What distance has the car traveled when it had reached a speed of 20m/s? Answer 41m Both images have background info for the practice problem. obtain a relationwe can combine Equations 20result is worth the effort!We first solve Equation 2.6 for r and then sulltion 2.10 and simplify:. (Dr. = Voc) + $as (ºr = var) ²2.aaxWe transfer the term to to the left side and multiply through by 2a,:U₂ - Dara.x = x + Uor2a, (x-xo) = 2000 - 200² + U² - 2uarvx + vaz.Finally, we combine terms and simplify to obtain the following equation:Velocity as a function of position for an object moving with constantaccelerationFor an object moving in a straight line with constant acceleration ax,v² = vo?² + 2ax(x - xo).Units: The units of each term in this equation reduce to m²/s².Notes:EXAMPLE 2.7 Entering the freewayNow we will apply Equation 2.11 to a problem in which time is neither given nor asked for. A sports car issitting 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. Whatdistance does the car travel in reaching a freeway speed of 30 m/s?SOLUTIONSET 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 +xdirection. Then Vox = 0, v, 30 m/s, and ax = 4.9 m/s².. This equation is generally used in problems that neither give a time t nor ask for one.• Vox is the initial velocity of the object when it is at its initial position xo.• U, is the final velocity of the object when it is at its final position x.SOLVE The acceleration is constant; the problem makes no mentionof time, so we can't use Equation 2.6 or Equation 2.10 by itself.(2.11)-v² = voz + 2ax(x − xo).x-xo.=the timer firstRearranging and substituting numerical values, we obtainv² - voz(30 m/s)² - 02ax2(4.9 m/s²)= 92 m.EquatAny kineacceleraVideo Tutor SolutionWe need a relationship between x and Ux, and this is provided byEquation 2.11:EquaCONTINUED multiplywe getilt we-nstantravel?ed toship,at thequa-OxoVor=01-a = 4.9 m/s²Ux - Voxt =axx = ?A FIGURE 2.19 A car accelerating on a ramp and merging ontoa freeway.30 m/s 04.9 m/s²Alternative Solution: We could have used Equation 2.6 to solve forthe time t first:30 m/s= 6.12 s.Equations of motion for constant accelerationUx= Vox + axtx = xo + voxt + 1/{axt²2v² = vo²² + 2ax(x − xo)Then Equation 2.10 gives the distance traveled:RIPO x = xo = Vot + a,²2.4 Motion with Constant AccelerationREFLECT We obtained the same result in one step when we usedEquation 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 giveconsistent results.The final speed of 30 m/s is about 67 mi/h, and 92 m is about01100 yd. Does this distance correspond to your own driving experience?Practice Problem: What distance has the car traveled when it hasreached a speed of 20 m/s? Answer: 41 m.madura susdt ob= 0 + (4.9 m/s²) (6.12 s)² = 92 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 constantacceleration can be solved by applying these equations. Here's a summary of the equations:vis(2.6)(2.10)hye (2.11)xo43oThe plot with constant acceleration:x = xo + vox² + 1a,1²The effect ofacceleration:a1²The plot we would getwith zero acceleration:x = xo +Vox¹ZOR

Name: What distance has the car traveled when it had reached a speed of 20m/
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Description: What distance has the car traveled when it had reached a speed of 20m/s? Answer 41m Both images have background info for the practice problem. obtain a relationwe can combine Equations 20result is worth the effort!We first solve Equation 2.6 for r a

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What distance has the car traveled when it had reached a speed of 20m/s? Answer 41m  Both images have background info for the practice problem. obtain a relationwe can combine Equations 20result is worth the effort!We first solve Equation 2.6 for r and then sulltion 2.10 and simplify:. (Dr. = Voc) + $as (ºr = var) ²2.aaxWe transfer the term to to the left side and multiply through by 2a,:U₂ - Dara.x = x + Uor2a, (x-xo) = 2000 - 200² + U² - 2uarvx + vaz.Finally, we combine terms and simplify to obtain the following equation:Velocity as a function of position for an object moving with constantaccelerationFor an object moving in a straight line with constant acceleration ax,v² = vo?² + 2ax(x - xo).Units: The units of each term in this equation reduce to m²/s².Notes:EXAMPLE 2.7 Entering the freewayNow we will apply Equation 2.11 to a problem in which time is neither given nor asked for. A sports car issitting 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. Whatdistance does the car travel in reaching a freeway speed of 30 m/s?SOLUTIONSET 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 +xdirection. Then Vox = 0, v, 30 m/s, and ax = 4.9 m/s².. This equation is generally used in problems that neither give a time t nor ask for one.• Vox is the initial velocity of the object when it is at its initial position xo.• U, is the final velocity of the object when it is at its final position x.SOLVE The acceleration is constant; the problem makes no mentionof time, so we can't use Equation 2.6 or Equation 2.10 by itself.(2.11)-v² = voz + 2ax(x − xo).x-xo.=the timer firstRearranging and substituting numerical values, we obtainv² - voz(30 m/s)² - 02ax2(4.9 m/s²)= 92 m.EquatAny kineacceleraVideo Tutor SolutionWe need a relationship between x and Ux, and this is provided byEquation 2.11:EquaCONTINUED multiplywe getilt we-nstantravel?ed toship,at thequa-OxoVor=01->a = 4.9 m/s²Ux - Voxt =axx = ?A FIGURE 2.19 A car accelerating on a ramp and merging ontoa freeway.30 m/s 04.9 m/s²Alternative Solution: We could have used Equation 2.6 to solve forthe time t first:30 m/s= 6.12 s.Equations of motion for constant accelerationUx= Vox + axtx = xo + voxt + 1/{axt²2v² = vo²² + 2ax(x − xo)Then Equation 2.10 gives the distance traveled:RIPO x = xo = Vot + a,²2.4 Motion with Constant AccelerationREFLECT We obtained the same result in one step when we usedEquation 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 giveconsistent results.The final speed of 30 m/s is about 67 mi/h, and 92 m is about01100 yd. Does this distance correspond to your own driving experience?Practice Problem: What distance has the car traveled when it hasreached a speed of 20 m/s? Answer: 41 m.madura susdt ob= 0 + (4.9 m/s²) (6.12 s)² = 92 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 constantacceleration can be solved by applying these equations. Here's a summary of the equations:vis(2.6)(2.10)hye (2.11)xo43oThe plot with constant acceleration:x = xo + vox² + 1a,1²The effect ofacceleration:a1²The plot we would getwith zero acceleration:x = xo +Vox¹ZOR

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