In this lesson, we will apply principles of vector addition to determine relative velocity.
<h2>Relative Velocity</h2>
If a person rows a boat across a rapidly flowing river and tries to head directly for the other shore, the boat instead moves diagonally relative to the shore, as in the figure below. The boat does not move in the direction in which it is pointed. The reason, of course, is that the river carries the boat downstream.
<div class="wp-caption alignnone" style="width: 671px"><img alt="A boat is trying to cross a river. Due to the velocity of river the path traveled by boat is diagonal. The velocity of boat v boat is in positive y direction. The velocity of river v river is in positive x direction. The resultant diagonal velocity v total which makes an angle of theta with the horizontal x axis is towards north east direction." height="484" src="https://data.ulearngo.com/assets/707dec26-40be-4af3-8434-c254783c76b0" width="671"/>A boat trying to head straight across a river will actually move diagonally relative to the shore as shown. Its total velocity (solid arrow) relative to the shore is the sum of its velocity relative to the river plus the velocity of the river relative to the shore.</div>
Similarly, if a small airplane flies overhead in a strong crosswind, you can sometimes see that the plane is not moving in the direction in which it is pointed, as illustrated in the figure below. The plane is moving straight ahead relative to the air, but the movement of the air mass relative to the ground carries it sideways.
<div class="wp-caption alignnone" style="width: 366px"><img alt="An airplane is trying to fly straight north with velocity v sub p. Due to wind velocity v sub w in south west direction making an angle theta with the horizontal axis, the plane’s total velocity is thirty eight point 0 meters per seconds oriented twenty degrees west of north." height="375" src="https://data.ulearngo.com/assets/801fe516-49df-4e78-a765-2c5b36b9002d" width="366"/>An airplane heading straight north is instead carried to the west and slowed down by wind. The plane does not move relative to the ground in the direction it points; rather, it moves in the direction of its total velocity (solid arrow).</div>
In each of these situations, an object has a velocity relative to a medium (such as a river) and that medium has a velocity relative to an observer on solid ground. The velocity of the object relative to the observer is the sum of these velocity vectors, as indicated in both figures above. These situations are only two of many in which it is useful to add velocities. In this module, we first re-examine how to add velocities and then consider certain aspects of what relative velocity means.
How do we add velocities? Velocity is a vector (it has both magnitude and direction); the rules of vector addition discussed in the <a href="https://app.ulearngo.com/lesson/vectors-in-two-dimensions/">graphical methods</a> and <a href="https://app.ulearngo.com/lesson/analytical-methods-of-vector-addition-and-subtraction/">analytical methods</a> apply to the addition of velocities, just as they do for any other vectors. In one-dimensional motion, the addition of velocities is simple—they add like ordinary numbers.
For example, if a field hockey player is moving at \(\text{5 m/s}\) straight toward the goal and drives the ball in the same direction with a velocity of \(\text{30 m/s}\) relative to her body, then the velocity of the ball is \(\text{35 m/s}\) relative to the stationary, profusely sweating goalkeeper standing in front of the goal.
In two-dimensional motion, either graphical or analytical techniques can be used to add velocities. We will concentrate on analytical techniques. The following equations give the relationships between the magnitude and direction of velocity (\(v\) and \(\theta \)) and its components (\({v}_{x}\) and \({v}_{y}\)) along the x- and y-axes of an appropriately chosen coordinate system:
\({v}_{x}=v\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \)
\({v}_{y}=v\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \)
\(v=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}}\)
\(\theta ={\text{tan}}^{-1}\left({v}_{y}/{v}_{x}\right).\)
<div class="wp-caption alignnone" style="width: 394px"><img alt="The figure shows components of velocity v in horizontal x axis v x and in vertical y axis v y. The angle between the velocity vector v and the horizontal axis is theta." height="275" src="https://data.ulearngo.com/assets/a7ca61c7-7bad-4776-b853-b0ddfb7080d3" width="394"/>The velocity, \(v\), of an object traveling at an angle \(\theta \) to the horizontal axis is the sum of component vectors \({\mathbf{\text{v}}}_{x}\) and \({\mathbf{\text{v}}}_{y}\).</div>
These equations are valid for any vectors and are adapted specifically for velocity. The first two equations are used to find the components of a velocity when its magnitude and direction are known. The last two are used to find the magnitude and direction of velocity when its components are known.
