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.

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter2: One Dimensional Motion

Section: Chapter Questions

Problem 49PQ: A driver uniformly accelerates his car such that a=6.851im/s2. a. Assuming he starts from rest, find...

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Concept explainers

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

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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