# READ ALL CHECKPOINTS IN THE CHAPTER !
# STUDY THE NUMBERED OR UNNUMBERED FIGURES, GRAPHS, AND TABLES. 
# IT DOES NOT HURT TO DO EXTRA END-OF-CHAPTER REVIEW QUESTIONS, PLUG AND CHUGS, RANKINGS , OR EXERCISES AND  PROBLEMS. 
#IF TIME PERMITS GET MORE PRACTICE WITH THE OUT OF SEQUENCE  SAMPLE MULTIPLE CHOICE QUIZZES AT VERTICAL LEFT HAND COLUMN  Q1, Q2, Q3, LINKS  ETC HERE~  http://www.nvaphysics.com/P11AU11.htm .
Quiz 3 (These problems will contribute to real quiz 2 next Thursday) Physics 11 Autumn 2011  --- Selected topics + exercises/problems from Ch.3
Quiz 3 Physics 11 Autumn 2011  --- Selected topics from Ch. 3
PLUG AND CHUG---2, 10, 14, 15. 16, 20 EVALUATE USING FORMULAS
RANKING---1, 2, 4
EXERCISES--1, 5, 6, 7. 11, 14, 15, 16, 17, 18, 19, 25, 26, 27, 29, 32, 36, 38, 40
PROBLEMS---1, 2
DISCUSSIONS
RANKING
1. Hint ~  the largest relative speed and velocity is given by D.
2. See figure  3.6 or figure 2.3. The ball will speed up as it descends down a slope and slow down when it rises up a slope. 
4.  The answer is independent of the mass; See figure 3.8. 
EXERCISES 
1.   See CheckPoint page 36.   Subtract the two speeds since the cars are moving in the same direction.  
5.  What is the acceleration of a particle moving in a straight line at constant speed?  Is the velocity changing?   See page 36. 
6. If the particle moves in a straight line and slows down,  the acceleration points in a direction opposite the velocity and direction of motion. 
7.  Compare with exercise 1. Would you add or subtract the speeds in the case of cars moving in opposite directions? 
11.  See figure 3.11
14.  See figure 3.11 
15.  See figure 3.8 at the top of the the ball's flight. 
16.  Remember acceleration is the rate of change in velocity and velocity has both magnitude ( the speed) and direction. If neither speed or direction changes, then what is the acceleration?  Is it possible  for the velocity to change while the speed remains  constant? See #11.
17. (a) See exercise5. (b) See exercise 15
18.  See exercise 6
19.  See figures 3.6 , 2.2 and 2.3. In the first image on the left, clearly the acceleration does not change. So your choices are from the second and third images.  
25. See figure 3.7
26.  See figure 3.7 and notice the distances between the rock.
27.  Does the acceleration change as the object falls? 
29. Figure 3.8
32. Does the rate of change of the velocity have anything to do with the start velocity ?  See figure 3.8. Would the velocity change by a different amount each second if you threw the rock upward with speed of 40 m/s instead of 30 m/s? Or would the rock just move to a higher maximum height at the same acceleration? Now, think about throwing the rock down instead of dropping it down ( with start speed zero.) 
36.  See Plug and Chug 14, 15 and 16.
38.   See figure 3.7. 
40.  See RANKING #2,  IN CASE B, the ball will descend down the second slope and increase its speed to a higher value which  will be maintained on the lowest flat part until the last slope, which is upward.  In A, the ball never receives this extra boost in speed by another slope.  Which ball has the largest average speed , A or B? The one with the largest speed takes the least time. 
PROBLEMS
1. NOTE IN FIGURE 3.8 THE VELOCITY IS POSITIVE ON THE WAY UP AND NEGATIVE ON THE WAY DOWN. To find the time it takes the object to reach its  maximum height  see figure 3.8 for guidance.  When the object is at the maximum height, the velocity is zero. In figure 3.8 it takes 3 seconds to reach the top since the velocity decreases  by about 
10 m/s every second and the starting velocity is + 30 m/s. So in figure 3.8 you could write the velocity as 
Vv = 30 m/s  - (10 m/s2)*t  where Vv represents the velocity in vertical  direction. If you set Vv = 0 and solve for t , you get get the time it takes to get to the top: 0 =  30 m/s  - (10 m/s2)*t . TRY THE SAME METHOD FOR A STARTING SPEED OF  40 m/s for practice. 


# IN ANY EVENT WHEN YOU GET THE TIME,   DOUBLE  IT     IN ORDER TO GET THE  TOTAL   TIME IN THE AIR.
# THE TIME FROM THE START TO THE TOP IS THE SAME AS THE TIME FROM THE TOP BACK TO THE STARING POINT. 

To find the distance d  traveled to the top, you can pretend the object falls from rest from the top and moves down for the same time period you computed above. The distance traveled and the time period on the way up is the same as the distance traveled  and time period on the way down to get back to the starting point. So just pretend you start at the top with zero velocity and fall from rest downward for the time period you computed above. In this case you can use  the formula d = (1/2)*g*t2. Plug in the time you computed above into this formula. 

 

HERE IS ANOTHER WAY TO GET THE DISTANCE USING CHECKPOINT ON PAGE 43: 

change in position = average velocity*time

average velocity = (start velocity + final velocity)/2 and the time to get from the starting point to the top is 3 seconds as stated above. Note : The start velocity is given and the final velocity at the top is ZERO.  
2. NOTE IN FIGURE 3.8 THE VELOCITY IS POSITIVE ON THE WAY UP AND NEGATIVE ON THE WAY DOWN.
(a) See Problem 1 discussions. 
(b) remember the object loses about 10 m/s of speed each second on the way up and at the top the speed is ZERO. So what must be the velocity 1.0 second before reaching the top? Your answer will be positive since you are giving me the velocity  on the way up.  
(c) see comment in part (b). Note the change in velocity will be negative since you are computing : 
velocity at top -  velocity one second before reaching the top. The velocity at the top is zero and the velocity one second before reaching the is positive. So your answer will be negative. 
(d)  See comment in part (b) and remember on the way down the velocity is negative.  The object picks up 10 m/s of speed after one second of falling from the top where the speed is zero. However your answer must include a negative sign to indicate the object moves down.     
(e) The change in velocity will be negative since you are computing:
velocity one second after the top - velocity at the top,
Note:  Velocity at top is zero and the velocity one second after the top is negative. So your answer is negative.
(f) Compute: Velocity one second after the top -  Velocity one second before the top. Your answer will be NEGATIVE.
(g) The acceleration is constant (never changes.) Since it points down, it is negative. What is the value of the acceleration at all points on the path, including the top where the velocity is ZERO?