EXAMPLE 2.9 Pursuit! In this example we will, for the first time, look at the constant accelerated motion of two objects moving simultaneously. Clearly, we will need to describe the position and velocity of each separately and then use the resulting equations to answer any questions about their motion. Suppose that a motorist is traveling at a con- stant velocity of 15 m/s and passes a school-crossing corner where the speed limit is 10 m/s (about 22 mi/h). A police officer on a motorcycle stopped at the corner then starts off in pursuit with a constant acceleration of 3.0 m/s² (Figure 2.22a). (a) How much time elapses before the officer catches up with the car? (b) What is the officer's speed at that point? (c) What is the total distance the officer has traveled at that point? SCHOOL O Xp ap= 3.0 m/s² XC (a) A FIGURE 2.22 (a) A diagram of the problem. (b) Graphs of position as a function of time for the police officer and the motorist. SOLUTION a SET UP Both objects move in a straight line, which we'll designate as the x axis. Then all positions, velocities, and accelerations have only x components. We'll omit the subscripts x in our solution, but always keep in mind that when we say "velocity," we really mean "x compo- nent of velocity," and so on. The motorcycle and the car both moua with UCO = 15 m/s alaration 2.4 Motion with Constant Acceleration 45 x (m) 160 120- 80- 40- 0 (b) 2 Video Tutor Solution Motorist, 1 4 Officer 1 1 6 8 10 12 1(s) There are two times when the two vehicles have the same x co- ordinate; the first is the time (r= 0) when the car passes the parked motorcycle at the corner, and the second is the time when the officer catches up. Part (b): From Equation 2.6, we know that the officer's velocity up at any time t is given by

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CAL CONTENT
EXAMPLE 2.9 Pursuit!
In this example we will, for the first time, look at the constant accelerated motion of two objects moving
simultaneously. Clearly, we will need to describe the position and velocity of each separately and then use the
resulting equations to answer any questions about their motion. Suppose that a motorist is traveling at a con-
stant velocity of 15 m/s and passes a school-crossing corner where the speed limit is 10 m/s (about 22 mi/h).
A police officer on a motorcycle stopped at the corner then starts off in pursuit with a constant acceleration
of 3.0 m/s² (Figure 2.22a). (a) How much time elapses before the officer catches up with the car? (b) What is
the officer's speed at that point? (c) What is the total distance the officer has traveled at that point?
BARCH
indoes
pinn
lamin sed
1340 25
2500 SCHOOL
CROSSING
O
or
Xp
ap = 3.0 m/s²
ved bas jas mail net 25 owi
adt as teong as ooiwt lev
(a)
A FIGURE 2.22 (a) A diagram of the problem. (b) Graphs of position as a function of time for
the police officer and the motorist.
xc
xp = xo + Upot + apt² = 0 + 0 + (3.0 m/s²),²
= (3.0 m/s²)r²,
xc = xo +ucot + act² = 0 + (15 m/s)t + 0 = (15 m/s)t.
Uco 15 m/s.
=
SOLUTION
SET UP Both objects move in a straight line, which we'll designate as
the x axis. Then all positions, velocities, and accelerations have only
x components. We'll omit the subscripts x in our solution, but always
keep in mind that when we say "velocity," we really mean "x compo-
nent of velocity," and so on.
th The motorcycle and the car both move with constant acceleration,
so we can use the formulas we have developed. We have two different
objects in motion; Figure 2.22a shows our choice of coordinate system.
The origin of coordinates is at the corner where the officer is stationed;
both objects are at this point at time t = 0, so xo = 0 for both. Let xp
be the police officer's position and xc be the car's position, at any time
t. At the instant when the officer catches the car, the two objects are at
the same position. We need to apply the constant-acceleration equations and the officer's position is
to each, and find the time when xp and xc are equal. We denote the
initial x components of velocities as Upo and Uco and the accelerations
0, Uco 15
as ap and ac. From the data given, we have Upo= 0, Uco = 15 m/s,
ap= 3.0 m/s², and ac = 0.
m/s,
Part (a): At the time the officer catches the car, both must be at the
same position, so at this time, xc = xp. Equating the preceding two
expressions, we have
(3.0 m/s²)² = (15 m/s)r,
t = 0, 10 s.
2.4 Motion with Constant Acceleration 45
OSAO
C
x (m)
160
120
80
40
0
ol
(b)
2
Video Tutor Solution
Motorist
Officer
1
8 10 12 (s)
1
46
sertal odi
There are two times when the two vehicles have the same x co-
ordinate; the first is the time (t = 0) when the car passes the parked
motorcycle at the corner, and the second is the time when the officer
catches up.
SOLVE Applying Equation 2.10 to each object, we find that Figure 2.22b shows graphs of xp and xc as functions of time. At first
REFLECT A graphical description of the motion is helpful.
xc is greater because the motorist is ahead of the officer. But the car
travels with constant velocity, while the officer accelerates, closing the
gap between the two. At the point where the curves cross, the officer has
caught up to the motorist. We see again that there are two times when
the two positions are the same. Note that the two vehicles don't have the
same speed at either of these times.
Part (b): From Equation 2.6, we know that the officer's velocity Up at
any time t is given by
Up = Upo + apt = 0 + (3.0 m/s²)r,
so when t = 10 s, Up= 30 m/s. When the officer overtakes the car, she
is traveling twice as fast as the motorist is.
