College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Distance and Speed
The speed, distance and time relation forms the basis of study of motion. You will need these concepts in the study of Mechanics. This concept is applicable for students who are studying the following courses –
Question
If the car in example 2.5 maintains its constant acceleration for a total time of 10s, what total distance does it travel? Answer 250m Both images have background info to solve the practice problem.
e of od- als Px t We call the position at time =0 the initial position, denoted by xo. The position at the later time is simply x. Thus, for the time interval A = 0 and the corresponding dis- placement x-xo. the general definition gives CAFE (2.9) Vav, x Finally, we equate Equations 2.8 and 2.9 and simplify the result to obtain the following equation: . x - хо t Position as a function of time for an object moving with constant acceleration For an object undergoing constant acceleration, its position at some time t is given by x = xo + vot + ½a,1². (2.10) Units: The units of each of the three terms on the right-hand side reduce to meters. Notes: xo is the position at t = 0. • Vor is the velocity at t = 0. • ay is the acceleration, which is assumed to be constant in time. x is the position at some later time t. 100 05/12 19 EXAMPLE 2.6 Passing distance Let's now revisit Example 2.5 and calculate how far the car travels during its 5.0 s of acceleration. SOLUTION SET UP We use the same coordinates as in Example 2.5. As before, Vax=+15 m/s and a = +2.0 m/s². SOLVE We want to solve for x-xo, the distance traveled by the car during the 5.0 s time interval. The acceleration is constant, so we can use Equation 2.10 to find x = xo = voxt + axt² Video Tutor Solution (15 m/s) (5.0 s) + (2.0 m/s²) (5.0 s)² = 75 m + 25 m = 100 m. CONTINUED
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Transcribed Image Text:e of od- als Px t We call the position at time =0 the initial position, denoted by xo. The position at the later time is simply x. Thus, for the time interval A = 0 and the corresponding dis- placement x-xo. the general definition gives CAFE (2.9) Vav, x Finally, we equate Equations 2.8 and 2.9 and simplify the result to obtain the following equation: . x - хо t Position as a function of time for an object moving with constant acceleration For an object undergoing constant acceleration, its position at some time t is given by x = xo + vot + ½a,1². (2.10) Units: The units of each of the three terms on the right-hand side reduce to meters. Notes: xo is the position at t = 0. • Vor is the velocity at t = 0. • ay is the acceleration, which is assumed to be constant in time. x is the position at some later time t. 100 05/12 19 EXAMPLE 2.6 Passing distance Let's now revisit Example 2.5 and calculate how far the car travels during its 5.0 s of acceleration. SOLUTION SET UP We use the same coordinates as in Example 2.5. As before, Vax=+15 m/s and a = +2.0 m/s². SOLVE We want to solve for x-xo, the distance traveled by the car during the 5.0 s time interval. The acceleration is constant, so we can use Equation 2.10 to find x = xo = voxt + axt² Video Tutor Solution (15 m/s) (5.0 s) + (2.0 m/s²) (5.0 s)² = 75 m + 25 m = 100 m. CONTINUED
42 CHAPTER 2 Motion Along a Straight Line REFLECT If the speed were constant and equal to the initial value Vo 15 m/s, the car would travel 75 m in 5.0 s. It actually travels farther because the speed is increasing. From Example 2.5, we know that the final velocity is u, 25 m/s, so the average velocity for the 5.0 s segment of motion is Vox + Ux 2 Uav,x= 15 m/s + 25 m/s 20 m/s. 2 An alternative way to obtain the distance traveled is to multiply the average velocity by the time interval. When we do this, we get x-xo=Uav.x = (20 m/s) (5.0 s) = 100 m, the same result we obtained using Equation 2.10. Practice Problem: If the car in Example 2.5 maintains its constant acceleration for a total time of 10 s, what total distance does it travel? Answer: 250 m. For some problems, we aren't given the time interval for the motion, and we need to obtain a relationship for x, Ux, and a, that doesn't contain t. To obtain such a relationship, we can combine Equations 2.6 and 2.10 to eliminate t. This involves a little algebra, but the result is worth the effort! t = We first solve Equation 2.6 for t and then substitute the resulting expression into Equa- tion 2.10 and simplify: UxVox ax x = xo + vox Ux - Vox ax + Vor) ². ar We transfer the term xo to the left side and multiply through by 2ax: 2ax(x xo) = 200xx - 200 + v² – 2000, + vo?. Finally, we combine terms and simplify to obtain the following equation: Velocity as a function of position for an object moving with constant acceleration uiqmie For an object moving in a straight line with constant acceleration a, v₂ = vor + 2ax(x - xo). Units: The units of each term in this equation reduce to m²/s². Notes: (2.11) • 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. • This equation is generally used in problems that neither give a time t nor ask for one.
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Transcribed Image Text:42 CHAPTER 2 Motion Along a Straight Line REFLECT If the speed were constant and equal to the initial value Vo 15 m/s, the car would travel 75 m in 5.0 s. It actually travels farther because the speed is increasing. From Example 2.5, we know that the final velocity is u, 25 m/s, so the average velocity for the 5.0 s segment of motion is Vox + Ux 2 Uav,x= 15 m/s + 25 m/s 20 m/s. 2 An alternative way to obtain the distance traveled is to multiply the average velocity by the time interval. When we do this, we get x-xo=Uav.x = (20 m/s) (5.0 s) = 100 m, the same result we obtained using Equation 2.10. Practice Problem: If the car in Example 2.5 maintains its constant acceleration for a total time of 10 s, what total distance does it travel? Answer: 250 m. For some problems, we aren't given the time interval for the motion, and we need to obtain a relationship for x, Ux, and a, that doesn't contain t. To obtain such a relationship, we can combine Equations 2.6 and 2.10 to eliminate t. This involves a little algebra, but the result is worth the effort! t = We first solve Equation 2.6 for t and then substitute the resulting expression into Equa- tion 2.10 and simplify: UxVox ax x = xo + vox Ux - Vox ax + Vor) ². ar We transfer the term xo to the left side and multiply through by 2ax: 2ax(x xo) = 200xx - 200 + v² – 2000, + vo?. Finally, we combine terms and simplify to obtain the following equation: Velocity as a function of position for an object moving with constant acceleration uiqmie For an object moving in a straight line with constant acceleration a, v₂ = vor + 2ax(x - xo). Units: The units of each term in this equation reduce to m²/s². Notes: (2.11) • 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. • This equation is generally used in problems that neither give a time t nor ask for one.
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