(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 thevelocity. Furthermore, if the object is slowing down, then it does not necessarilyfollow that the acceleration is negative. Similarly, if the object is speeding up, it doesnot necessarily follow that it has positive acceleration.EXAMPLE 2.3 Acceleration in a space walkIn this example we will use Equation 2.3 to calculate the acceleration of an astronaut on a space walk. Theastronaut has left the space shuttle on a tether to test a new personal maneuvering device. She moves alonga straight line directly away from the shuttle. Her onboard partner measures her velocity before and aftercertain maneuvers, and obtains these results:(a) U1xV2x =(b) UlxV2x =(c) UlxV2x =-1.0 m/s (speeding up);(d) UlxV2x = -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,Quet1.2 m/s (speeding up);1.2 m/s (slowing down); 36CHAPTER 2 Motion Along a Straight LineSOLUTIONSOLVE To find the astronaut's average acceleration in each case, we useon 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 inPart (a)Part (b)Part (c)Part (d)BEFOREVLx = 0.8 m/sx= 1.6 m/sx = -0.4 m/sUlx= -1.6 m/sHxAFTERU2x = 1.2 m/sU2x = 1.2 m/sU2 -1.0 m/sU2x-0.8 m/sPart (b): dav, xal boogs so delityPart (c): aav, xA FIGURE 2.11 We can use a sketch and table to organize theinformation given in the problem.each case is Aux = U2x - U1x.noPart (a): aav,x1.2 m/s 0.8 m/s4s2s1.2 m/s 1.6 m/s4s - 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 downin (b) and (d), but the average acceleration is positive in (a) and (d) andnegative in (b) and (c). So negative acceleration does not necessarilyindicate a slowing down.Practice Problem: For a fifth set of maneuvers, the astronaut's partnercalculates 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 finalvelocity? Did she speed up or slow down over the 2 s interval? Answers:-1.4 m/s, speed up.

Given data: Time interval (Δt) = 2 s Initial velocity (u) = -0.4 m/s Acceleration (a) = -0.5 m/s2…

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

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter2: Kinematics

Section: Chapter Questions

Problem 30CQ: Consider the velocity vs. time graph of a person in an elevator shown in Figure 2.58. Suppose the...

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