I just need the practice problem solved with work shown. The rest of the image provides the information needed to solve the problem. The correct answer is also provided to check work. The info starts at example 2.3 Units: m/s²Notes:dav, xV2xUlx12 - 11AUxAt|S(2.3)Average acceleration is a vector.• 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.U2x = 1.2 m/s (speeding up);(a) U₁x = 0.8 m/s,(b) Ulx = 1.6 m/s,(c) U₁x = -0.4 m/s,U2x =1.2 m/s (slowing down);U2x =-1.0 m/s (speeding up);(d) v₁x = -1.6m/s,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.close to 200 mthe average vis because, frdisplacementEXAMPLE 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: 36CHAPTER 2 Motion Along a Straight LineSOLUTIONSET UP We use the diagram in Figure 2.11 to organize our data.Part (a)Part (b)Part (c)Part (d)BEFOREVlx = 0.8 m/sVlx = 1.6 m/sU1xVlx = -0.4 m/sUlx = -1.6 m/sAFTERU2x→→→→x1.2 m/sdanslaU2x = 1.2 m/sU2x = -0.8 m/sat boong saA FIGURE 2.11 We can use a sketch and table to organize theinformation given in the problem.SOLVE To find the astronaut's average acceleration in each case, we useAux/At.the definition of average acceleration (Equation 2.3): Aav,xThe time interval is At = 2.0 s in all cases; the change in velocity ineach case is AUx = U2x - Ulx.Δυχ1.2 m/s 0.8 m/sImorPart (a): dav, xPart (b): dav, xPart (c): dav, xPart (d): dav, x===-4s2s1.2 m/s - 1.6 m/s= -0.2 m/s²;4s2s agabova sil= +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;=-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) andU2x = -1.0 m/s oor negative in (b) and (c). So negative acceleration does not necessarilyindicate a slowing down.-362201420+0.4 m/s². VASto bolovadi nodwPractice 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.pony10000Example 2.3 raises the question of what the sign of acceleration means and howrelates to the signs of displacement and velocity, Recouistallex component) is defined

Name: I just need the practice problem solved with work shown. The rest of t
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Description: I just need the practice problem solved with work shown. The rest of the image provides the information needed to solve the problem. The correct answer is also provided to check work. The info starts at example 2.3 Units: m/s²Notes:dav, xV2xUlx12 -