Average Velocity and Instantaneous Velocity

If a particle moves from one point to another, we might need to know how fast it moves. Just as in Chapter 2, we can define two quantities that deal with “how fast”: average velocity and instantaneous velocity. However, here we must consider these quantities as vectors and use vector notation.

If a particle moves through a displacement Δ in a time interval Δt, then its average velocity _avg is

This tells us that the direction of _avg (the vector on the left side of Eq. 4-8) must be the same as that of the displacement Δ (the vector on the right side). Using Eq. 4-4, we can write Eq. 4-8 in vector components as

For example, if the particle in Sample Problem 4-1 moves from its initial position to its later position in 2.0 s, then its average velocity during that move is

That is, the average velocity (a vector quantity) has a component of 6.0 m/s along the x axis and a component of 1.5 m/s along the z axis.

When we speak of the velocity of a particle, we usually mean the particle’s instantaneous velocity  at some instant. This  is the value that _avg approaches in the limit as we shrink the time interval Δt to 0 about that instant. Using the language of calculus, we may write  as the derivative

Figure 4-4 shows the path of a particle that is restricted to the xy plane. As the particle travels to the right along the curve, its position vector sweeps to the right. During time interval Δt, the position vector changes from ₁ to ₂ and the particle’s displacement is Δ.

To find the instantaneous velocity of the particle at, say, instant t₁ (when the particle is at position 1), we shrink interval Δt to 0 about t₁. Three things happen as we do so. (1) Position vector ₂ in Fig. 4-4 moves toward ₁ so that Δ shrinks toward zero. (2) The direction of Δ/Δt (and thus of _avg) approaches the direction of the line tangent to the particle’s path at position 1. (3) The average velocity _avg approaches the instantaneous velocity  at t₁.

In the limit as Δt → 0, we have _avg →  and, most important here, _avg takes on the direction of the tangent line. Thus,  has that direction as well:

 The direction of the instantaneous velocity  of a particle is always tangent to the particle’s path at the particle’s position.

The result is the same in three dimensions:  is always tangent to the particle’s path.

To write Eq. 4-10 in unit-vector form, we substitute for  from Eq. 4-1:

Fig. 4-4   The displacement Δ of a particle during a time interval Δt, from position 1 with position vector ₁ at time t₁ to position 2 with position vector ₂ at time t₂. The tangent to the particle’s path at position 1 is shown.

Fig. 4-5   The velocity  of a particle, along with the scalar components of .

This equation can be simplified somewhat by writing it as

where the scalar components of  are

For example, dx/dt is the scalar component of  along the x axis. Thus, we can find the scalar components of  by differentiating the scalar components of .

Figure 4-5 shows a velocity vector  and its scalar x and y components. Note that  is tangent to the particle’s path at the particle’s position. Caution: When a position vector is drawn, as in Figs. 4-1 through 4-4, it is an arrow that extends from one point (a “here”) to another point (a “there”). However, when a velocity vector is drawn, as in Fig. 4-5, it does not extend from one point to another. Rather, it shows the instantaneous direction of travel of a particle at the tail, and its length (representing the velocity magnitude) can be drawn to any scale.

CHECKPOINT 2 The figure shows a circular path taken by a particle. If the instantaneous velocity of the particle is  = (2 m/s) − (2 m/s), through which quadrant is the particle moving at that instant if it is traveling (a) clockwise and (b) counterclockwise around the circle? For both cases, draw  on the figure.

Sample Problem 4-3

For the rabbit in Sample Problem 4-2, find the velocity  at time t = 15 s, in unit-vector notation and in magnitude-angle notation.

Solution: There are two Key Ideas here: (1) We can find the rabbit’s velocity  by first finding the velocity components. (2) We can find those components by taking derivatives of the components of the rabbit’s position vector. Applying the v_x part of Eq. 4-12 to Eq. 4-5, we find the x component of  to be

At t = 15 s, this gives v_x = −2.1 m/s. Similarly, applying the v_y part of Eq. 4-12 to Eq. 4-6, we find that the y component is

At t = 15 s, this gives v_y = −2.5 m/s. Equation 4-11 then yields

which is shown in Fig. 4-6, tangent to the rabbit’s path and in the direction the rabbit is running at t = 15 s.

To get the magnitude and angle of , either we use a vector-capable calculator or we follow Eq. 3-6 to write

Fig. 4-6   The rabbit’s velocity  at t = 15 s. The velocity vector is tangent to the path at the rabbit’s position at that instant. The scalar components of  are shown.

(Although 50° has the same tangent as −130°, inspection of the signs of the velocity components indicates that the desired angle is in the third quadrant, given by 50° − 180° = −130°.)