Part (c): When t= 10 s, the car's position is
xc= (15 m/s) (10 s) = 150 m,
bas xp = (3.0 m/s²) (10 s)² = 150 m.
This result verifies that, at the time the officer catches the car, they have
gone equal distances and are at the same position.
Practice Problem: If the officer's acceleration is 5.0 m/s², what dis-
tance does she travel before catching up with the car? What is her veloc-
ity when she has caught up? During what time interval is she moving
more slowly than the car? Answers: 90 m, 30 m/s, 0 to 3 s.
Transcribed Image Text:CAL CONTENT EXAMPLE 2.9 Pursuit! In this example we will, for the first time, look at the constant accelerated motion of two objects moving simultaneously. Clearly, we will need to describe the position and velocity of each separately and then use the resulting equations to answer any questions about their motion. Suppose that a motorist is traveling at a con- stant velocity of 15 m/s and passes a school-crossing corner where the speed limit is 10 m/s (about 22 mi/h). A police officer on a motorcycle stopped at the corner then starts off in pursuit with a constant acceleration of 3.0 m/s² (Figure 2.22a). (a) How much time elapses before the officer catches up with the car? (b) What is the officer's speed at that point? (c) What is the total distance the officer has traveled at that point? BARCH indoes pinn lamin sed 1340 25 2500 SCHOOL CROSSING O or Xp ap = 3.0 m/s² ved bas jas mail net 25 owi adt as teong as ooiwt lev (a) A FIGURE 2.22 (a) A diagram of the problem. (b) Graphs of position as a function of time for the police officer and the motorist. xc xp = xo + Upot + apt² = 0 + 0 + (3.0 m/s²),² = (3.0 m/s²)r², xc = xo +ucot + act² = 0 + (15 m/s)t + 0 = (15 m/s)t. Uco 15 m/s. = SOLUTION SET UP Both objects move in a straight line, which we'll designate as the x axis. Then all positions, velocities, and accelerations have only x components. We'll omit the subscripts x in our solution, but always keep in mind that when we say "velocity," we really mean "x compo- nent of velocity," and so on. th The motorcycle and the car both move with constant acceleration, so we can use the formulas we have developed. We have two different objects in motion; Figure 2.22a shows our choice of coordinate system. The origin of coordinates is at the corner where the officer is stationed; both objects are at this point at time t = 0, so xo = 0 for both. Let xp be the police officer's position and xc be the car's position, at any time t. At the instant when the officer catches the car, the two objects are at the same position. We need to apply the constant-acceleration equations and the officer's position is to each, and find the time when xp and xc are equal. We denote the initial x components of velocities as Upo and Uco and the accelerations 0, Uco 15 as ap and ac. From the data given, we have Upo= 0, Uco = 15 m/s, ap= 3.0 m/s², and ac = 0. m/s, Part (a): At the time the officer catches the car, both must be at the same position, so at this time, xc = xp. Equating the preceding two expressions, we have (3.0 m/s²)² = (15 m/s)r, t = 0, 10 s. 2.4 Motion with Constant Acceleration 45 OSAO C x (m) 160 120 80 40 0 ol (b) 2 Video Tutor Solution Motorist Officer 1 8 10 12 (s) 1 46 sertal odi There are two times when the two vehicles have the same x co- ordinate; the first is the time (t = 0) when the car passes the parked motorcycle at the corner, and the second is the time when the officer catches up. SOLVE Applying Equation 2.10 to each object, we find that Figure 2.22b shows graphs of xp and xc as functions of time. At first REFLECT A graphical description of the motion is helpful. xc is greater because the motorist is ahead of the officer. But the car travels with constant velocity, while the officer accelerates, closing the gap between the two. At the point where the curves cross, the officer has caught up to the motorist. We see again that there are two times when the two positions are the same. Note that the two vehicles don't have the same speed at either of these times. Part (b): From Equation 2.6, we know that the officer's velocity Up at any time t is given by Up = Upo + apt = 0 + (3.0 m/s²)r, so when t = 10 s, Up= 30 m/s. When the officer overtakes the car, she is traveling twice as fast as the motorist is. Part (c): When t= 10 s, the car's position is xc= (15 m/s) (10 s) = 150 m, bas xp = (3.0 m/s²) (10 s)² = 150 m. This result verifies that, at the time the officer catches the car, they have gone equal distances and are at the same position. Practice Problem: If the officer's acceleration is 5.0 m/s², what dis- tance does she travel before catching up with the car? What is her veloc- ity when she has caught up? During what time interval is she moving more slowly than the car? Answers: 90 m, 30 m/s, 0 to 3 s.
Name:
G#
Page 44-Practice Problem 2.9:
If the officer's acceleration is 5.0m/s2, what distance does she travel before catching up
with the car? What is her velocity when she has caught up? During what time interval is
she moving more slowly than the car? Answer: 90m, 30m/s, 0 to 3s.
Transcribed Image Text:Name: G# Page 44-Practice Problem 2.9: If the officer's acceleration is 5.0m/s2, what distance does she travel before catching up with the car? What is her velocity when she has caught up? During what time interval is she moving more slowly than the car? Answer: 90m, 30m/s, 0 to 3s.
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