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<h3 data-type="title">Optional Take-Home Experiment: Relative Velocity of a Boat</h3>
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Fill a bathtub half-full of water. Take a toy boat or some other object that floats in water. Unplug the drain so water starts to drain. Try pushing the boat from one side of the tub to the other and perpendicular to the flow of water. Which way do you need to push the boat so that it ends up immediately opposite? Compare the directions of the flow of water, heading of the boat, and actual velocity of the boat.
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<h3>Example on Adding Velocities: A Boat on a River</h3>
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<div class="wp-caption alignnone" style="width: 400px"><img alt="A boat is trying to cross a river. Due to the velocity of the river the path traveled by the boat is diagonal. The velocity of the boat, v boat, is equal to zero point seven five meters per second and is in positive y direction. The velocity of the river, v-river, is equal to one point two meters per second and is in positive x direction. The resultant diagonal velocity v total, which makes an angle of theta with the horizontal x axis, is towards north east direction." height="287" src="https://data.ulearngo.com/assets/6bb7f20d-552a-4c92-ba1e-8049a2f440f7" width="400"/>A boat attempts to travel straight across a river at a speed 0.75 m/s. The current in the river, however, flows at a speed of 1.20 m/s to the right. What is the total displacement of the boat relative to the shore?</div>
Refer to the figure above, which shows a boat trying to go straight across the river. Let us calculate the magnitude and direction of the boat’s velocity relative to an observer on the shore, \({\mathbf{\text{v}}}_{\text{tot}}\). The velocity of the boat, \({\mathbf{\text{v}}}_{\text{boat}}\), is 0.75 m/s in the \(y\)-direction relative to the river and the velocity of the river, \({\mathbf{\text{v}}}_{\text{river}}\), is 1.20 m/s to the right.
Strategy
We start by choosing a coordinate system with its \(x\)-axis parallel to the velocity of the river, as shown in the figure above. Because the boat is directed straight toward the other shore, its velocity relative to the water is parallel to the \(y\)-axis and perpendicular to the velocity of the river. Thus, we can add the two velocities by using the equations \({v}_{\text{tot}}=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}}\) and \(\theta ={\text{tan}}^{-1}\left({v}_{y}/{v}_{x}\right)\) directly.
Solution
The magnitude of the total velocity is
\({v}_{\text{tot}}=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}}\text{,}\)
where
\({v}_{x}={v}_{\text{river}}=1\text{.}\text{20 m/s}\)
and
\({v}_{y}={v}_{\text{boat}}=0\text{.}\text{750 m/s.}\)
Thus,
\({v}_{\text{tot}}=\sqrt{\left(1.20\text{ m/s})\right)^{2}+\left(0.750\text{ m/s}\right)^{2}}\)
yielding
\({v}_{\text{tot}}=1\text{.}\text{42 m/s.}\)
The direction of the total velocity \(\theta \) is given by:
\(\theta ={\text{tan}}^{-1}\left({v}_{y}/{v}_{x}\right)={\text{tan}}^{-1}\left(0\text{.}\text{750}/1\text{.}\text{20}\right)\text{.}\)
This equation gives
\(\theta =\text{32}\text{.}0º\text{.}\)
Discussion
Both the magnitude \(v\) and the direction \(\theta \) of the total velocity are consistent with the figure above. Note that because the velocity of the river is large compared with the velocity of the boat, it is swept rapidly downstream. This result is evidenced by the small angle (only \(32.0º\)) the total velocity has relative to the riverbank.
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<h3>Example on Calculating Velocity: Wind Velocity Causes an Airplane to Drift</h3>
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Calculate the wind velocity for the situation shown in the figure below. The plane is known to be moving at 45.0 m/s due north relative to the air mass, while its velocity relative to the ground (its total velocity) is 38.0 m/s in a direction \(\text{20}\text{.0º}\) west of north.
<div class="wp-caption alignnone" style="width: 610px"><img alt="An airplane is trying to fly north with velocity v p equal to forty five meters per second at angle of one hundred and ten degrees but due to wind velocity v w in south west direction making an angle theta with the horizontal axis it reaches a position in north west direction with resultant velocity v total equal to thirty eight meters per second and the direction is twenty degrees west of north." height="615" src="https://data.ulearngo.com/assets/bb1720f5-4f1b-4354-9b7f-11b7254907d9" width="610"/>An airplane is known to be heading north at 45.0 m/s, though its velocity relative to the ground is 38.0 m/s at an angle west of north. What is the speed and direction of the wind?</div>
Strategy
In this problem, somewhat different from the previous example, we know the total velocity \({\mathbf{\text{v}}}_{\text{tot}}\) and that it is the sum of two other velocities, \({\mathbf{\text{v}}}_{\text{w}}\) (the wind) and \({\mathbf{\text{v}}}_{\text{p}}\) (the plane relative to the air mass). The quantity \({\mathbf{\text{v}}}_{\text{p}}\) is known, and we are asked to find \({\mathbf{\text{v}}}_{\text{w}}\).
