Example 2-4 Calculating Average Acceleration
The object described in Example 2-3 has velocity vx = -10 m>s at t = 0, vx = 0 at t = 10 s, vx = +10 m>s at t = 20 s,
vx = +1.4 m>s at t = 30 s, and vx = 0 at t = 40 s. Find the object’s average acceleration for the time intervals (a) t = 0 to
10 s, (b) t = 10 s to 20 s, (c) t = 20 s to 30 s, and (d) t = 30 s to 40 s.
Set Up
The figures in Example 2-3 show the x–t
graph and motion diagram for this object.
For each time interval we’ll use the definition
of average acceleration, Equation 2-3.
Average acceleration for linear
motion:
aaverage, x =
x (m)
60
Point 4
40
Dvx
v2x - v1x
=
Dt
t2 - t1
20
(2-3)
0
Point 1
10
–20
Point 3
20
30
Point 5
t (s)
40
–40
Point 2
–60
Solve
(a) For the time interval t1 = 0 to t2 = 10 s
(from point 1 to point 2 in the x–t graph),
we have vx1 = -10 m>s and vx2 = 0.
Because the velocity becomes more positive
(it goes from negative to zero), the average x
acceleration is positive.
aaverage, x =
(b) For the time interval t2 = 10 s to t3 = 20 s
(from point 2 to point 3 in the x–t graph),
we have vx2 = 0 and vx3 = +10 m>s. The
velocity again becomes more positive (it goes
from zero to a positive value), so the average
x acceleration is again positive.
aaverage, x =
(c) For the time interval t3 = 20 s to t4 = 30 s
(from point 3 to point 4 in the x–t graph), we
have vx3 = +10 m>s and vx4 = +1.4 m>s.
The velocity becomes more negative (less
positive), so the average x acceleration is
negative.
(d) For the time interval t4 = 30 s to t5 =
40 s (from point 4 to point 5 in the x–t graph),
we have vx4 = +1.4 m>s and vx5 = 0. Again
the velocity becomes more negative and the
average acceleration is negative.
=
v2x - v1x
t2 - t1
0 - 1 -10 m>s2
10 s - 0
= +1.0 m>s 2
=
v3x - v2x
t3 - t2
1 +10 m>s2 - 0
20 s - 10 s
= +1.0 m>s 2
aaverage, x =
=
=
=
+10 m>s
10 s
+10 m>s
10 s
v4x - v3x
t4 - t3
1 +1.4 m>s2 - 1 +10 m>s2
30 s - 20 s
= -0.86 m>s 2
aaverage, x =
=
v5x - v4x
t5 - t4
0 - 1 +1.4 m>s2
40 s - 30 s
= -0.14 m>s 2
=
=
-8.6 m>s
10 s
-1.4 m>s
10 s
Reflect
Let’s check these results to see how they agree with the general rule about
the algebraic signs of velocity vx and average acceleration aaverage, x. From
t1 = 0 to t2 = 10 s, vx is negative (the x–t graph slopes downward) but
aaverage, x = +1.0 m>s 2 is positive. Because the signs are different, the speed
must decrease during this interval, and indeed it does (from 10 m>s to zero).
From t2 = 10 s to t3 = 20 s, vx is positive (the x–t graph slopes upward) and
aaverage,x = +1.0 m>s 2 is also positive. During this interval the signs of
vx and aaverage, x are the same and the speed increases (from 0 to 10 m>s),
in accordance with the rule. During the intervals from t3 = 20 s to t4 = 30 s
(for which aaverage,x = -0.86 m>s 2) and from t4 = 30 s to t5 = 40 s (for which
aaverage,x = -0.14 m>s 2), vx and aaverage, x have opposite signs (vx is positive and
aaverage, x is negative) and the speed decreases (from 10 m>s to 1.4 m>s to zero) as
it should. When looking at changes in speed, the sign of the acceleration alone
isn’t what’s important—what matters is how the signs of velocity and acceleration
compare.
x (m)
60
40
20
0
–20
–40
–60
vx < 0 vx > 0
ax > 0 ax > 0
10
vx > 0
ax < 0
20
30
t (s)
40