Since the train starts from rest, its change in velocity is \(\Delta v=+30.0 \mathrm{~km} / \mathrm{h}\), where the plus sign means velocity to the right.
3. Plug in known values and solve for the unknown, \(\bar{a}\).
\[\bar{a}=\frac{\Delta v}{\Delta t}=\frac{+30.0 \mathrm{~km} / \mathrm{h}}{20.0 \mathrm{~s}} \nonumber\]
4. Since the units are mixed (we have both hours and seconds for time), we need to convert everything into SI units of meters and seconds.
\[\bar{a}=\left(\frac{+30 \mathrm{~km} / \mathrm{h}}{20.0 \mathrm{~s}}\right)\left(\frac{10^{3} \mathrm{~m}}{1 \mathrm{~km}}\right)\left(\frac{1 \mathrm{~h}}{3600 \mathrm{~s}}\right)=0.417 \mathrm{~m} / \mathrm{s}^{2} \nonumber\]
Discussion
The plus sign means that acceleration is to the right. This is reasonable because the train starts from rest and ends up with a velocity to the right (also positive). So acceleration is in the same direction as the change in velocity, as is always the case.
Now suppose that at the end of its trip, the train in Figure \(\PageIndex{6}\) (a) slows to a stop from a speed of 30.0 km/h in 8.00 s. What is its average acceleration while stopping?
Strategy
In this case, the train is slowing down and its acceleration is negative because it is toward the left. As in the previous example, we must find the change in velocity and the change in time and then solve for acceleration.
Solution
1. Identify the knowns. \(v_{0}=30.0 \mathrm{~km} / \mathrm{h}\), \(v_{\mathrm{f}}=0 \mathrm{~km} / \mathrm{h}\) (the train is stopped, so its velocity is 0), and \(\Delta t=8.00 \mathrm{~s}\).
2. Solve for the change in velocity, \(\Delta v\).
\[\Delta v=v_{\mathrm{f}}-v_{0}=0-30.0 \mathrm{~km} / \mathrm{h}=-30.0 \mathrm{~km} / \mathrm{h} \nonumber\]
3. Plug in the knowns, \(\Delta v\) and \(\Delta t\), and solve for \(\bar{a}\).
\[\bar{a}=\frac{\Delta v}{\Delta t}=\frac{-30.0 \mathrm{~km} / \mathrm{h}}{8.00 \mathrm{~s}}\nonumber\]
4. Convert the units to meters and seconds.
\[\bar{a}=\frac{\Delta v}{\Delta t}=\left(\frac{-30.0 \mathrm{~km} / \mathrm{h}}{8.00 \mathrm{~s}}\right)\left(\frac{10^{3} \mathrm{~m}}{1 \mathrm{~km}}\right)\left(\frac{1 \mathrm{~h}}{3600 \mathrm{~s}}\right)=-1.04 \mathrm{~m} / \mathrm{s}^{2} \nonumber.\]
Discussion
The minus sign indicates that acceleration is to the left. This sign is reasonable because the train initially has a positive velocity in this problem, and a negative acceleration would oppose the motion.
The graphs of position, velocity, and acceleration vs. time for the trains in Example \(\PageIndex{4}\) and Example \(\PageIndex{5}\) are displayed in Figure \(\PageIndex{9}\). (We have taken the velocity to remain constant from 20 to 40 s, after which the train slows down.)
What is the average velocity of the train in part b of Example \(\PageIndex{2}\), and shown again below, if it takes 5.00 min to make its trip?
Strategy
Average velocity is displacement divided by time. It will be negative here, since the train moves to the left and has a negative displacement.
Solution
1. Identify the knowns. \(x_{\mathrm{f}}^{\prime}=3.75 \mathrm{~km}, \ x_{0}^{\prime}=5.25 \mathrm{~km}, \ \Delta t=5.00 \mathrm{~min}\).
2. Determine displacement, \(\Delta x^{\prime}\). We found \(\Delta x^{\prime}\) to be −1.5 km in Example \(\PageIndex{2}\).
3. Solve for average velocity.
\[\bar{v}=\frac{\Delta x^{\prime}}{\Delta t}=\frac{-1.50 \mathrm{~km}}{5.00 \mathrm{~min}} \nonumber\]
4. Convert units.
\[\bar{v}=\frac{\Delta x^{\prime}}{\Delta t}=\left(\frac{-1.50 \mathrm{~km}}{5.00 \mathrm{~min}}\right)\left(\frac{60 \mathrm{~min}}{1 \mathrm{~h}}\right)=-18.0 \mathrm{~km} / \mathrm{h} \nonumber\]
Discussion
The negative velocity indicates motion to the left.
Finally, suppose the train in Figure \(\PageIndex{10}\) slows to a stop from a velocity of 20.0 km/h in 10.0 s. What is its average acceleration?
Strategy
Once again, let’s draw a sketch:
As before, we must find the change in velocity and the change in time to calculate average acceleration.
Solution
1. Identify the knowns. \(v_{0}=-20 \mathrm{~km} / \mathrm{h}, \ v_{\mathrm{f}}=0 \mathrm{~km} / \mathrm{h}, \ \Delta t=10.0 \mathrm{~s}\).
2. Calculate \(\Delta v\). The change in velocity here is actually positive, since
\[\Delta v=v_{\mathrm{f}}-v_{0}=0-(-20 \mathrm{~km} / \mathrm{h})=+20 \mathrm{~km} / \mathrm{h} \nonumber\]
3. Solve for \(\bar{a}\).
\[\bar{a}=\frac{\Delta v}{\Delta t}=\frac{+20.0 \mathrm{~km} / \mathrm{h}}{10.0 \mathrm{~s}} \nonumber\]
4. Convert units.
\[\bar{a}=\left(\frac{+20.0 \mathrm{~km} / \mathrm{h}}{10.0 \mathrm{~s}}\right)\left(\frac{10^{3} \mathrm{~m}}{1 \mathrm{~km}}\right)\left(\frac{1 \mathrm{~h}}{3600 \mathrm{~s}}\right)=+0.556 \mathrm{~m} / \mathrm{s}^{2} \nonumber\]
Discussion
The plus sign means that acceleration is to the right. This is reasonable because the train initially has a negative velocity (to the left) in this problem and a positive acceleration opposes the motion (and so it is to the right).
Sign and Direction
Perhaps the most important thing to note about these examples is the signs of the answers. In our chosen coordinate system, plus means the quantity is to the right and minus means it is to the left. This is easy to imagine for displacement and velocity, but it is a little less obvious for acceleration. Most people interpret negative acceleration as the slowing of an object. This was not the case in Example \(\PageIndex{7}\), where a positive acceleration slowed a negative velocity. The crucial distinction was that the acceleration was in the opposite direction from the velocity. In fact, a negative acceleration will increase a negative velocity. For example, the train moving to the left in Figure \(\PageIndex{10}\) is sped up by an acceleration to the left. In that case, both \(v\) and \(a\) are negative. The plus and minus signs give the directions of the accelerations. If acceleration has the same sign as the change in velocity, the object is speeding up. If acceleration has the opposite sign of the change in velocity, the object is slowing down.
Section Summary
- Consider which values are provided in a motion problem and how they are related to one another.
Contributors
Curated from resources found in Introduction to Physics published by OpenStax.