{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

PHY302K-Sai-HW1-solutionspg1

# PHY302K-Sai-HW1-solutionspg1 - Homework 1 — sai — p 1...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Homework 1 — sai — _ p 1 This print—out should have 33 questions. Multiple—choice questions may continue on the next column or page — ﬁnd all choices before answering. 001 (part 1 of 2) 10.0 points While John is traveling along a straight inter- state highway, he notices that the mile marker reads 241 km. John travels until he reaches the 148 km marker and then retraces his path to the 172 km marker. What is John’s resultant displacement from the 241 km marker? Correct answer: —69 km. Explanation: Let the initial position be so = a = 241 km, the second position b = 148 km, and his ﬁnal position sf = c = 172 km. Thus As=sf—SO=172km—241km2—69km 002 (part 2 of 2) 10.0 points How far has he traveled? Correct answer: 117 km. Explanation: The distance traveled is given by d: |b—a|+|c—b| = |148 km — 241 kml + |172 km — 148 km| 2117 km 003 (part 1 of 2) 10.0 points Consider a moving object whose position a: is plotted as a function of the time t on the following ﬁgure: ll 2 4 6 Clearly, the object moved in different ways during the time intervals denoted I, II and III on the ﬁgure. During these three intervals, when was the object’s speed highest? Caution: Do not confuse the speed with the velocity. 1. During interval I. 2. Same speed during each of the three in- tervals. 3. During interval III. correct 4. During interval II. 5. During intervals II and III (same speed during those two intervals). Explanation: The velocity V is the slope of the a:(t) curve; the magnitude 1) = |v| of this slope is the speed. Looking at the picture we see that the curve is steepest (in absolute magnitude) during the interval III and that is when the object had the highest speed. 004 (part 2 of 2) 10.0 points During which interval(s) did the object’s ve— locity remain constant? 1. During each of the three intervals. cor- rect 2. During interval I only. 3. During none of the three intervals. 4. During interval III only. 5. During interval II only. Explanation: For each of the three intervals I, II or III, the 11:05) curve is linear, so its slope — the velocity V — is constant during each interval. Between the intervals, the velocity did change in a rather abrupt manner — but it did remain constant during each interval. 005 (part 1 of 6) 5.0 points Consider the following graph of motion. ...
View Full Document

{[ snackBarMessage ]}