2.4.1 Signs and Units

An important aspect of acceleration is its sign. Acceleration a⃗, like position r⃗ and velocity v ⃗, is a vector. For motion in one dimension, the sign of ax (or ay) is positive if the vector a⃗ points to the right (or up), negative if it points to the left (or down). This was illustrated in 《 Figure 1.18 and the very important 《 Tactics Box 1.3, which you may wish to review. It’s particularly important to emphasize that positive and negative values of as do not correspond to “speeding up” and “slowing down.”

EXAMPLE 2.8 Relating acceleration to velocity

  1. A bicyclist has a velocity of 6 m/s and a constant acceleration of 2 (m/s)/s. What is her velocity 1 s later? 2 s later?

  2. A bicyclist has a velocity of −6 m/s and a constant acceleration of 2 (m/s)/s. What is his velocity 1 s later? 2 s later?

SOLVE

  1. An acceleration of 2 (m/s)/s means that the velocity increases by 2 m/s every 1 s. If the bicyclist’s initial velocity is 6 m/s, then 1 s later her velocity will be 8 m/s. After 2 s, which is 1 additional second later, it will increase by another 2 m /s to 10 m/s. After 3 s it will be 12 m/s. Here a positive ax is causing the bicyclist to speed up.

  2. If the bicyclist’s initial velocity is a negative −6 m/s but the acceleration is a positive +2 (m/s)/s, then 1 s later his velocity will be −4 m/s. After 2 s it will be −2 m/s, and so on. In this case, a positive ax is causing the object to slow down (decreasing speed v). This agrees with the rule from Tactics Box 1.3: An object is slowing down if and only if vx and ax have opposite signs.

NOTE   It is customary to abbreviate the acceleration units (m/s)/s as m/s2. For example, the bicyclists in Example 2.8 had an acceleration of 2 m/s2. We will use this notation, but keep in mind the meaning of the notation as “(meters per second) per second.”

EXAMPLE 2.9 Running the court

A basketball player starts at the left end of the court and moves with the velocity shown in FIGURE 2.20. Draw a motion diagram and an acceleration-versus-time graph for the basketball player.

FIGURE 2.20 Velocity-versus-time graph for the basketball player of Example 2.9.

The figure illustrates the velocity-verses-time graph of a basketball player in motion.

VISUALIZE The velocity is positive (motion to the right) and increasing for the first 6 s, so the velocity arrows in the motion diagram are to the right and getting longer. From t=6 s to 9 s the motion is still to the right (vx is still positive), but the arrows are getting shorter because vx is decreasing. There’s a turning point at t=9 s, when vx=0 m/s, and after that the motion is to the left (vx is negative) and getting faster. The motion diagram of FIGURE 2.21a shows the velocity and the acceleration vectors.

FIGURE 2.21 Motion diagram and acceleration graph for Example 2.9.

The figure consists of two parts labeled ("a") and (b) that depict the motion diagram and acceleration-versus-time graph of a basketball player in motion.

SOLVE Acceleration is the slope of the velocity graph. For the first 6 s, the slope has the constant value

ax=ΔvxΔt=6.0  m/s6.0 s=1.0  m/s2

The velocity then decreases by 12 m/s during the 6 s interval from t=6 s to t=12 s, so

ax=ΔvxΔt=−12 m/s6.0 s=−2.0 m/s2

The acceleration graph for these 12 s is shown in FIGURE 2.21b. Notice that there is no change in the acceleration at t=9 s, the turning point.

REVIEW The sign of ax does not tell us whether the object is speeding up or slowing down. The basketball player is slowing down from t=6 s to t=9 s, then speeding up from t=9 s to t=12 s. Nonetheless, his acceleration is negative during this entire interval because his acceleration vector, as seen in the motion diagram, always points to the left.