(a) The situation (b) Our sketch Vehicle Uav, x Vehicle At time = 2.00 s, the cheetah's position x₂ is The displacement during this interval is x₂ = 20.0 m+ (5.00 m/s²) (2.00 s)² = 40.0 m. We choose to place the origin at the vehicle. We draw an axis. We point it in the direction the cheetah runs, so that our values will be positive. (3) We mark the initial positions of the cheetah and the antelope. (We won't use the antelope's position-but we don't know that yet.) A FIGURE 2.10 (a) The situation for this problem; (b) the sketch we draw. SOLUTION SET UP Figure 2.10b shows the diagram we sketch. First we define a coordinate system, orienting it so that the cheetah runs in the +x direc- tion. We add the points we are interested in, the values we know, and the values we will need to find for parts (a) and (b). Ax=x₂x₁ = 40.0 m 25.0 m = 15.0 m. - Δ.r 40.0 m 25.0 m Ar 2.00 s-1.00 s Cheetah SOLVE Part (a): To find the displacement Ax, we first find the so that cheetah's positions (the values of x) at time t₁ = 1.00 s and at time = 12 2.00 s by substituting the values of t into the given equation. At time f 1.00 s, the cheetah's position x, is x₁ = 20.0 m+ (5.00 m/s²) (1.00 s)² = 25.0 m. Cheetah starts Xo-20.0m +-0 15.0 m 1.00 s Part (b): Knowing the displacement from 1.00 s to 2.00 s, we can now find the average velocity for that interval: VIX - ? x₁-? 15.0 m/s. ₁-1.00 s Ax-? Vav,x=? X2=? 1₂ -2.00 s Antelope We're interested in the cheetah's motion between 1s and 2 s after it begins running. We place dots to represent those points. Ax At Antelope 26.05 m 1.10 s 50.0 m Part (c): The instantaneous velocity at 1.00 s is approximately exactly) equal to the average velocity in the interval from ₁ t₂ = 1.10 s (i.e., At = 0.10 s). At t = 1.10 s, x₂ = 20.0 m+ (5.00 m/s) (1.10 s)² = 26.05 m, →x (m) 6 We add symbols for known and unknown quantities. We use subscripts 1 and 2 for the points at ris and r = 2 s. 25.0 m 1.00 s (but not 1.00 s to 10.5 m/s. Uav,x= 8.53 If you use the same procedure to find the average velocities for time intervals of 0.01 s and 0.001 s, you get 10.05 m/s and 10.005 m/s, respectively. As Ar gets smaller and smaller, the average velocity gets closer and closer to 10.0 m/s. We conclude that the instantaneous velocity at time t 1.0 s is 10.0 m/s. REFLECT As the time interval Ar approaches zero, the average veloc ity in the interval is closer and closer to the limiting value 10.0 m/s which we call the instantaneous velocity at time t = 1.00 s. Note tha when we calculate an average velocity, we need to specify two times- the beginning and end times of the interval-but for instantaneo velocity at a particular time, we specify only that one time. Practice Problem: What the cheetah's average speed over (a) first second of the attack and (b) the first two seconds of the attac Answers: (a) 5.00 m/s, (b) 10.0 m/s.

Practice problem 2.2 What is the cheetahs average speed over (a) the first second of the attack and (b). The first two seconds of the attack? Answers (a) 5.00m/s (b) 10.0 m/s Both photos have background info to solve the practice problem.

