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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.

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.
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. www 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 09:0 An Video Tutor Demo Video Tutor Solut CONTI
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. www 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 09:0 An Video Tutor Demo Video Tutor Solut CONTI
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
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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