Example 2-5 Decoding a vx–t Graph
Figure 2-12 shows both the x–t graph and the
vx–t graph for the motion of the same object
that we examined in ­Examples 2-3 and 2-4.
Using these graphs alone, determine whether
the object’s acceleration ax is positive, negative,
or zero at (a) t = 0, (b) t = 10 s, (c) t = 20 s,
(d) t = 30 s, and (e) t = 40 s.
vx–t graph for the object from Examples 2-3 and 2-4.
vx (m/s)
x (m)
15
60
10
40
5
20
0
–20
Figure 2-12 ​Two graphs to interpret The x–t graph and
(b) vx-t graph
(a) x-t graph
–40
–60
10
20
30
40
t (s)
0
–5
10
20
30
40
t (s)
–10
–15
Set Up
We use two ideas: (i) acceleration is the rate of change of velocity, and (ii) the value of the acceleration equals the slope
of the vx–t graph (Table 2-2).
Solve
(a) At t = 0, the vx–t graph has a positive (upward) slope, which means that vx is increasing and hence that ax is positive.
The x–t graph shows the same thing. This graph has a steep negative slope at t = 0, which means vx is negative, but as
time increases the slope becomes shallower as the slope and vx become closer to zero. Because vx is changing from a
negative value toward zero, the velocity is increasing and ax is positive.
(b) At t = 10 s, the vx–t graph has the same upward slope as at t = 0, which means that the acceleration ax is again positive. The velocity is zero at this instant, so the object is momentarily at rest. It’s still accelerating, however; the object has
a negative velocity just before t = 10 s, and the velocity is positive just after t = 10 s. This is the time at which the object
turns around. You can also see this from the x–t graph, the slope of which is changing from negative to zero to positive
around t = 10 s.
(c) At t = 20 s, the vx–t graph has zero slope, so the acceleration is zero. At this instant, the velocity is neither increasing
nor decreasing. That’s why the x–t graph at this time is nearly a straight line of constant slope, indicating that the velocity isn’t changing at this instant.
(d) At t = 30 s, the vx–t graph has a negative (downward) slope, so ax is negative. The x–t graph has a positive slope at
this instant (vx is positive) but is flattening out, so the slope and vx are both becoming more negative, which is another
way of saying that the acceleration is negative.
(e) At t = 40 s, the vx–t graph is horizontal, so its slope is zero and the acceleration is likewise zero. At this time, the
velocity is not changing. The x–t graph also reveals that the velocity is not changing, because the slope of this curve is
constant at t = 40 s. (The slope of the x–t curve is also zero, which means that the constant value of vx is zero. In other
words, the object is at rest and is remaining at rest.)
Reflect
We can check our answers by seeing what they tell us about how the speed of the object is changing. At t = 0, the
­object has negative velocity and positive acceleration. Because vx and ax have opposite signs, the object is slowing
down. At t = 20 s, the velocity is positive but the acceleration is zero; the velocity is not changing at this instant, so the
object is neither speeding up nor slowing down. At t = 30 s, the velocity is positive and the acceleration is negative, so
the ­object is once again slowing down. Finally, at t = 40 s, the object is at rest (vx = 0) and is remaining at rest, so it is
neither speeding up nor slowing down. You can confirm these conclusions by comparing with the motion diagram in
Example 2-3. Are they consistent?