Velocity
Objective: Understand the definition of displacement and velocity; know how to calculate displacement
and velocity, understand the idea of instantaneous velocity as a derivative.
Displacement
The displacement of an object is the change in its position.
displacement ∆~r = ~r2 − ~r1
(1)
Velocity
velocity ~v =
~r2 − ~r1
(t2 − t1 )
(2)
(The magnitude of velocity, |~v | is called the speed.)
Example
Suppose that a hockey puck travels as shown in the time-elapsed image below. It starts at the lower lefthand corner at t=0 s. Each subsequent image is at t=1 s, t=2 s, etc. Use the image to answer the following
questions. You will need to define a coordinate system and an origin. Let’s define the origin at the first
image (t=0) with +x to the right and +y upward.
Figure 1: Each gridline represents 0.05 m.
What is the displacement of the puck between t = 1 s and t = 4 s? First, sketch the initial position
vector ~r1 . Then sketch the second position vector ~r2 . Now, sketch the vector (~r2 -~r1 . Note that this is the
vector that when added to ~r1 gives you ~r2 . That’s because ~r2 = ~r1 + (~r2 − ~r − 1).
Now, calculate it algebraically.
displacement of puck = ~r2 − ~r1 =
(3)
Do you get the same thing? The numerical result should be consistent with your graphical result.
The velocity of the puck is the displacement divided by the time interval. Remember that dividing a
vector by a scalar gives results in a vector. Therefore, what is the velocity vector?
velocity of puck ~v =
~r2 − ~r1
=
t2 − t1
(4)
Note that is points in the direction of the motion as if it traveled along a straight line from point 1 to
point 2.
What is the speed of the puck?
speed of puck |~v | =
(5)
What is the direction of the puck’s motion expressed as a unit vector?
direction of velocity of puck v̂ =
~v
=
|~v |
(6)
Put the pieces back together to check your work:
|~v |v̂ =
(7)
~r2 − ~r1 = ~v (t2 − t1 )
(8)
~r2 = ~r1 + ~v (t2 − t1 )
(9)
Predicting a new position
The definition of velocity can be rewritten as
which can be used to calculate
This says that if we know the starting position of an object and we know the velocity of the object, we
can predict the new position of the object. In component form, this equation looks like
< x2 , y2 , z2 >=< x1 , y1 , z1 > + < vx , vy , vz > (t2 − t1 )
(10)
Since the x-component of the vector on the left of the equal sign must equal the x-component of the
vector on the right of the equal sign (and likewise for the other components), then we can write
x2 = x1 + vx (t2 − t1 )
(11)
y2 = y1 + vy (t2 − t1 )
(12)
y2 = y1 + vy (t2 − t1 )
(13)
For example, at time t1 = 12.18s, a ball’s position vector is ~r1 =< 20, 8, −12 > m. The ball’s velocity
is ~v =< 9, −4, 6 > m/s. At time t2 = 12.21s, where is the ball, assuming that its velocity hardly changes
during this short time interval?
~r2 = ~r1 + ~v (t2 − t1 ) =
(14)
Note that if the velocity changes significantly during the time interval, in either magnitude or direction,
our prediction for the new position may not be very accurate.
A more compact notation
We use the symbol ∆ to mean “change of” or “take the final value of something – the initial value of
something”. For example,
displacement of puck, or change of position, ∆~r = ~r2 − ~r1
(15)
change of time interval ∆t = t2 − t1
(16)
change of time interval ∆t = t2 − t1
(17)
velocity of puck ~v =
∆~r
∆t
(18)
Derivative
When the velocity of an object is not constant, such as the example of the puck, it is not accurate to calculate
the velocity of the object during a large time interval. As a result, we calculate the velocity during a very
small time interval (the limit as ∆ t approaches zero). Then, our predictions will be more accurate. When
the velocity is calculated in this way, it is the velocity at an instant of time. Then,
velocity ~v =
d~r
dt
(19)