# Relative velocity one dimension Virtual Experiment

#### Quiz

Q1) Car A is moving at $\ v_{A}$= 60 km/h to the right with respect to the ground.  Car B is moving at $\ v_{B}$= 80 km/h to the left with respect to the ground.  What is the velocity of Car A with respect to Car B (the velocity of Car A as measured by the passenger in Car B)?
A. 20 km/h, left                 B. 20 km/h, right               C. 60 km/h, left
D. 140 km/h, left               E. 140 km/h, right

The velocity of Car A with respect to Car B ($\ v_{AB}$) is given by: $\ v_{AB} = v_{A} – v_{B}$. $\ v_{AB}$ = 60 km/h [right] – 80 km/h [left]) = 60 km/h –(– 80 km/h) = 140 km/h [right]

Q2) Car A is moving at $\ v_{A}$ = 60 km/h to the right with respect to the ground.  Car B is moving at $\ v_{B}$ = 80 km/h to the right with respect to the ground.  What is the velocity of Car A with respect to Car B (the velocity of Car A measured by Car B)?
A. 20 km/h, left                 B. 20 km/h, right               C. 60 km/h, left
D. 140 km/h, left               E. 140 km/h, right

:  The velocity of Car A with respect to Car B ($\ v_{AB}$) is given by: $\ v_{AB} = v_{A} – v_{B}$. $\ v_{AB}$= (60 km/h [right]) – (80 km/h [right]) = (60 km/h – 80 km/h) = -20 km/h [left]

#### Relative velocity one dimension Simulation(Virtual Experiment)

When the next simulation is not visible, please refer to the following link.

Relative velocity
The case of the faster car overtaking your car was easy to solve with a minimum of thought and effort, but you will encounter many situations in which a more systematic method of solving such problems is beneficial. To develop this method, write down all the information that is given and that you want to know in the form of velocities with subscripts appended.

$\ v_{se}$ = +80 km/h, north
(Here the subscript se means the velocity of the slower car with respect to Earth.)
$\ v_{fe}$ = +90 km/h, north
(The subscript fe means the velocity of the fast car with respect to Earth.)

We want to know $\ v_{fs}$, which is the velocity of the fast car with respect to the slower car. To find this, we write an equation for $\ v_{fs}$ in terms of the other velocities, so on the right side of the equation the subscripts start with f and eventually end with s. Also, each velocity subscript starts with the letter that ended the preceding velocity subscript.

$\ v_{fs} = v_{fe} + v_{es}$

The boldface notation indicates that velocity is a vector quantity. This approach to adding and monitoring subscripts is similar to vector addition, in which vector arrows are placed head to tail to find a resultant.

We know that $\ v_{es} = −v_{se}$ because an observer in the slow car perceives Earth as moving south at a velocity of 80 km/h while a stationary observer on the ground (Earth) views the car as moving north at a velocity of 80 km/h.

Thus, this problem can be solved as follows:

$\ v_{fs} = v_{fe} + v_{es} = v_{fe} - v_{se}$

$\ v_{fs}$ = (+90 km/h north) − (+80 km/h north) = +10 km/h, north

When solving relative velocity problems, follow the above technique for writing subscripts. The particular subscripts will vary depending on the problem, but the method for ordering the subscripts does not change. A general form of the relative velocity equation is $\ v_{ac} = v_{ab} + v_{bc}$. This general form may help you remember the technique for writing subscripts.

Example 1.
Illustrates the concept of relative velocity by showing a passenger walking toward the front of a moving train. The people sitting on the train see the passenger walking with a velocity of +2.0 m/s, where the plus sign denotes a direction to the right. Suppose the train is moving with a velocity of +9.0 m/s relative to an observer standing on the ground. Then the ground-based observer would see the passenger moving with a velocity of +11 m/s, due in part to the walking motion and in part to the train’s motion. As an aid in describing relative velocity, let us define the following symbols:

Figure 1  The velocity of the passenger relative to the ground-based observer is vPG. It is the vector sum of the velocity $\ v_{PT}$ of the passenger relative to the train and the velocity $\ v_{TG}$ of the train relative to the ground: $\ v_{PG}= v_{PT}+ v_{TG}$.

$\ v_{PT}$ = velocity of the Passenger relative to the Train = +2.0 m/s
$\ v_{TG}$= velocity of the Train relative to the Ground = +9.0 m/s
$\ v_{PG}$ = velocity of the Passenger relative to the Ground = +11.0 m/s

The vector addition problem this illustrates is

$\ v_{PG}= v_{PT}+ v_{TG}$

If we define the forward direction to be the positive direction,

then, because the vectors we are adding are both in the same direction, we are indeed dealing with that very special case in which the magnitude of the resultant is just the sum of the magnitudes of the vectors we are adding:

$\ v_{PG}= v_{PT}+ v_{TG}$

$\ v_{PG}$ = (2.0 m/s) + (9.0 m/s)= +11.0 m/s
$\ v_{PG}$ = 11.0 m/s in the direction in which the train is traveling

You already know all the concepts you need to know to solve relative velocity problems (you know what velocity is and you know how to do vector addition) so the best we can do here is to provide you with some more worked examples. We’ve just addressed the easiest kind of relative velocity problem, the kind in which all the velocities are in one and the same direction. The second easiest kind is the kind in which the two velocities to be added are in opposite directions.

Example 2.
A train is moving  along a ground at a constant 9.0 m/s. a passenger that has a velocity of 2m/s straight backward, (toward the back of the train). Find the velocity of the passenger, relative to the ground.

$\ v_{PG}= v_{PT}+ v_{TG}$

$\ v_{PG}$=  -$\left | v_{PT} \right |$+ $\ v_{TG}$

$\ v_{PG}$= (-2.0 m/s) + (9.0 m/s)= +7.0 m/s
$\ v_{PG}$ = 7.0 m/s in the direction in which the train is traveling