# Relative Velocity

by Jason "Mussolini" Crandall, Doug "Castro" Antoni, April 1998

Here are the quantities you will use:

• v = Velocity
• t = Time
• x = Distance

## New Quantities

### Defining Relative Velocity

I hope by now you have become very comfortable with vectors because solving relative velocity problems requires an extensive understanding of vectors. Relative velocity means the speed of a moving object with respect to another moving object. This is less complicated than it sounds. For example, consider two cars headed opposite directions on I-5, both traveling at 60 mph. As they approach each other, the speed of one car relative to the other is 120 mph (60 mph + 60 mph = 120 mph). Specifically, to an observer on one train, the other train seems to be approaching at 120 mph. Alternately, if one car traveling at 60 mph passes a second car traveling at 45 mph, the speed of the first car relative to the second would be 15 mph (60 mph -45 mph = 15 mph).
If both objects (or velocities) are in the same direction, relative velocity is nothing more than addition or subtraction. But if they are not along the same line, we must use vectors. It is also very important to define the frame of reference (that is, whether the specified velocity is with respect to another object, the Earth or something else entirely). Thus, each velocity will use not one, but two subscripts: the first will refer to the object and the second to the reference frame in which the object has this velocity. For example, suppose a Jet-ski (yes, it's a Jet-ski … trust me) were trying to cross the Tualatin River as in Fig. 1.1. For the purposes of this example, assume the Tualatin River actually moves. Vector V(ws) is the Velocity of the Water with respect to the Shore. V(js) is the Velocity of the Jet-ski with respect to the Shore, and V(jw) is the Velocity of the Jet-ski with respect to the Water. Here, the boat is pulled downstream by the current. Questions of this type will generally ask you to determine how far downstream the boat will drift before reaching the other side. A second type of question will ask at what angle into the stream must the boat be headed in order to travel directly across the stream. Examples of each of these types of the problems will follow.

## Formulas

Here is a list of the all of the formulas we will be using:

• v = x / t

## General Problem Solving Strategy:

2. Go through the problem and figure out what is given or implied
Make a list, and identify the quantities you know.
3. Draw a diagram. Label each vector correctly and make sure to add them tip to tail.
4. Use the velocity formula and trigonometric equations to solve for the unknown. Make sure your calculator is in degree mode, not radian.
5. Add what you just found in the last step to your list of knowns.
6. Check to see if you have found the answer. If not, repeat the
previous two steps until you are done.

## Example problem 1

Our Jet-ski can travel 2.6 m/s in still water. (a) If our Jet-ski heads (heads, not travels) directly across the Tualatin (whose current is .4 m/s), what is the velocity (magnitude and angle) of the Jet-ski relative to the shore? (b) If the Tualatin is 50 meters wide, what be the Jet-ski's horizontal position, relative to its starting point, when it reaches the other side?

(a) As always, draw a diagram. Because the velocity of the Jet-ski and the Tualatin River are perpendicular to each other, we can use the Pythagorean Theorem (a2 + b2 = c2) find the magnitude and direction of the boat's velocity relative to the shore. We label the two legs of our right triangle V(jw) and V(ws). Our hypotenuse is V(js). Thus:

V(js)2 = V(jw)2 + V(ws)2

So we square the velocity of the Jet-ski and the river, then add them together and get:

V(js)2 = 6.92 m/s

We take the square root and find that the magnitude of the Jet-ski's velocity with respect to the shore is 2.63 m/s. But what about its direction?

Because we know all three sides of the triangle now, we could use either cosine, sine or tangent to find our angle. However, let's assume that we haven't found the hypotenuse yet. In that case, the angle at which the boat is traveling would be the inverse tangent of V(ws) / V(jw). Numerically:

tan q = (.4 m/s) / (2.6 m/s)

Solving for q, we get 8.8 degrees downstream.

(b) Despite the current, the boat is approaching the opposite shore at 2.6 m/s. If the river is 50 m wide, it will take the boat 19.2 seconds to reach the other shore. But during this time, the boat is also drifting downstream. The current is still .4 m/s, so after 19.2 seconds the boat will have drifted 7.7 meters downstream (just use the velocity formula).

## Example problem 2

A plane traveling north at 150 m/s is flying toward a runway. A cross-wind of 30 m/s begins to blow from the east. The pilot realizes he is being blown off course. At what angle must he head in order to travel directly north? What will be the plane's velocity relative to the runway?

First, draw a diagram. It's easy to see that the plane must adjust its course to head into the wind in order to continue traveling north. But what angle should the pilot begin heading? The hypotenuse of our triangle should be 150 m/s and the opposite side is 50 m/s. So sinq = (50 m/s) / (150 m/s). Solving for q, we get 19.47 degrees east of north.

There are several ways to find the plane's velocity with respect to the runway (the second leg of our triangle). We could use the angle we've just calculated, but I prefer the Pythagorean Theorem. We have b and c but not a, so our equation will be:

c2 - b2 = a2

Solving for c2 - b2, we get:

20,000 = a2

So a is 141.4 m/s.

## Sample Problems

link to a solution to the problem. Try the problem, check your
answer, and go to the solution if you do not understand.

