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is the sum of two length at equal to the change in velocity EXAMPLE 2.5 Passing speed In this first example of constant accelerated motion, we will simply consider a car that is initially traveling along a straight stretch of highway at 15 m/s. At t = 0 the car begins to accelerate at 2.0 m/s² in order to pass a truck. What is the velocity of the car after 5.0 s have elapsed? SOLUTION SET UP Figure 2.18 shows what we draw. We take the origin of coor- dinates to be at the initial position of the car, where U = Vox and t = 0, and we let the +x direction be the direction of the car's initial velocity. With this coordinate system, Vox = +15 m/s and a, = +2.0 m/s². SOLVE The acceleration is constant during the 5.0 s time interval, so we can use Equation 2.6 to find v: Ux= Vox + axt= 15 m/s + (2.0 m/s2) (5.0 s) = 15 m/s + 10 m/s = 25 m/s. REFLECT The velocity and acceleration are in the same direction, so the speed increases. An acceleration of 2.0 m/s² means that the velocity -15 m/s Vox =6 ax=2.0 m/s² OVKO KBorstk Video Tutor Soluti Vx 01 += 0 t=5.0 s A FIGURE 2.18 The diagram we draw for this problem. increases by 2.0 m/s every second, so in 5.0 s the velocity increases by 10 m/s. Practice Problem: If the car maintains its constant acceleration, how much additional time does it take the car to reach a velocity of 35 m/s? Answer: 5.0 s. Units: Notes Xo V a

is the sum of two length at equal to the change in velocity EXAMPLE 2.5 Passing speed In this first example of constant accelerated motion, we will simply consider a car that is initially traveling along a straight stretch of highway at 15 m/s. At t = 0 the car begins to accelerate at 2.0 m/s² in order to pass a truck. What is the velocity of the car after 5.0 s have elapsed? SOLUTION SET UP Figure 2.18 shows what we draw. We take the origin of coor- dinates to be at the initial position of the car, where U = Vox and t = 0, and we let the +x direction be the direction of the car's initial velocity. With this coordinate system, Vox = +15 m/s and a, = +2.0 m/s². SOLVE The acceleration is constant during the 5.0 s time interval, so we can use Equation 2.6 to find v: Ux= Vox + axt= 15 m/s + (2.0 m/s2) (5.0 s) = 15 m/s + 10 m/s = 25 m/s. REFLECT The velocity and acceleration are in the same direction, so the speed increases. An acceleration of 2.0 m/s² means that the velocity -15 m/s Vox =6 ax=2.0 m/s² OVKO KBorstk Video Tutor Soluti Vx 01 += 0 t=5.0 s A FIGURE 2.18 The diagram we draw for this problem. increases by 2.0 m/s every second, so in 5.0 s the velocity increases by 10 m/s. Practice Problem: If the car maintains its constant acceleration, how much additional time does it take the car to reach a velocity of 35 m/s? Answer: 5.0 s. Units: Notes Xo V a

College Physics
College Physics
1st Edition
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:Paul Peter Urone, Roger Hinrichs
Chapter2: Kinematics
Section: Chapter Questions
Problem 20PE: An Olympic-class sprinter starts a race with an acceleration of 4.50 m/s2. (a) What is her speed...
