g A to bolas ton avis torfjon i amil old ai maldongs of S noEXAMPLE 2.6 Passing distanceLet's now revisit Example 2.5 and calculate how far the car travels during its 5.0 s of acceleration.SOLUTIONSET UP We use the same coordinates as in Example 2.5. As before,Vox = +15 m/s and ax +2.0 m/s².=SOLVE We want to solve for x - xo, the distance traveled by the carduring the 5.0 s time interval. The acceleration is constant, so we can27use Equation 2.10 to findx = xo = voxt + 1⁄2axt²Q口口Video Tutor Solution= (15 m/s) (5.0 s) + (2.0 m/s²) (5.0 s)²= 75 m + 25 m = 100 m.(s) ow102CONTINUED 42CHAPTER 2 Motion Along a Straight LineVOREFLECT If the speed were constant and equal to the initial valueVo = 15 m/s, the car would travel 75 m in 5.0 s. It actually travelsfarther because the speed is increasing. From Example 2.5, we knowthat the final velocity is Ux = 25 m/s, so the average velocity for the5.0 s segment of motion isVav, xVox + Vx215 m/s + 25 m/s2=20 m/s.An alternative way to obtain the distance traveled is to multiplyaverage velocity by the time interval. When we do this, we gettheWex - x0 = Uav, xt = (20 m/s) (5.0 s) = 100 m, the same resultobtained using Equation 2.10.Practice Problem: If the car in Example 2.5 maintains its constantacceleration for a total time of 10 s, what total distance does it travel?Answer: 250 m.

Given data: Initial velocity (v0x) = +15 m/s Acceleration (ax) = +2.0 m/s2 Time (t) = 10 s…

Answered: Practice Problem: If the car in Example…

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.

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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Distance and Speed

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Question

g A to bolas ton avis torfjon i amil old ai maldongs of S no
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,
Vox = +15 m/s and ax +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
27
use Equation 2.10 to find
x = xo = voxt + 1⁄2axt²
Q
口口
Video Tutor Solution
= (15 m/s) (5.0 s) + (2.0 m/s²) (5.0 s)²
= 75 m + 25 m = 100 m.
(s) ow
102
CONTINUED

42
CHAPTER 2 Motion Along a Straight Line
VO
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 Ux = 25 m/s, so the average velocity for the
5.0 s segment of motion is
Vav, x
Vox + Vx
2
15 m/s + 25 m/s
2
=
20 m/s.
An alternative way to obtain the distance traveled is to multiply
average velocity by the time interval. When we do this, we get
the
We
x - x0 = Uav, xt = (20 m/s) (5.0 s) = 100 m, the same result
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.

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