None of the velocities are perpendicular, but it is possible to find their components along a common set of perpendicular axes. If we can find the components of \({\mathbf{\text{v}}}_{\text{w}}\), then we can combine them to solve for its magnitude and direction. As shown in the figure above, we choose a coordinate system with its x-axis due east and its y-axis due north (parallel to \({\mathbf{\text{v}}}_{\text{p}}\)). (You may wish to look back at the <a href="https://app.ulearngo.com/lesson/resolving-a-vector-into-perpendicular-components/">discussion of the addition of vectors using perpendicular components</a>.)
Solution
Because \({\mathbf{\text{v}}}_{\text{tot}}\) is the vector sum of the \({\mathbf{\text{v}}}_{\text{w}}\) and \({\mathbf{\text{v}}}_{\text{p}}\), its x- and y-components are the sums of the x- and y-components of the wind and plane velocities. Note that the plane only has vertical component of velocity so \({v}_{px}=0\) and \({v}_{py}={v}_{\text{p}}\). That is,
\({v}_{\text{tot}x}={v}_{\text{w}x}\)
and
\({v}_{\text{tot}y}={v}_{\text{w}y}+{v}_{\text{p}}\text{.}\)
We can use the first of these two equations to find \({v}_{\text{w}x}\):
\({v}_{\text{w}x}={v}_{\text{tot}x}={v}_{\text{tot}}\text{cos 110º}\text{.}\)
Because \({v}_{\text{tot}}=\text{38.0 m/s}\) and \(\text{cos 110º}=–0.342\) we have
\({v}_{\text{w}x}=\left(\text{38.0 m/s}\right)\left(\text{–0.342}\right)=\text{–13 m/s.}\)
The minus sign indicates motion west which is consistent with the diagram.
Now, to find \({v}_{\text{w}\text{y}}\) we note that
\({v}_{\text{tot}y}={v}_{\text{w}y}+{v}_{\text{p}}\)
Here \({v}_{\text{tot}y}={v}_{\text{tot}}\text{sin 110º}\); thus,
\(v_{wy}=\left(\text{38.0 m/s}\right)\left(0.940\right)-45.0\text{ m/s}=-9.29\text{ m/s.}\)
This minus sign indicates motion south which is consistent with the diagram.
Now that the perpendicular components of the wind velocity \({v}_{\text{w}x}\) and \({v}_{\text{w}y}\) are known, we can find the magnitude and direction of \({\mathbf{\text{v}}}_{\text{w}}\). First, the magnitude is
\(\begin{array}{lll}{v}_{\text{w}}&amp; =&amp; \sqrt{{v}_{\text{w}x}^{2}+{v}_{\text{w}y}^{2}}\\ &amp; =&amp; \sqrt{\left(-\text{13}\text{.0 m/s}\right)^{2}+\left(-9\text{.}\text{29 m/s}\right)^{2}}\end{array}\)
so that
\({v}_{\text{w}}=\text{16}\text{.}0\text{ m/s.}\)
The direction is:
\(\theta ={\text{tan}}^{-1}\left({v}_{\text{w}y}/{v}_{\text{w}x}\right)={\text{tan}}^{-1}\left(-9\text{.}\text{29}/-\text{13}\text{.}0\right)\)
giving
\(\theta =\text{35}\text{.}6º\text{.}\)
Discussion
The wind’s speed and direction are consistent with the significant effect the wind has on the total velocity of the plane, as seen in the figure above. Because the plane is fighting a strong combination of crosswind and head-wind, it ends up with a total velocity significantly less than its velocity relative to the air mass as well as heading in a different direction.
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Note that in both of the last two examples, we were able to make the mathematics easier by choosing a coordinate system with one axis parallel to one of the velocities. We will repeatedly find that choosing an appropriate coordinate system makes problem solving easier. For example, in projectile motion we always use a coordinate system with one axis parallel to gravity.

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Relative Velocity

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Explore the principles of two-dimensional kinematics, including mastering vector addition and subtraction, resolving a vector into components, and analyzing projectile motion and relative velocities.


Relative Velocity

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