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marcute 10mg 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.

marcute 10mg 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.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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(Just the one practice problem worked out)
. It describes how the velocity is changing with time. 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. 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: (a) U1x V2x = (b) Ulx V2x = (c) Ulx V2x = -1.0 m/s (speeding up); (d) Ulx V2x = -0.8 m/s (slowing down). If t₁ = 2 s and t₂ = 4s in each case, find the average acceleration for each set of data. = = = 0.8 m/s, 1.6 m/s, -0.4 m/s, = -1.6m/s, Quet 1.2 m/s (speeding up); 1.2 m/s (slowing down);
Transcribed Image Text:. It describes how the velocity is changing with time. 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. 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: (a) U1x V2x = (b) Ulx V2x = (c) Ulx V2x = -1.0 m/s (speeding up); (d) Ulx V2x = -0.8 m/s (slowing down). If t₁ = 2 s and t₂ = 4s in each case, find the average acceleration for each set of data. = = = 0.8 m/s, 1.6 m/s, -0.4 m/s, = -1.6m/s, Quet 1.2 m/s (speeding up); 1.2 m/s (slowing down);
36 CHAPTER 2 Motion Along a Straight Line SOLUTION SOLVE To find the astronaut's average acceleration in each case, we use on the definition of average acceleration (Equation 2.3): dav,x= Aux/At. SET UP We use the diagram in Figure 2.11 to organize our data.toy The time interval is At = 2.0 s in all cases; the change in velocity in Part (a) Part (b) Part (c) Part (d) BEFORE VLx = 0.8 m/s x= 1.6 m/s x = -0.4 m/s Ulx = -1.6 m/s Hx AFTER U2x = 1.2 m/s U2x = 1.2 m/s U2 -1.0 m/s U2x -0.8 m/s Part (b): dav, x al boogs so delity Part (c): aav, x A FIGURE 2.11 We can use a sketch and table to organize the information given in the problem. each case is Aux = U2x - U1x. no Part (a): aav,x 1.2 m/s 0.8 m/s 4s2s 1.2 m/s 1.6 m/s 4s - 2 s +0.2 m/s²; Part (d): dav, x = how brail diw -0.2 m/s²; -1.0 m/s - (-0.4 m/s) 4s - 2s -0.8 m/s - (-1.6 m/s) 4s2s = -0.3 m/s²; = +0.4 m/s². 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.
Transcribed Image Text:36 CHAPTER 2 Motion Along a Straight Line SOLUTION SOLVE To find the astronaut's average acceleration in each case, we use on the definition of average acceleration (Equation 2.3): dav,x= Aux/At. SET UP We use the diagram in Figure 2.11 to organize our data.toy The time interval is At = 2.0 s in all cases; the change in velocity in Part (a) Part (b) Part (c) Part (d) BEFORE VLx = 0.8 m/s x= 1.6 m/s x = -0.4 m/s Ulx = -1.6 m/s Hx AFTER U2x = 1.2 m/s U2x = 1.2 m/s U2 -1.0 m/s U2x -0.8 m/s Part (b): dav, x al boogs so delity Part (c): aav, x A FIGURE 2.11 We can use a sketch and table to organize the information given in the problem. each case is Aux = U2x - U1x. no Part (a): aav,x 1.2 m/s 0.8 m/s 4s2s 1.2 m/s 1.6 m/s 4s - 2 s +0.2 m/s²; Part (d): dav, x = how brail diw -0.2 m/s²; -1.0 m/s - (-0.4 m/s) 4s - 2s -0.8 m/s - (-1.6 m/s) 4s2s = -0.3 m/s²; = +0.4 m/s². 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.
Expert Solution
Step 1

Given data: 

Time interval (Δt) = 2 s

Initial velocity (u) = -0.4 m/s

Acceleration (a) = -0.5 m/s2

Required:

The final velocity (v) 

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