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Practice Problem: What distance has the car traveled when it has reached a speed of 20 m/s? Answer: 41 m.

Practice Problem: What distance has the car traveled when it has reached a speed of 20 m/s? Answer: 41 m.

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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(I just need the practice problem worked out. The other picture has the background info needed to solve the answer. The correct answer is also provided to be able to check work.) 

Finally, we Velocity as a function of position for an acceleration For an object moving in a straight line with constant acceleration ax, - 2 2 v² = voz² + 2ax(x − xo). Ux Units: The units of each term in this equation reduce to m²/s². Notes: is the initial velocity of the object when it is at its initial position xo. • Vox • Ux This equation is generally used in problems that neither give a time t nor ask for o U is the final velocity of the object when it is at its final position x. EXAMPLE 2.7 Entering the freeway is Now we will apply Equation 2.11 to a problem in which time is neither given nor asked for. A sports car sitting at rest in a freeway entrance ramp. The driver sees a break in the traffic and floors the car's accelera- tor, so that the car accelerates at a constant 4.9 m/s² as it moves in a straight line onto the freeway. What distance does the car travel in reaching a freeway speed of 30 m/s? SOLUTION SET UP As shown in Figure 2.19, we place the origin at the initial posi- tion of the car and assume that the car travels in a straight line in the +x direction. Then Vox = 0, Ux = 30 m/s, and ax 4.9 m/s². SOLVE The acceleration is constant; the problem makes no mention of time, so we can't use Equation 2.6 or Equation 2.10 by itself. 2 2 V v² = v₁² + 2ax(x − xo). x - xo. We need a relationship between x and Ux, and this is provided by Equation 2.11: Rearranging and substituting numerical values, we obtain (30 m/s)² - 0 2(4.9 m/s²) V 2 X - 2 VOX (2.11 2ax = Video Tutor Solut = Equati pos Ec If S 92 m. CONT
Transcribed Image Text:Finally, we Velocity as a function of position for an acceleration For an object moving in a straight line with constant acceleration ax, - 2 2 v² = voz² + 2ax(x − xo). Ux Units: The units of each term in this equation reduce to m²/s². Notes: is the initial velocity of the object when it is at its initial position xo. • Vox • Ux This equation is generally used in problems that neither give a time t nor ask for o U is the final velocity of the object when it is at its final position x. EXAMPLE 2.7 Entering the freeway is Now we will apply Equation 2.11 to a problem in which time is neither given nor asked for. A sports car sitting at rest in a freeway entrance ramp. The driver sees a break in the traffic and floors the car's accelera- tor, so that the car accelerates at a constant 4.9 m/s² as it moves in a straight line onto the freeway. What distance does the car travel in reaching a freeway speed of 30 m/s? SOLUTION SET UP As shown in Figure 2.19, we place the origin at the initial posi- tion of the car and assume that the car travels in a straight line in the +x direction. Then Vox = 0, Ux = 30 m/s, and ax 4.9 m/s². SOLVE The acceleration is constant; the problem makes no mention of time, so we can't use Equation 2.6 or Equation 2.10 by itself. 2 2 V v² = v₁² + 2ax(x − xo). x - xo. We need a relationship between x and Ux, and this is provided by Equation 2.11: Rearranging and substituting numerical values, we obtain (30 m/s)² - 0 2(4.9 m/s²) V 2 X - 2 VOX (2.11 2ax = Video Tutor Solut = Equati pos Ec If S 92 m. CONT
Vox = 0 t = Ux D - otxo ax = 4.9 m/s² A FIGURE 2.19 A car accelerating on a ramp and merging onto a freeway. ax Alternative Solution: We could have used Equation 2.6 to solve for the time t first: VOX x = ? 30 m/s-0 4.9 m/s² 6 Ux = 30 m/s = 6.12 s. X t acceleration e A 2.4 Motion with Constant Acceleration 43 Then Equation 2.10 gives the distance traveled: x = xo = Voxt + a₂² - хо = 0 + (4.9 m/s²) (6.12 s)² = 92 m. REFLECT We obtained the same result in one step when we used Equation 2.11 and in two steps when we used Equations 2.6 and 2.10. When we use them correctly, Equations 2.6, 2.10, and 2.11 always give consistent results. The final speed of 30 m/s is about 67 mi/h, and 92 m is about 100 yd. Does this distance correspond to your own driving experience? Practice Problem: What distance has the car traveled when it has reached a speed of 20 m/s? Answer: 41 m. Equations 2.6, 2.10, and 2.11 are the equations of motion with constant acceleration. any kinematic problem involving the motion of an object in a straight line with constant cceleration can be solved by applying these equations. Here's a summary of the equations: The plot with constant acceleration: ^ x = xo + vaxt + a 1² The effect of acceleration: a,t²
Transcribed Image Text:Vox = 0 t = Ux D - otxo ax = 4.9 m/s² A FIGURE 2.19 A car accelerating on a ramp and merging onto a freeway. ax Alternative Solution: We could have used Equation 2.6 to solve for the time t first: VOX x = ? 30 m/s-0 4.9 m/s² 6 Ux = 30 m/s = 6.12 s. X t acceleration e A 2.4 Motion with Constant Acceleration 43 Then Equation 2.10 gives the distance traveled: x = xo = Voxt + a₂² - хо = 0 + (4.9 m/s²) (6.12 s)² = 92 m. REFLECT We obtained the same result in one step when we used Equation 2.11 and in two steps when we used Equations 2.6 and 2.10. When we use them correctly, Equations 2.6, 2.10, and 2.11 always give consistent results. The final speed of 30 m/s is about 67 mi/h, and 92 m is about 100 yd. Does this distance correspond to your own driving experience? Practice Problem: What distance has the car traveled when it has reached a speed of 20 m/s? Answer: 41 m. Equations 2.6, 2.10, and 2.11 are the equations of motion with constant acceleration. any kinematic problem involving the motion of an object in a straight line with constant cceleration can be solved by applying these equations. Here's a summary of the equations: The plot with constant acceleration: ^ x = xo + vaxt + a 1² The effect of acceleration: a,t²
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