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course Phy 241
The randomly selected physics assignments for the quiz practice.11:50pm 9/25/12
These are my randomly selected problems from
Assignment 1
Problem 1 Version 25
If the velocity of the object changes from 2 cm / sec to 12 cm / sec in 4 seconds, then at what average rate is the velocity changing?
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My Answer:
aAve = dv/dt. (12-2)/4 = 10/4 = 2.5 cm/s^2.
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Starting from rest, a ball rolls down a constant incline. It requires 3.7 seconds to roll the 65 centimeter length of the incline. What is its average velocity?
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My Answer:
vAve = ds/dt. vAve = 65/3.7 = 17.6 cm/s.
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An object which accelerates uniformly from rest will attain a final velocity which is double its average velocity.
What therefore is the final velocity of this ball?
What average rate is the velocity of the ball therefore changing?
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My Answer:
Assuming this is still pertaining to the ball covering 65 cm in 3.7 seconds from the previous scenario.
vAve = (vf+ v0)/2. We know it started from rest, so v0 = 0. vf/2 = 17.6 cm/s. vf = 35.2 cm/s.
Average rate of change of velocity: acceleration.
aAve = dv/dt. (vf-v0)/dt.
dt= 3.7.
v0 = 0
vf= 35.2 cm/s.
aAve = 35.2/ 3.7.
aAve = 9.5 cm/s^2.
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In 12 seconds, a bicycle accelerating uniformly from rest covers a distance of 148 meters. At what average rate is its velocity changing?
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Looking for aAve.
vAve = ds/dt. ds = 148. dt = 12.
vAve = 12.33 m/s.
vAve = (vf+v0)/2. v0 = 0. vf = 2* 12.33 = 24.66.
aAve = dv/dt. (vf - v0)/dt.
aAve = 24.66/12 = 2 m/s^2.
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Assignment 2
What are the average acceleration and final velocity of an object which accelerates uniformly from rest, traveling a distance of 69 cm from start to finish and ending up with a velocity of 18.1 cm/sec?
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ds = 69 cm.
v0 = 0 (from rest).
vf = 18.1.
vAve = (vf+v0)/2.
vAve = 18.1/2 = 9.05 cm/s.
vAve = ds/dt.
9.05 = 69/dt.
dt = 7.6 s.
aAve = dv/dt.
aAve = (18.1)/7.6 = 2.4 cm/s^2.
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Assignment 3
A bee is making a beeline for its hive. Its velocity is measured at a distance of 83 meters from the observer, when clock time is t = 3 sec and its velocity is 4 m/s, and again at clock time t = 8 sec. It it accelerates uniformly at .9 m/s/s during the time interval, then what is its velocity at t = 8 sec, and its average velocity during this time interval? How far is the bee from the observer at t = 8 sec?
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We know dt = 8-3 = 5s.
a = dv/dt.
a = .9 m/s^2.
vf = ?
v0 = 4 m/s.
. 9 = (vf - 4)/5. vf - 4 = 4.5.
vf = 8.5 m/s at t = 8 s.
vAve = (vf+v0)/2. (8.5 + 4)/2.
vAve = 6.3 m/s.
vAve = ds/dt.
6.25 * 5 = 31.25 m towards the observer.
83 - 31.25 = 51.8 m away from the observer.
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Week 2 Quiz 1
The velocity of an object increases at a uniform rate from 5 m/s to 24 m/s in 14 seconds.
How far does it travel and what is its acceleration?
Sketch the corresponding velocity vs. clock time graph.
Explain the meaning of the slope of the graph and its area.
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v0 = 5
vf = 24
dt = 14.
vAve = (vf+v0)/2. 29/2 = 14.5 m/s.
vAve = ds/dt.
14.5 = ds/ 14.
ds = 14.5*14 = 203 meters.
Travels 203 meters.
aAve = dv/dt. (24-5)/14.
Average acceleration is 1.4 m/s^2.
On a velocity vs time graph, the slope is the acceleration because its the rate at which velocity is changing, and the area is the distance traveled because the integral of the function is area of the graph and the integral of velocity with respect to time is position.
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Everything looks good. You should be in good shape for the Major Quiz.
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