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 ~r − ~r velocity ~v = 2 1 (2) (t2 − t1) (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 left- hand 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?

~r − ~r velocity of puck ~v = 2 1 = (4) t2 − t1 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?

~v direction of velocity of puckv ˆ = = (6) |~v| Put the pieces back together to check your work:

|~v|vˆ = (7)

Predicting a new position The definition of velocity can be rewritten as

~r2 − ~r1 = ~v(t2 − t1) (8) which can be used to calculate

~r2 = ~r1 + ~v(t2 − t1) (9) 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)

∆~r velocity of puck ~v = (18) ∆t 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,

d~r velocity ~v = (19) dt