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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 he moving 5.0s after he passes the signpost if x 5.0m and vox 15m/s? Answer: 105m, 25m/s.

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 he moving 5.0s after he passes the signpost if x 5.0m and vox 15m/s? Answer: 105m, 25m/s.

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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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 he moving 5.0s after he passes the signpost if x 5.0m and vox 15m/s? Answer: 105m, 25m/s.
Transcribed Image Text: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 he moving 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 motorcycle Here 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 passes a signpost at x = 0 marking the city limits. His acceleration is constant: a = 4.0 m/s². At time t = 0, he is 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? SOLUTION SET UP Figure 2.21 shows our diagram. The problem tells us that x = 0 at the signpost, so that is the origin of coordinates. We point the xaxis east, in the direction of motion. The (constant) acceleration is a, = 4.0 m/s². At the initial time r = 0, the position is xo = 5.0 m and the 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 2.10 gives the position x as a function of time 1: I 0 x = 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. REFLECT 6. Take a hard look at your results to see whether they make sense. Are they within the general 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 m t=0 A FIGURE 2.21 unknown. Carry consistency check. a = 4.0 m/s² Vor 15 m/s x = ? t = 2.0 s Ux = ? →→x (east) OY (25 m/s)² VAK Part (b): We want to know the value of x when u = 25 m/s. Note that this occurs at a time later than 2.0 s and at a point farther than 43 m from the signpost. From Equation 2.11, we have v² = 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 when Ux = 25 m/s: Ux = Vox + axt, od zla soittube 25 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 from 15 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 the signpost if xo = 5.0 m and vox = 15 m/s? Answers: 105 m, 25 m/s.
Transcribed Image Text:EXAMPLE 2.8 Constant acceleration on a motorcycle Here 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 passes a signpost at x = 0 marking the city limits. His acceleration is constant: a = 4.0 m/s². At time t = 0, he is 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? SOLUTION SET UP Figure 2.21 shows our diagram. The problem tells us that x = 0 at the signpost, so that is the origin of coordinates. We point the xaxis east, in the direction of motion. The (constant) acceleration is a, = 4.0 m/s². At the initial time r = 0, the position is xo = 5.0 m and the 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 2.10 gives the position x as a function of time 1: I 0 x = 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. REFLECT 6. Take a hard look at your results to see whether they make sense. Are they within the general 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 m t=0 A FIGURE 2.21 unknown. Carry consistency check. a = 4.0 m/s² Vor 15 m/s x = ? t = 2.0 s Ux = ? →→x (east) OY (25 m/s)² VAK Part (b): We want to know the value of x when u = 25 m/s. Note that this occurs at a time later than 2.0 s and at a point farther than 43 m from the signpost. From Equation 2.11, we have v² = 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 when Ux = 25 m/s: Ux = Vox + axt, od zla soittube 25 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 from 15 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 the signpost if xo = 5.0 m and vox = 15 m/s? Answers: 105 m, 25 m/s.
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