### 1.

A river has a current of 1.5 m/s, and a swimmer has a velocity of .89 m/s.
a. If they head straight across the river, what will be their speed (hypotenuse) with respect to the shore?
b. If the river is 45 m wide, how much time will it take them to cross?
c. How far downstream will they drift in that time?
(a. 1.74 m/s, b. 50.6 s, c. 75.8 m)

### 2.

A canoe heads straight across a river with a current of .80 m/s. It takes the canoe 35 seconds to cross the 75 m wide river.
a. What is the speed of the canoe with respect to the water?
b. How far does the canoe drift downstream?
c. What is the speed of the canoe with respect to the shore?
(a. 2.14 m/s, b. 28 m, c. 2.29 m/s)

### 3.

A motorboat can go 2.4 m/s on a river where the current is 1.8 m/s. The motorboat must go 240 m upstream, and then back.
a. What is the speed of the boat with respect to the shore as it travels upstream?
b. Downstream?
c. How much time does it take it to go upstream?
d. Downstream?
(a. .6 m/s, b. 4.2 m/s, c. 400 s, d. 57.14 s)

### 4.

The current in a river 117 m wide is 1.45 m/s, and your boat can go 3.67 m/s.
a. What time will it take you to cross the river if you point straight across?
b. What Speed would you go if you pointed straight in to the current?
c. What angle upstream of straight across must you point your boat to actually go straight across?
(a. 31.9 s, b. 2.22 m/s, c. 23.3 degrees)

### 5.

A ferry boat points upstream at some angle to go straight across a river. The river current is 1.8 m/s, and it takes the boat 30 seconds to cross the 60 m wide river.
a) What is the speed of the boat with respect to the shore?
b) What is the speed of the boat on still water?
c) What angle upstream of straight across does the boat point?
(a. 2.0 m/s, b. 2.7 m/s, c. 42 degrees)

## 1.

A river has a current of 1.5 m/s, and a swimmer has a velocity of .89 m/s.
a. If they head straight across the river, what will be their speed (hypotenuse) with respect to the shore?
b. If the river is 45 m wide, how much time will it take them to cross?
c. How far downstream will they drift in that time?
(a. 1.74 m/s, b. 50.6 s, c. 75.8 m)

• x = 45 m
• V of river = 1.5 m/s
• V of Swimmer = .89 m/s

First, we have to find their total speed in respect to the shore. As the problem hints, this is the hypotenuse of the triangle formed by the swimmer's velocity across the river, and the river's current downstream. This is found simply by:

a^2+b^2=c^2

We simply put the swimmer's velocity and the river's current in for "a" and "b." Therefore, the answer is 1.74 m/s. Now, for part "b." This is a simple problem that you have already learned how to do. You simply divide distance by velocity and you get 50.6 s. In part "c" you use the same equation, but you use the river's velocity and the time you just found. Multiply the speed of the river by the time, and you get 75.8 m.

## 2.

A canoe heads straight across a river with a current of .80 m/s. It takes the canoe 35 seconds to cross the 75 m wide river. a. What is the speed of the canoe with respect to the water? b. How far does the canoe drift downstream? c. What is the speed of the canoe with respect to the shore? (a. 2.14 m/s, b. 28 m, c. 2.29 m/s)

Here's what we know:

• V of river = .8 m/s
• t = 35
• x = 75 m

First, we find the speed of the canoe across the river. This is simply distance divided by time and you get 2.14 m/s. Part "b" is also more of the same, simply multiply the V of the river by the time the canoe was out there and you get 28 m. Part "c" is like the first part of problem 1. We take the two velocities we know and construct a triangle and find it's hypotenuse. Using the V of the river and the speed of the canoe across the river that we found in part "a," we plug into a^2+b^2=c^2 and get 2.29 m/s.

## 3.

A motorboat can go 2.4 m/s on a river where the current is 1.8 m/s. The motorboat must go 240 m upstream, and then back. a. What is the speed of the boat with respect to the shore as it travels upstream? b. Downstream? c. How much time does it take it to go upstream? d. Downstream? (a. .6 m/s, b. 4.2 m/s, c. 400 s, d. 57.14 s)

All right, here we go. This problem is actually very simple!! All you have to do is subtract the velocities for part "a" and get .6 m/s. In part "b" you add them and get 4.2 m/s. Then, use the same old formula and divide the distance by the velocities to get 400 s upstream and 57.14 s downstream.

## 4.

The current in a river 117 m wide is 1.45 m/s, and your boat can go 3.67 m/s. a. What time will it take you to cross the river if you point straight across? b. What Speed would you go if you pointed straight in to the current? c. What angle upstream of straight across must you point your boat to actually go straight across? (a. 31.9 s, b. 2.22 m/s, c. 23.3 degrees)

The first one is stuff we have done already. Divide the distance across the river by the speed the boat can go and you get 31.9 m/s. Then, we repeat stuff from the previous problem and subtract velocities to get 2.22 m/s.

Now for something new. First, we set up our famous triangle, only this time we use the V of the river as the short side and the speed of the boat as the hypotenuse. Then, we use SohCahToa to find the angle. Specifically:

Sin=1.45/3.67

and you get 23.3 degrees.

## 5.

A ferry boat points upstream at some angle to go straight across a river. The river current is 1.8 m/s, and it takes the boat 30 seconds to cross the 60 m wide river. a) What is the speed of the boat with respect to the shore? b) What is the speed of the boat on still water? c) What angle upstream of straight across does the boat point? (a. 2.0 m/s, b. 2.7 m/s, c. 42 degrees)