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9.4: Uniform motion problems

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Oct 3, 2021
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- 9.3: Work-rate problems
- 9.5: Revenue problems

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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

Example 9.4.1

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.

Table 9.4.1
	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.

Note

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.

Example 9.4.2

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.

Table 9.4.2
	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

Exercise 9.4.1

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?

Exercise 9.4.2

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?

Exercise 9.4.3

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?

Exercise 9.4.4

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?

Exercise 9.4.5

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?

Exercise 9.4.6

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.

Exercise 9.4.7

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.

Exercise 9.4.8

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?

Exercise 9.4.9

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?

Exercise 9.4.10

By going $15$ miles per hour faster, a train would have required $1$ hour less to travel $180$ miles. How fast did it travel?

Exercise 9.4.11

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?

Exercise 9.4.12

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.

Exercise 9.4.13

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?

Uniform Motion Problems

Uniform Motion Problems with Streams and Winds

Uniform Motion Problems Homework

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