Solving Uniform Motion Applications

We used a table to organize the information in uniform motion problems when we introduced them earlier. We’ll continue using the table here. The basic equation was D = rt where D is the distance travelled, r is the rate, and t is the time.

Our first example of a uniform motion application will be for a situation similar to some we have already seen, but now we can use two variables and two equations.

Example

Translate to a system of equations and then solve:

Joni left St. Louis on the interstate, driving west towards Denver at a speed of 65 miles per hour. Half an hour later, Kelly left St. Louis on the same route as Joni, driving 78 miles per hour. How long will it take Kelly to catch up to Joni?

Solution

A diagram is useful in helping us visualize the situation.

This figure shows a diagram. Denver is on the left and St. Louis is on the right. There is a ray stretching from St. Louis to Denver. It is labeled “Joni” and “65 m p h.” There is another ray stretching from St. Louis to Denver. It is labeled “Kelly (1/2 hour later)” and “78 m p h.”

Identify and name what we are looking for. A chart will help us organize the data. We know the rates of both Joni and Kelly, and so we enter them in the chart.
We are looking for the length of time Kelly, k, and Joni, j, will each drive. Since \(D=r·t\) we can fill in the Distance column.
Translate into a system of equations. To make the system of equations, we must recognize that Kelly and Joni will drive the same distance. So, \(65j=78k.\) Also, since Kelly left later, her time will be \(\frac{1}{2}\) hour less than Joni’s time. So, \(k=j-\frac{1}{2}.\)
Now we have the system.
Solve the system of equations by substitution.
Substitute \(k=j-\frac{1}{2}\) into the second equation, then solve for j.



To find Kelly’s time, substitute j = 3 into the first equation, then solve for k.


Check the answer in the problem. Joni 3 hours (65 mph) = 195 miles. Kelly \(2\frac{1}{2}\) hours (78 mph) = 195 miles. Yes, they will have traveled the same distance when they meet.
Answer the question.	Kelly will catch up to Joni in \(2\frac{1}{2}\) hours. By then, Joni will have traveled 3 hours.

Many real-world applications of uniform motion arise because of the effects of currents—of water or air—on the actual speed of a vehicle. Cross-country airplane flights in the United States generally take longer going west than going east because of the prevailing wind currents.

Let’s take a look at a boat travelling on a river. Depending on which way the boat is going, the current of the water is either slowing it down or speeding it up.

The two figures below show how a river current affects the speed at which a boat is actually travelling. We’ll call the speed of the boat in still water b and the speed of the river current c.

In the figure below the boat is going downstream, in the same direction as the river current. The current helps push the boat, so the boat’s actual speed is faster than its speed in still water. The actual speed at which the boat is moving is b + c.

This figure shows a boat floating in water. On the right, there is an arrow pointing towards the boat. It is labeled “c.” On the left, there is an arrow pointing away from the boat. It is labeled “b.”

In the figure below the boat is going upstream, opposite to the river current. The current is going against the boat, so the boat’s actual speed is slower than its speed in still water. The actual speed of the boat is \(b-c\).

This figure shows a boat floating in water. To the left is an arrow pointing away from the boat labeled “b,” and an arrow pointing towards the boat labeled “c.”

We’ll put some numbers to this situation in the example below.

Example

Translate to a system of equations and then solve:

A river cruise ship sailed 60 miles downstream for 4 hours and then took 5 hours sailing upstream to return to the dock. Find the speed of the ship in still water and the speed of the river current.

Solution

Read the problem.

This is a uniform motion problem and a picture will help us visualize the situation.

This figure shows an arrow labeled “c” which continues to the right, representing the wave. Under the wave is a ray that points to the right and is labeled “four hours.” Under this ray is another ray pointing to the left labeled “five hours.” It is the same length as the ray labeled “four hours.” There is a bracket under the ray labeled “five hours.” The bracket is labeled “60 miles.”

Identify what we are looking for.	We are looking for the speed of the ship in still water and the speed of the current.
Name what we are looking for.	Let \(s=\) the rate of the ship in still water. \(\phantom{\rule{1.5em}{0ex}}c=\) the rate of the current
A chart will help us organize the information. The ship goes downstream and then upstream. Going downstream, the current helps the ship; therefore, the ship’s actual rate is s + c. Going upstream, the current slows the ship; therefore, the actual rate is s − c.
Downstream it takes 4 hours. Upstream it takes 5 hours. Each way the distance is 60 miles.
Translate into a system of equations. Since rate times time is distance, we can write the system of equations.
Solve the system of equations. Distribute to put both equations in standard form, then solve by elimination.
Multiply the top equation by 5 and the bottom equation by 4. Add the equations, then solve for s.
Substitute s = 13.5 into one of the original equations.




Check the answer in the problem. The downstream rate would be 13.5 + 1.5 = 15 mph. In 4 hours the ship would travel 15 · 4 = 60 miles. The upstream rate would be 13.5 − 1.5 = 12 mph. In 5 hours the ship would travel 12 · 5 = 60 miles.
Answer the question.	The rate of the ship is 13.5 mph and the rate of the current is 1.5 mph.

Wind currents affect airplane speeds in the same way as water currents affect boat speeds. We’ll see this in the example below. A wind current in the same direction as the plane is flying is called a tailwind. A wind current blowing against the direction of the plane is called a headwind.

Example

Translate to a system of equations and then solve:

A private jet can fly 1095 miles in three hours with a tailwind but only 987 miles in three hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Solution

Read the problem.

This is a uniform motion problem and a picture will help us visualize.

This figure shows an arrow labeled “3 hours” which continues to the right, representing the wind. Under the wave is a ray that points to the right and is labeled “j plus w equals 365” and “1,095 miles”. Under this ray is another ray pointing to the left labeled “j minus w equals 329” and “987 miles.”

Identify what we are looking for.	We are looking for the speed of the jet in still air and the speed of the wind.
Name what we are looking for.	Let \(j=\) the speed of the jet in still air. \(\phantom{\rule{1.4em}{0ex}}w=\) the speed of the wind
A chart will help us organize the information. The jet makes two trips-one in a tailwind and one in a headwind. In a tailwind, the wind helps the jet and so the rate is j + w. In a headwind, the wind slows the jet and so the rate is j − w.
Each trip takes 3 hours. In a tailwind the jet flies 1095 miles. In a headwind the jet flies 987 miles.
Translate into a system of equations. Since rate times time is distance, we get the system of equations.
Solve the system of equations. Distribute, then solve by elimination.
Add, and solve for j. Substitute j = 347 into one of the original equations, then solve for w.




Check the answer in the problem. With the tailwind, the actual rate of the jet would be 347 + 18 = 365 mph. In 3 hours the jet would travel 365 · 3 = 1095 miles. Going into the headwind, the jet’s actual rate would be 347 − 18 = 329 mph. In 3 hours the jet would travel 329 · 3 = 987 miles.
Answer the question.	The rate of the jet is 347 mph and the rate of the wind is 18 mph.