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Wize High School Grade 12 Physics Textbook > Kinematics

Relative Velocity

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Relative Velocity


Velocities are always measured relative to a reference frame. This reference frame could be a moving car or someone standing still in street. The concept of relative velocity lets us consider speeds from different frames of reference. We use vectors representing the velocity of the reference frame.


  • The relative velocity of an object A with respect to another object B is the velocity of object A from the point of view of someone sitting on object B. This velocity is usually shown by v⃗A/B\vec{v}_{A/B}.

  • Relative velocity is essentially the vector difference of the two velocities v⃗A\vec{v}_A and v⃗B\vec{v}_Bwhere these two velocities are measured from a same reference frame such as ground.

vA/B⃗=vA⃗−vB⃗\boxed{\vec{v_{A/B}}=\vec{v_A}-\vec{v_B}}










Wize Tip
Remember that velocities are vectors. So, we will have to use vector addition and subtraction when working with above equation.


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General strategy

  1. Write out your velocities with the subscripts in the order of object-respect to-frame of reference.
  2. Write out the relation between them using the above equation.
  3. Sometimes we can draw a diagram showing velocity vectors to guess what the relationship between the velocities described in the problem are
  4. Solve for what you want.
Exam Tip
Usually problems involving a boat, swimmer crossing a river, or air plane flying in a windy condition could be solved using this method!



















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Example: Relative Velocity of a Ship


A ship sets sail from Rotterdam, heading due north at 7.00 m/s relative to the water. The local ocean current is 1.50 m/s in a direction 40.0° north of east. What is the velocity of the ship relative to Earth?

Solution:

It is good to draw a diagram and show velocities given in this problem and label them properly:

V⃗SE=V⃗SW+V⃗WE\vec{V}_{SE} = \vec{V}_{SW}+\vec{V}_{WE}

We can write down above vector equation in terms of its components:

(VSE)x=(VSW)x+(VWE)x({{V}_{SE}})_x = ({{V}_{SW}})_x+({V}_{WE})_x
(VSE)x=0+(1.50)cos(40)=1.15m/s({{V}_{SE}})_x= 0+ (1.50)cos(40)= 1.15m/s

(VSE)y=(VSW)y+(VWE)y({{V}_{SE}})_y = ({{V}_{SW}})_y+({V}_{WE})_y
(VSE)y=7+(1.50)sin(40)=7.96m/s({{V}_{SE}})_y = 7+ (1.50) sin (40)=7.96m/s

To find the speed of ship respect to earth we have to find the magnitude of velocity vector

VSE=(VSE)x2+(VSE)y2{{V}_{SE}} = \sqrt{({{V}_{SE}})_x^2+({{V}_{SE}})_y^2}
VSE=(1.14)2+(7.96)2=8.04m/s{{V}_{SE}} = \sqrt{(1.14)^2+ (7.96)^2}= 8.04m/s

In addition, the direction of velocity vector could be found using below equation:

tan(θ)=(VSE)y(VSE)x=7.961.15tan(\theta) = \dfrac{(V_{SE})_y}{(V_{SE})_x}= \dfrac{7.96}{1.15}

θ=81.78°\theta= 81.78\degree

This angle is measured counter-clockwise respect to positive x-axis. So, this vector is 81.78 degrees north of east.

Example: The Plane in Spain

A plane is flying over Spain with the wind blowing to the East at 50.0 km/h.

The instruments in the plane say that it is flying at 880.0 km/h at 48o north of West. What is its speed relative to the ground?

Because the plane is flying in the air, its speed is relative to the air. We can break up it's vector into the components. First let's draw the vector, and the coordinate.



The components are
v⃗PA,x = −880.0 cos⁡48° km/hi^=−588.8 km/hi^ v⃗PA,y= 880.0 sin⁡48° km/hj^=654.0 km/hj^\begin{array}{c} \vec{v}_{PA,x}\ =\ -880.0\ \cos48\degree\ km/h \hat{i} =-588.8\ km/h \hat{i}\\ ~\\ \vec{v}_{PA,y} =\ 880.0\ \sin 48\degree\ km/h \hat{j} =654.0\ km/h \hat{j} \end{array}

We know that the air is moving (that's the wind speed). So the velocity of air relative to the ground is

v⃗AG,x=50.0km/h i^ v⃗AG,y=0km/h j^\begin{array} {c} \vec{v}_{AG,x} = 50.0 km/h \ \hat{i} \\~\\ \vec{v}_{AG,y} = 0 km/h \ \hat{j} \end{array}

To get the velocity of the plane relative to the ground, we have this equation

v⃗PG=v⃗PA+v⃗AG\vec{v}_{PG} = \vec{v}_{PA} + \vec{v}_{AG}

So we just have to add the two vectors together. Since the wind is not in the vertical y direction, this addition is only in the x direction.

v⃗PG,x=v⃗PA,x+v⃗AG,x=(−588.8+50.0)km/h i^=−538.8km/h i^ v⃗PG,y=v⃗PA,y=654.0km/h/j^\begin{array}{c} \vec{v}_{PG,x} = \vec{v}_{PA,x} + \vec{v}_{AG,x} = ( -588.8 + 50.0 ) km/h \ \hat{i} = -538.8 km/h \ \hat{i} \\~\\ \vec{v} _{PG,y} = \vec{v}_{PA,y} = 654.0 km/h / \hat{j} \end{array}

For the speed, we would need the magnitude.

∣v⃗PG∣=(−538.8km/h)2+(654.0km/h)2=847.4km/h|\vec{v}_{PG}| = \sqrt{ (-538.8km/h)^2 + (654.0 km/h)^2 } = 847.4 km/h

We are travelling a little bit slower because of the wind. This makes sense! The plane is travelling against the wind.

Practice: Droning on

Say you are on the ground, outside, controlling a drone video camera. In your video, it seems like a bird flew by at 5.0 m/s southwest. From your point of view on the ground, the bird flew 1.0 m/s northwest. What is the velocity of your drone relative to you on the ground? Give your answer in terms of magnitude and the angle relative to the +x axis.



A pilot in a Boeing 747 is beginning a 1500km flight. The plane's speed is 1000km/h and air traffic control says that he will have to head 15o15^owest of south to maintain a southward course. If the flight takes 100min, what is the wind speed?