9.4: Uniform motion problems
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)We can recall uniform motion problems in the word problems chapter. We used the formula \(r\cdot t = d\) and organized the given information in a table. Now, we use the equation as
\[t=\dfrac{d}{r}\nonumber\]
We apply the same method in this section only the equations will be rational equations.
Uniform Motion Problems
Greg went to a conference in a city \(120\) miles away. On the way back, due to road construction he had to drive \(10\) mph slower which resulted in the return trip taking \(2\) hours longer. How fast did he drive on the way to the conference?
Solution
First, we can make a table to organize the given information and then create an equation. Let \(r\) represent the rate in which he drove to the conference.
| rate | time | distance | |
|---|---|---|---|
| To the conference | \(r\) | \(t\) | \(120\) |
| From the conference | \(r-10\) | \(t+2\) | \(120\) |
Now we can set up each equation.
\[\begin{array}{ll} t_{to}=\dfrac{120}{r}& t_{from}+2=\dfrac{120}{r-10} \\ t_{to}=\dfrac{120}{r}& t_{from}=\dfrac{120}{r-10}-2\end{array}\nonumber\]
Since we solved for \(t\) in each equation, we can set the \(t\)’s equal to each other and solve for \(r\):
\[\begin{array}{rl} t_{to}=t_{from}&\text{Set }t's\text{ equal to each other} \\ \dfrac{120}{r}=\dfrac{120}{r-10}-2&\text{Multiply by the LCD} \\ \color{blue}{r(r-10)}\color{black}{}\cdot\dfrac{120}{r}=\color{blue}{r(r-10)}\color{black}{}\cdot\dfrac{120}{r-10}-\color{blue}{r(r-10)}\color{black}{}\cdot 2&\text{Clear denominators} \\ 120(r-10)=120r-2r(r-10)&\text{Distribute} \\ 120r-1200=120r-2r^2+20r&\text{Combine like terms} \\ 120r-1200=140r-2r^2 &\text{Notice the }2r^2 \text{ term; solve by factoring} \\ 2r^2-20r-1200=0&\text{Reduce all terms by a factor of }2 \\ r^2-10r-600=0&\text{Factor} \\ (r+20)(r-30)=0&\text{Apply zero product rule} \\ r+20=0\text{ or }r-30=0&\text{Isolate variable terms} \\ r=-20\text{ or }r=30& \text{Solutions}\end{array}\nonumber\]
Since the rate of the car is always positive, we omit the solution \(r = −20\). Thus, Greg drove at a rate of \(30\) miles per hour to the conference.
The world’s fastest man (at the time of printing) is Jamaican Usain Bolt who set the record of running \(100\) meters in \(9.58\) seconds on August 16, 2009 in Berlin. That is a speed of over \(23\) miles per hour.
Uniform Motion Problems with Streams and Winds
Another type of uniform motion problem is where a boat is traveling in a river with the current or against the current (or an airplane flying with the wind or against the wind). If a boat is traveling downstream, the current will push it or increase the rate by the speed of the current. If a boat is traveling upstream, the current will pull against it or decrease the rate by the speed of the current.
A man rows down stream for \(30\) miles then turns around and returns to his original location, the total trip took \(8\) hours. If the current flows at \(2\) miles per hour, how fast would the man row in still water?
Solution
First, we can make a table to organize the given information and then create an equation. Let \(r\) represent the rate in which the man would row in still water.
| rate | time | distance | |
|---|---|---|---|
| Downstream | \(r+2\) | \(t\) | \(30\) |
| Upstream | \(r-2\) | \(8-t\) | \(30\) |
Now we can set up each equation.
