Got a fifth set of the maneuvers the astronaut’s partner calculates her average acceleration over the 2 s interval to be -0.5m/s^2. If the initial velocity of the astronaut was -0.4m/s what was her final? Did she speed up or slow down over the 2s interval? Answer (-1.4m/s, speed up) Both images have background info to help solve the practice problem.
Got a fifth set of the maneuvers the astronaut’s partner calculates her average acceleration over the 2 s interval to be -0.5m/s^2. If the initial velocity of the astronaut was -0.4m/s what was her final? Did she speed up or slow down over the 2s interval? Answer (-1.4m/s, speed up) Both images have background info to help solve the practice problem.
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Got a fifth set of the maneuvers the astronaut’s partner calculates her average acceleration over the 2 s interval to be -0.5m/s^2. If the initial velocity of the astronaut was -0.4m/s what was her final? Did she speed up or slow down over the 2s interval?
Answer (-1.4m/s, speed up)
Both images have background info to help solve the practice problem.

Transcribed Image Text:Units: m/s²
Notes:
.
dav, x
U2x Ulx
12-11
AUX
At
• Average acceleration is a vector.
• It describes how the velocity is changing with time.
(2.3)
The sign of the average acceleration is not necessarily the same as the sign of the
velocity. Furthermore, if the object is slowing down, then it does not necessarily
follow that the acceleration is negative. Similarly, if the object is speeding up, it does
not necessarily follow that it has positive acceleration.
U2x 1.2 m/s (speeding up);
(a) ULx= 0.8 m/s,
(b) U₁x = 1.6 m/s,
(c) ULx= -0.4 m/s,
U2x =
1.2 m/s (slowing down);
U2x =
-1.0 m/s (speeding up);
(d) ULx = -1.6m/s,
V2x = -0.8 m/s (slowing down).
If t₁ = 2s and t2 = 4s in each case, find the average acceleration for each set of data.
displacement of the race car is zero.
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EXAMPLE 2.3 Acceleration in a space walk
In this example we will use Equation 2.3 to calculate the acceleration of an astronaut on a space walk. The
astronaut has left the space shuttle on a tether to test a new personal maneuvering device. She moves along
a straight line directly away from the shuttle. Her onboard partner measures her velocity before and after
certain maneuvers, and obtains these results:
OSO
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Transcribed Image Text:36 CHAPTER 2 Motion Along a Straight Line
SOLUTION
SET UP We use the diagram in Figure 2.11 to organize our data.
Part (a)
Part (b)
Part (c)
Part (d)
Direction of
motion
U is positive
Ux
BEFORE
V, is negative
Ux
P=0.8 m/s
Vlx
= 1.6 m/s
= -0.4 m/s
x= -1.6 m/s
U2x
T
A FIGURE 2.11 We can use a sketch and table to organize the
information given in the problem.
AFTER
Vlx
he direction of the axis determines
e signs of velocity and acceleration.
U2r 1.2 m/s
-
=
U₂, 1.2 m/s
U2x
-
U2r-1.0 m/s
-
U20
= -0.8 m/s
+
Speeding up
dav, x (+)
-
=
U2x Ulx > 0, so aav, x > 0
dav..x (-)
--
AU,/At.
SOLVE To find the astronaut's average acceleration in each case, we use
the definition of average acceleration (Equation 2.3): dav.x = AU/Ar
The time interval is Ar 2.0 s in all cases; the change in velocity in
each case is Av, U2x
0.8 m/s
1.2 m/s
4s-2s
1.2 m/s 1.6 m/s
Ule
Part (a): av.x
= +0.2 m/s²;
= -0.2 m/s²;
Part (b): dav,x
Example 2.3 raises the question of what the sign of acceleration means and how
relates to the signs of displacement and velocity. Because average acceleration (i.e.,
x component) is defined as aav.x = Av/At, and At is always positive, the sign of accele
tion is the same as the sign of Aux. Figure 2.12 shows the four possible cases; make
you understand why the acceleration is positive or negative in each case.
The term deceleration is sometimes used for a decrease in speed; such a decrease
correspond to either positive or negative acceleration. Because of this ambiguity, we w
use the term in this text. Instead, we recommend careful attention to the interpretati
the algebraic sign of a, in relation to that of Ux.
VIX
U2x UL.x < 0, so aav, x < 0
-
Part (c): aay.x =
4s-2s
-1.0 m/s - (-0.4 m/s)
4s2s
Part (d): dav,x=
-0.8 m/s (-1.6 m/s)
4s-2s
= +0.4 m/s².A E
REFLECT The astronaut speeds up in cases (a) and (c) and slows down
in (b) and (d), but the average acceleration is positive in (a) and (d) and
negative in (b) and (c). So negative acceleration does not necessarily
indicate a slowing down.
Practice Problem: For a fifth set of maneuvers, the astronaut's partner
calculates her average acceleration over the 2 s interval to be -0.5 m/s².
If the initial velocity of the astronaut was -0.4 m/s, what was her final
velocity? Did she speed up or slow down over the 2 s interval? Answers:
-1.4 m/s, speed up.
U2x
= -0.3 m/s²;
Notice that the direction of the acceleration is not
necessarily the same as the direction of travel!
U2x
Slowing down
Ulx
aav, x (-)
U2x U1x < 0, so dav, x < 0
aav. x (+)
Ulx
U2xULx > 0, so dav, x > 0
U2x
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