The Acceleration Vector

b / A change in the magnitude of the velocity vector implies an acceleration.

c / A change in the direction of the velocity vector also produces a nonzero Δv vector, and thus a nonzero acceleration vector, Δv/Δt.

When all three acceleration components are constant, i.e., when the v_x - t, v_y - t, and v_z - t graphs are all linear, we can define the acceleration vector as

a = Δv/Δt , [only for constant acceleration]

which can be written in terms of initial and final velocities as

a = (v_f - v_i)/Δt . [only for constant acceleration]

If the acceleration is not constant, we define it as the vector made out of the a_x, a_y, and a_z components found by applying the slope-of- the-tangent-line technique to the v_x-t, v_y -t, and v_z -t graphs.

Now there are two ways in which we could have a nonzero acceleration. Either the magnitude or the direction of the velocity vector could change. This can be visualized with arrow diagrams as shown in figures b and c. Both the magnitude and direction can change simultaneously, as when a car accelerates while turning. Only when the magnitude of the velocity changes while its direction stays constant do we have a Δv vector and an acceleration vector along the same line as the motion.

Self-Check	(1) In figure b, is the object speeding up, or slowing down? (2) What would the diagram look like if vi was the same as vf ? (3) Describe how the Δv vector is different depending on whether an object is speeding up or slowing down.
Answer	(1) It is speeding up, because the final velocity vector has the greater magnitude. (2) The result would be zero, which would make sense. (3) Speeding up produced a Δv vector in the same direction as the motion. Slowing down would have given a Δv that bointed backward.

If this all seems a little strange and abstract to you, you're not alone. It doesn't mean much to most physics students the first time someone tells them that acceleration is a vector, and that the acceleration vector does not have to be in the same direction as the velocity vector. One way to understand those statements better is to imagine an object such as an air freshener or a pair of fuzzy dice hanging from the rear-view mirror of a car. Such a hanging object, called a bob, constitutes an accelerometer. If you watch the bob as you accelerate from a stop light, you'll see it swing backward. The horizontal direction in which the bob tilts is opposite to the direction of the acceleration. If you apply the brakes and the car's acceleration vector points backward, the bob tilts forward.

After accelerating and slowing down a few times, you think you've put your accelerometer through its paces, but then you make a right turn. Surprise! Acceleration is a vector, and needn't point in the same direction as the velocity vector. As you make a right turn, the bob swings outward, to your left. That means the car's acceleration vector is to your right, perpendicular to your velocity vector. A useful definition of an acceleration vector should relate in a systematic way to the actual physical effects produced by the acceleration, so a physically reasonable definition of the acceleration vector must allow for cases where it is not in the same direction as the motion.

Self-Check	In projectile motion, what direction does the acceleration vector have?
Answer	As we have already seen, the projectile has ax = 0 and ay = -g, so the acceleration vector is pointing straight down.

Rappelling.

The galloping horse.

Discussion Questions

A

When a car accelerates, why does a bob hanging from the rearview mirror swing toward the back of the car? Is it because a force throws it backward? If so, what force? Similarly, describe what happens in the other cases described above.

B

The following is a question commonly asked by students: "Why does the force vector always have to point in the same direction as the acceleration vector? What if you suddenly decide to change your force on an object, so that your force is no longer pointing the same direction that the object is accelerating?" What misunderstanding is demonstrated by this question? Suppose, for example, a spacecraft is blasting its rear main engines while moving forward, then suddenly begins firing its sideways maneuvering rocket as well. What does the student think Newton's laws are predicting?