Physics 211
10/17/2012
Practice Exam #2 - Solutions
Do not ip the page until told to do so.
Name:
Problem
1
2
3
4
5
6
Total
Grade
5
5
5
15
15
15
60
Points Possible
5
5
5
15
15
15
60
Useful Equations
1
x(t) = x0 + v0x t + ax t2
2
12
y (t) = y0 + v0y t + ay

Physics 211
9/14/2012
Practice Exam #1 - Solutions
Do not ip the page until told to do so.
Name:
Problem
1
2
3
4
5
6
Total
Grade
5
5
5
15
15
15
60
Points Possible
5
5
5
15
15
15
60
Useful Equations
1
x(t) = x0 + v0x t + ax t2
2
12
y (t) = y0 + v0y t + ay

Physics 211
Due: 10/17/2012
Homework Assignment #5 Solutions
Working in a group is encouraged but everyone must turn in their own version of the assignment.
Problem 1
If we want the car to be twice as fast at the bottom we can replace v0 by 2v0 and say th

Physics 211
Due: 10/08/2012
Homework Assignment #4 Solutions
Working in a group is encouraged but everyone must turn in their own version of the assignment.
Problem 1
The maximum acceleration is vf = v0 + at a = (vf v0 )/t = (0 12.5)/3.5 = 3.57 m/s2 . If

Physics 211
Due: 10/01/2012
Homework Assignment #3 Solutions
Working in a group is encouraged but everyone must turn in their own version of the assignment.
Problem 1 We know that, in one dimension, Fx = max . To nd the force, we rst nd the acceleration,

Physics 211
Due: 9/17/2012
Homework Assignment #2 Solutions
Working in a group is encouraged but everyone must turn in their own version of the assignment.
Problem 1
(a) The four-wheeler starts at the origin and then
moves on, making two turns. The positi

Physics 211
Due: 9/7/2012
Homework Assignment #1 Solutions
Working in a group is encouraged but everyone must turn in their own version of the assignment.
Problem 1
Velocity
Position
Time
Time
Problem 2
0
Symbol
xi
xf
vi
vf
a
ti
tf
2
Denition
initial posi