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. OSAGEOEXAMPLE 2.8 Constant acceleration on a motorcycleHere we will look at a more complex problem for which we will use all three of our primary constant-acceleration equations. A motorcyclist heading east through a small Iowa town accelerates after he passesa signpost at x = 0 marking the city limits. His acceleration is constant: ax =4.0 m/s². At time t = 0, heis 5.0 m east of the signpost and has a velocity of vox = 15 m/s. (a) Find his position and velocity at time1 = 2.0 s. (b) Where is the motorcyclist when his velocity is 25 m/s?SOLUTIONSET UP Figure 2.21 shows our diagram. The problem tells us thatx = 0 at the signpost, so that is the origin of coordinates. We point thex axis east, in the direction of motion. The (constant) acceleration isa = 4.0 m/s². At the initial time t = 0, the position is xo = 5.0 m andthe initial velocity is vox = 15 m/s.SOLVE Part (a): We want to know the position and velocity (the val-ues of x and Ux, respectively) at the later time t =.2.0 s. Equation 2.10gives the position x as a function of time t:x = xo + voxt + 1/a,t²= 5.0 m+ (15 m/s) (2.0 s) + (4.0 m/s²) (2.0 s)² = 43 m.Equation 2.6 gives the velocity Ux as a function of time t:Ux = Vox + axt= 15 m/s + (4.0 m/s²) (2.0 s) = 23 m/s.Voxxo = 5.0 mt = 0FIGURE 2.21=ax = 4.0 m/s²gendo the results change15 m/sX = ?t = 2.0 sTOUx = ?→ x (east)Video Tutor Solut25 m/s. Note thatthis occurs at a time later than 2.0 s and at a point farther than 43 m fromPart (b): We want to know the value of x when uxthe signpost. From Equation 2.11, we have=2v² = voz² + 2ax(x − xo),Ux(25 m/s)² = (15 m/s)2 + 2(4.0 m/s²)(x - 5.0 m),x = 55 m.Alternatively, we may use Equation 2.6 to find first the time whenUx = 25 m/s:Ux = Vox + axt,25 m/s = 15 m/s + (4.0 m/s²)(t),t = 2.5 s.Then, from Equation 2.10,x = xo + voxt + 1⁄2 axt²== 5.0 m+ (15 m/s) (2.5 s) + (4.0 m/s²) (2.5 s)²= 55 m.sors Rigodi poloniworlaA FIGUEthe policSOLUTIONSET UP Bthe x axisx comporkeep in mnent ofTheso we cobjectsThe orbothbe tht. Attheto EiniasaFREFLECT Do the results make sense? The cyclist accelerates from15 m/s (about 34 mi/h) to 23 m/s (about 51 mi/h) in 2.0 s while travel-ing a distance of 38 m (about 125 ft). This is fairly brisk acceleration,but well within the realm of possibility for a high-performance bike.Practice Problem: If the acceleration is 2.0 m/s² instead of 4.0 m/s.where is the cyclist, and how fast is he moving, 5.0 s after he passes thesignpost if xo = 5.0 m and vox = 15 m/s? Answers: 105 m, 25 m/s.

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Given data: 
Acceleration (ax) = 2.0 m/s2
Time (t) = 5.0 s
Initial position (x0) = 5.0 m
Initial…

Practice Problem: If the acceleration is 2.0 m/s² instead of 4.0 m/s. where is the cyclist, and how fast is he moving, 5.0 s after he passes the signpost if xo = 5.0 m and vox 15 m/s? Answers: 105 m, 25 m/s. =