\[\begin{array}{ll} t_{ds}=\dfrac{30}{r+2}& 8-t_{us}=\dfrac{30}{r-2} \\ t_{ds}=\dfrac{30}{r+2}& t_{us}=8-\dfrac{30}{r-2}\end{array}\nonumber\]
Since we solved for \(t\) in each equation, we can set the \(t\)’s equal to each other and solve for \(r\):
\[\begin{array}{rl} t_{ds}=t_{us}&\text{Set }t\text{'s equal to each other} \\ \dfrac{30}{r+2}=8-\dfrac{30}{r-2}&\text{Multiply by the LCD} \\ \color{blue}{(r+2)(r-2)}\color{black}{}\cdot\dfrac{30}{r+2}=\color{blue}{(r+2)(r-2)}\color{black}{}\cdot 8-\color{blue}{(r+2)(r-2)}\color{black}{}\cdot\dfrac{30}{r-2}&\text{Clear denominators} \\ 30(r-2)=8(r+2)(r-2)-30(r+2)&\text{Distribute} \\ 30r-60=8r^2-32-30r-60&\text{Combine like terms} \\ 30r-60=8r^2-30r-92&\text{Notice the }8r^2\text{ term; solve by factoring} \\ 8r^2-60r-32=0&\text{Reduce all terms by a factor of }4 \\ 2r^2-15r-8=0&\text{Factor} \\ (2r+1)(r-8)=0&\text{Apply zero product rule} \\ 2r+1=0\text{ or }r-8=0&\text{Isolate the variable terms} \\ r=-\dfrac{1}{2}\text{ or }r=8&\text{Solutions}\end{array}\nonumber\] Since the rate of the boat is always positive, we omit the solution \(r = −\dfrac{1}{2}\). Thus, the man rowed at a rate of \(8\) miles per hour in still water.
Uniform Motion Problems Homework
A train traveled \(240\) kilometers at a certain speed. When the engine was replaced by an improved model, the speed was increased by \(20\) km/hr and the travel time for the trip was decreased by \(1\) hour. What was the rate of each engine?
The rate of the current in a stream is \(3\) km/hr. A man rowed upstream for \(3\) kilometers and then returned. The round trip required \(1\) hour and \(20\) minutes. How fast was he rowing?
A pilot flying at a constant rate against a headwind of \(50\) km/hr flew for \(750\) kilometers, then reversed direction and returned to his starting point. He completed the round trip in \(8\) hours. What was the speed of the plane?
Two drivers are testing the same model car at speeds that differ by \(20\) km/hr. The one driving at the slower rate drives \(70\) kilometers down a speedway and returns by the same route. The one driving at the faster rate drives \(76\) kilometers down the speedway and returns by the same route. Both drivers leave at the same time, and the faster car returns \(\dfrac{1}{2}\) hour earlier than the slower car. At what rates were the cars driven?
An athlete plans to row upstream a distance of \(2\) kilometers and then return to his starting point in a total time of \(2\) hours and \(20\) minutes. If the rate of the current is \(2\) km/hr, how fast should he row?
An automobile goes to a place \(72\) miles away and then returns, the round trip occupying \(9\) hours. His speed in returning is \(12\) miles per hour faster than his speed in going. Find the rate of speed in both going and returning.
An automobile made a trip of \(120\) miles and then returned, the round trip occupying \(7\) hours. Returning, the rate was increased \(10\) miles an hour. Find the rate of each.
The rate of a stream is \(3\) miles an hour. If a crew rows downstream for a distance of \(8\) miles and then back again, the round trip occupying \(5\) hours, what is the rate of the crew in still water?
The railroad distance between two towns is \(240\) miles. If the speed of a train were increased \(4\) miles an hour, the trip would take \(40\) minutes less. What is the usual rate of the train?
By going \(15\) miles per hour faster, a train would have required \(1\) hour less to travel \(180\) miles. How fast did it travel?
Mr. Jones visits his grandmother who lives \(100\) miles away on a regular basis. Recently a new freeway has opend up and, although the freeway route is \(120\) miles, he can drive \(20\) mph faster on average and takes \(30\) minutes less time to make the trip. What is Mr. Jones’ rate on both the old route and on the freeway?
If a train had traveled \(5\) miles an hour faster, it would have needed \(1 \dfrac{1}{2}\) hours less time to travel \(150\) miles. Find the rate of the train.
A traveler having \(18\) miles to go, calculates that his usual rate would make him one-half hour late for an appointment; he finds that in order to arrive on time he must travel at a rate one-half mile an hour faster. What is his usual rate?