34 CHAPTER 2 Motion Along a Straight Line
(a) The situation
(b) Our sketch
Uav,x=
Ax
At
Vehicle
=
Vehicle
of
At time = 2.00 s, the cheetah's position x₂ is
1₂
x₂ = 20.0 m+ (5.00 m/s²) (2.00 s)² = 40.0 m.
The displacement during this interval is
SOLUTION
SET UP Figure 2.10b shows the diagram we sketch. First we define a
coordinate system, orienting it so that the cheetah runs in the +x direc-
tion. We add the points we are interested in, the values we know, and the
values we will need to find for parts (a) and (b).
40.0 m - 25.0 m
2.00 s-1.00 s
Cheetah
We draw an axis.
We point it in the
direction the
cheetah runs, so
that our values
will be positive.
A FIGURE 2.10 (a) The situation for this problem; (b) the sketch we draw.
SOLVE Part (a): To find the displacement Ax, we first find the
cheetah's positions (the values of x) at time t₁ = 1.00 s and at time
12= 2.00 s by substituting the values of t into the given equation. At
time = 1.00 s, the cheetah's position x₁ is
x₁ = 20.0 m+ (5.00 m/s²) (1.00 s)² = 25.0 m.
Cheetah
starts
Xo-20.0m
+-0
2 We choose 3 We mark the initial
to place the
origin at the
vehicle.
A.x = x₂x₁ = 40.0 m 25.0 m = 15.0 m.
Part (b): Knowing the displacement from 1.00 s to 2.00 s, we can now
find the average velocity for that interval:
=
V₁x = ?
X₁₂-?
1₁-1.00 s
15.0 m
= 15.0 m/s.
1.00 s
positions of the cheetah
and the antelope. (We
won't use the antelope's
position-but we don't
know that yet.)
Ax=?
Vav,x=?
so that
X₂=?
t₂ = 2.00 s
We're interested in the
cheetah's motion
between 1s and 2 s after
it begins running. We
place dots to represent
those points.
Uav,x=
Antelope
Ax
At
Antelope
Part (c): The instantaneous velocity at 1.00 s is approximately (but not
exactly) equal to the average velocity in the interval from ₁= 1.00 s to
12 1.10 s (i.e., Ar= 0.10 s). At t₂ = 1.10 s,
x₂ = 20.0 m+ (5.00 m/s) (1.10 s)² = 26.05 m,
26.05 m
1.10 s
50.0 m
→x (m)
65 We add symbols for
known and unknown
quantities. We use
subscripts 1 and 2 for
the points at f= 1 s
and r = 2 s.
25.0 m
1.00 s
= 10.5 m/s.
If you use the same procedure to find the average velocities for time
intervals of 0.01 s and 0.001 s, you get 10.05 m/s and 10.005 m/s,
respectively. As At gets smaller and smaller, the average velocity gets
closer and closer to 10.0 m/s. We conclude that the instantaneous
velocity at time t = 1.0 s is 10.0 m/s.
REFLECT As the time interval Ar approaches zero, the average veloc-
ity in the interval is closer and closer to the limiting value 10.0 m/s,
which we call the instantaneous velocity at time t = 1.00 s. Note that
when we calculate an average velocity, we need to specify two times-
the beginning and end times of the interval-but for instantaneous
velocity at a particular time, we specify only that one time.
Practice Problem: What is the cheetah's average speed over (a) th
first second of the attack and (b) the first two seconds of the attack
Answers: (a) 5.00 m/s, (b) 10.0 m/s.

a
velocity hom graphs UI position versus time
Figure 2.9 shows graphs of an object's position along the x axis versus
time for four separate trials. Which of these statements is or are correct?
A FIGURE 2.9
XX
A. The velocity is greater in trial 2 than in trial 3.
B. The velocity is not constant during trial 1.
C. During trial 4, the object changes direction as it passes through the
point x = 0.
D. The velocity is constant during trials 2, 3, and 4.
(Remember that velocity means "x component of velocity.")
SOLUTION The graph of the motion during trial I has a changing
slope and therefore v, isn't constant. Trials 2, 3, and 4 all have graphs
with constant slope and thus correspond to motion with constant veloc-
ity. (The velocity is zero during trial 2, as indicated by the slope of zero,
but this value is still constant.) Trial 3 has a positive velocity, while the
velocity in trial 4 is negative. Thus, answers B and D are correct.
EXAMPLE 2.2 Average and instantaneous velocities
Here we will look at the difference between average and instantaneous velocities via a specific example of
a cheetah chasing its prey. The cheetah is crouched in ambush 20.0 m to the east of an observer's vehicle,
as shown in Figure 2.10a. At time t = 0, the cheetah charges an antelope in a clearing 50.0 m east of the
observer. The cheetah runs along a straight line; the observer estimates that, during the first 2.00 s of the
attack, the cheetah's coordinate x varies with time t according to the equation
x = 20.0 m+ (5.00 m/s²)t².
(a) Find the displacement of the cheetah during the interval between t₁ = 1.00 s and t₂ = 2.00 s. (b) Find
the average velocity during this time interval. (c) Estimate the instantaneous velocity at time t₁ = 1.00 s by
taking At = 0.10 s.
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