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Pap with constant acceleration v Then E Velocity as a function of time For an object that has x component of velocil constant acceleration a,, the velocity U, at any later time Ux= Vor + at. Units: m/s uct of the change in velocity per unit time, ar, and the time interval t. Therefore, it equal change of velocity-that is, the change in velocity per unit time. The term at is the prod We can interpret Equation 2.6 as follows: The acceleration a, is the constant rate at any time is the sum of the initial velocity Vox (at t= 0) and the change in velocity a the total change in velocity from the initial time t= 0 to the later time t. The velocity Figure 2.17b shows this analysis in graphical terms. The height u, of the graph at any time is the sum of two segments: one with length vox equal to the initial velocity, the other with length at equal to the change in velocity during time t. EXAMPLE 2.5 Passing speed In this first example of constant accelerated motion, we will simply consider a car that is initially traveling along a straight stretch of highway at 15 m/s. At t=0 the car begins to accelerate at 2.0 m/s² in order to pass a truck. What is the velocity of the car after 5.0 s have elapsed? SOLUTION SET UP Figure 2.18 shows what we draw. We take the origin of coor- dinates to be at the initial position of the car, where u = Vox and t = 0, and we let the +x direction be the direction of the car's initial velocity. With this coordinate system, Vox = +15 m/s and a, = +2.0 m/s². SOLVE The acceleration is constant during the 5.0 s time interval, so we can use Equation 2.6 to find vx: Ux = Vox + axt= 15 m/s + (2.0 m/s²) (5.0 s) = 15 m/s + 10 m/s = 25 m/s. REFLECT The velocity and acceleration are in the same direction, so the speed increases. An acceleration of 2.0 m/s² means that the velocity Yox = 15 m/s ano ax=2.0 m/s² ote onts: ori We can also get an We call the position later time t is simp placement x-xe Video Tutor Solution Finally, we equ equation: Position as For an obje Units: Th Notes: Ot=0 t = 5.0 s 16 A FIGURE 2.18 The diagram we draw for this problem. increases by 2.0 m/s every second, in 5.0 s the velocity increases by 10 m/s. Hab oft of Practice Problem: If the car maintains its constant acceleration, how much additional time does it take the car to reach a velocity of 35 m/s? Answer: 5.0 s. • xo is • Vax • ax xi L
Transcribed Image Text:Pap with constant acceleration v Then E Velocity as a function of time For an object that has x component of velocil constant acceleration a,, the velocity U, at any later time Ux= Vor + at. Units: m/s uct of the change in velocity per unit time, ar, and the time interval t. Therefore, it equal change of velocity-that is, the change in velocity per unit time. The term at is the prod We can interpret Equation 2.6 as follows: The acceleration a, is the constant rate at any time is the sum of the initial velocity Vox (at t= 0) and the change in velocity a the total change in velocity from the initial time t= 0 to the later time t. The velocity Figure 2.17b shows this analysis in graphical terms. The height u, of the graph at any time is the sum of two segments: one with length vox equal to the initial velocity, the other with length at equal to the change in velocity during time t. EXAMPLE 2.5 Passing speed In this first example of constant accelerated motion, we will simply consider a car that is initially traveling along a straight stretch of highway at 15 m/s. At t=0 the car begins to accelerate at 2.0 m/s² in order to pass a truck. What is the velocity of the car after 5.0 s have elapsed? SOLUTION SET UP Figure 2.18 shows what we draw. We take the origin of coor- dinates to be at the initial position of the car, where u = Vox and t = 0, and we let the +x direction be the direction of the car's initial velocity. With this coordinate system, Vox = +15 m/s and a, = +2.0 m/s². SOLVE The acceleration is constant during the 5.0 s time interval, so we can use Equation 2.6 to find vx: Ux = Vox + axt= 15 m/s + (2.0 m/s²) (5.0 s) = 15 m/s + 10 m/s = 25 m/s. REFLECT The velocity and acceleration are in the same direction, so the speed increases. An acceleration of 2.0 m/s² means that the velocity Yox = 15 m/s ano ax=2.0 m/s² ote onts: ori We can also get an We call the position later time t is simp placement x-xe Video Tutor Solution Finally, we equ equation: Position as For an obje Units: Th Notes: Ot=0 t = 5.0 s 16 A FIGURE 2.18 The diagram we draw for this problem. increases by 2.0 m/s every second, in 5.0 s the velocity increases by 10 m/s. Hab oft of Practice Problem: If the car maintains its constant acceleration, how much additional time does it take the car to reach a velocity of 35 m/s? Answer: 5.0 s. • xo is • Vax • ax xi L
Name: G# Page 40-Practice Problem 2.5: If the car maintains its constant accelration, how much additional time does it take the car to reach a velocity of 35m/s? Answer: 5.0s
Transcribed Image Text:Name: G# Page 40-Practice Problem 2.5: If the car maintains its constant accelration, how much additional time does it take the car to reach a velocity of 35m/s? Answer: 5.0s
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