QUIZ 2 (ANSWERS)

CHAPTER 2
Discussion questions: 1, 2, 3,4, 5, 9, 10,11,12,13
(THE DISCUSSION QUESTIONS WILL BE EXPLORED  IN THE LAB SECTIONS AND NOT TURNED IN online. THESE QUESTIONS ALONG WITH THE PROBLEMS BELOW WILL BE THE BASIS FOR REVIEW AND ICQ'S DURING LAB SECTIONS. THESE QUESTIONS  REINFORCE THE EXERCISES/PROBLEMS BELOW.)

Turn in the following exercises/problems next week:  4,9, 10, 12, 16,19, 28, 30, 31, 32, 70, 72, 73, 77
Discussions will be provided below as needed over time.  For example note  #72 is not difficult if you find the time it takes the two cars to collide.   You can think of one car traveling a distance of 200 m at  20 m/s before it collides with a brick wall.  Assume the grasshopper travels back and forth between  car and  wall.   Once you find the time it takes to collide, find the distance the bug travels from its speed.
 
EXERCISE/PROBLEMS
Problems labeled "mup" discussed in class.
mup1 . Think of two cars starting in opposite directions 200 m apart.| v1|*t + | v2| *t = 200 .
Assuming v1 > v2, angle with vertical  =  180 - [360 - | v1 |*360*/(| v1 | + | v2 |)] =  | v1 |*360*/(| v1 | + | v2 |)] - 180 =
360*{| v1 |/(| v1 | + | v2 |)  - 1/2}. When |v1 | = |v2|, angle = 0.
mup2. (a) and (b) The final version of this problem includes the distance from the starting point . You can find the clockwise distance  for simplicity--find  the total dance traveled, then divide by 200 m for the number  of revolutions. Say you get 22.654. Find the product of the fractional part 0.654 and circumference . In the current example multiply the total time for the fast car  in part (a) by speed 6.20 m/s. This will give you the total distance fast car travels. Then divide  by 200 m to get the number of revolutions--between 8.00 and 9.00.  Subtract 8.00 to get the fractional part f. Find product 
f*200 m = the number of meters from  starting point  clockwise. Note final answers are two sig. figs.
10. The  slope of the tangent at each time t equals the instantaneous velocity. Note: The velocity is constant during time intervals in which the x vs t curve is a straight line segment, whether flat or inclined.

9. See section 2.6 for an illustration of  displacement  as the area under the velocity curve. When you divide that area by the time interval, you get the average velocity. The average speed on the other hand is  total distance/ total time.
(a) The velocity is positive and  constant over  each of the two intervals indicating the particle moves to the right, assuming rightward is positive,  and never turns around.  Thus,  the average speed equals the average velocity in this special case. The average velocity is found by dividing the area under the curve by 3.0 seconds. This is a concept from Math 1--The average of a function over interval [a, b] equals the integral over this range divided by b - a. In this case,  b = 3.0 seconds and a = 0 seconds. The total area is the sum of two rectangular areas.
(b) For the average velocity, same concept as in  (a) but now the area under the curve is negative between 2.0 and 3.0 seconds. The  total area is the sum of a positive and negative number. For the average speed you will get a larger number since  total distance  is greater than total displacement.

12. The  slope of the tangent at each time t equals the instantaneous acceleration.. Note: The acceleration is constant during time intervals in which the v vs t curve is a straight line segment, whether flat or inclined. Average acceleration between any two times =
(change in v) /(change in t).
16. The signs of  average accelerations are (a)  negative (b) negative (c) negative. Find magnitudes.
mup3.  You can plot v vs t from  the derivative of x vs t. For x vs t, the  slope of the tangent at each time t equals the instantaneous velocity. Note: The velocity is constant during time intervals in which the x vs t curve is a straight line segment.

There are 5 intervals:

[0,5]: The slope increases from zero to a value equal to the slope of the line for interval [5,15]

[5,15]: The slope is constant and positive

[15,25]: The slope decreases to zero at t = 20 seconds and from zero decreases  to a value equal to the slope of the line for
interval   [25,35].

[25,35]: The slope is constant and negative.

[35, 40]: The slope increases to zero.

You can plot   a vs t    from  the derivative of    v vs t. For  v vs t , the  slope of the tangent at each time t equals the instantaneous acceleration. Note:  The acceleration is constant during time intervals in which the v vs t curve is a straight line segment, whether flat or inclined.

There are 5 intervals:

[0,5]:  v increases linearly from zero; the slope is positive and constant

[5,15]: v is constant; the slope is zero.

[15,25]: v decreases linearly; the slope is negative and constant

[25,35]: v is constant; the slope is zero.

[35, 40]: v increases linearly to zero; the slope is positive and constant

 
30. Acceleration is constant and negative.  To get the total displacement find the sum of the areas under the curve; the areas are negative for t > 6.0 seconds. To get the total distance find the sum of the absolute values of  the  areas under the curve.
31.. The accelerations are just the slopes of the line segments. To get the total displacement find the sum of the areas under the curve; the areas are never negative.  The total distance in this case equals the total displacement since all motions are positive.
mup4. The sketches of "a vs t" , "v vs t" and "x vs t" graphs look surprisingly like those of problem 19 .
32.  Two vehicles moving  in the same direction. Indeed the graphs are faintly  similar to those of #70 if  the passenger train and  freight train collide. The first crossing represents that collision , the smallest root of the quadratic equation like that described below.  This problem's situation is different though:  a car with constant velocity passes an accelerating car; the accelerating car later passes the car that has constant velocity.
70. Solve  200 + 15*t = 25*t - (1/2)*(0.1)t2. The left hand side of the equation represents the freight train's position, the right the passenger train. If they collide, compare the passenger train's and freight train's instantaneous velocity at the collision  point. Explain the  meaning of the velocity difference. 
72.. The cars' relative speed is 20 m/s.  (See section 3.5)  
73..Another case of finding the intersection of two graphs: D + (1/2)*a'*t2 = (1/2)*a*t2 , where "a' " represent the truck's acceleration,  "a " the car's acceleration,  D  the unknown initial difference in distance, and  40.0 m = (1/2)*a'*t2 ; also compare the truck's and car's   instantaneous velocities as needed.
77. The exercise belongs to a problem class defined by  73 and 32: The unifying theme is two objects moving relative to each other.  It  demonstrates something important for all motion problems: Identify  the  motion of a specific  point on the each extended object. For ease of computations, 1-D motion   uses point idealizations. For this problem, the key points are  car's front and truck's  rear,  all computations  based on  distances between these two points which you can deduce from car's and truck's physical   lengths.
Lets' write the equation for  car's  front displacement after time t relative to the origin (x = 0):  
(20.0 m/s)*t + (1/2)*a*t2,  where the car's acceleration is given.
Let's write the equation for the truck's  rear displacement  relative to zero (x = 0)  after  the same time t:  24.0 m + (20.0 m/s)*t.
[staff notes: edit below~] 
We are given  the displacement of car front = displacement of truck rear + 26.0 m + 21.0 m + 4.5 m =
displacement of truck front + 51.5 m.  In other words,   (20.0 m/s)*t + (1/2)*a*t2 = 24.0 m + 51.5  m + (20.0 m/s)*t  or
(20.0 m/s)*t + (1/2)*a*t2 = 75.5  m + (20.0 m/s)*t.  Solving this  you can answer all questions.