Second photo had background info for the practice problem Name:G#Page 44-Practice Problem 2.8:If the acceleration is 2.0m/s² instead of 4.0m/s2, where is the cyclist, and how fast is hemoving 5.0s after he passes the signpost if x 5.0m and vox 15m/s? Answer: 105m,25m/s. EXAMPLE 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: a = 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 time✓= 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 thexaxis east, in the direction of motion. The (constant) acceleration isa, = 4.0 m/s². At the initial time r = 0, the position is xo = 5.0 m andthe initial velocity is vor= 15 m/s.SOLVE Part (a): We want to know the position and velocity (the val-ues of x and U, respectively) at the later time t = 2.0 s. Equation 2.10gives the position x as a function of time 1:I0x = xo + vaxt + a₂²= 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 U, as a function of time t:Ux = Vox + axt= 15 m/s + (4.0 m/s²) (2.0 s) = 23 m/s.REFLECT6. Take a hard look at your results to see whether they make sense. Are they within thegeneral range of magnitudes you expected? If you change one of the given quantities,do the results change in a way you can predict?Xo = 5.0 mt=0A FIGURE 2.21unknown. Carryconsistency check.a = 4.0 m/s²Vor 15 m/sx = ?t = 2.0 sUx = ?→→x (east)OY(25 m/s)²VAKPart (b): We want to know the value of x when u = 25 m/s. Note thatthis occurs at a time later than 2.0 s and at a point farther than 43 m fromthe signpost. From Equation 2.11, we havev² = voz + 2ax(x - xo),x = 55 m.Video Tutor Solution(15 m/s)² + 2(4.0 m/s²)(x - 5.0 m),Alternatively, we may use Equation 2.6 to find first the time whenUx = 25 m/s:Ux = Vox + axt, od zlasoittube25 m/s = 15 m/s + (4.0 m/s²) (t),t = 2.5 s.Then, from Equation 2.10,x = xo + voxt + a²= 5.0 m+ (15 m/s) (2.5 s) + (4.0 m/s²) (2.5 s)²= 55 m.REFLECT 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/s²,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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Description: Second photo had background info for the practice problem Name:G#Page 44-Practice Problem 2.8:If the acceleration is 2.0m/s² instead of 4.0m/s2, where is the cyclist, and how fast is hemoving 5.0s after he passes the signpost if x 5.0m and vox 15m/s?