QUIZ 4 (ANSWERS),  CHAPTER 3: READ ALL BOOK EXAMPLES SINCE SOME ARE LIKE HINTS TO QUIZ PROBLEMS BELOW !

VIRTUAL LAB SIMULATION
Discussions will be provided below as needed. 
CH. 3: DISCUSSION QUESTIONS  FOR EXAM PREPARATION: 2, 3, 4, 5, 6, 7, 9, 10, 12, 13, 14, 18
CH. 3: EXERCISES/PROBLEMS: GENERAL POSITION AND MOTION VECTORS: 2, 3, 8 
(ALSO TRY 7); PROJECTILE , CIRCULAR AND RELATIVE MOTION: 12, 14, 18, 20, 21, 33, 36, 39, 41& 42 ( ALSO TRY 76 AND 81 TO COMPARE), 54, 58, 60 (ALSO TRY 63), 75, 80; FINALLY DO #30, CHAPTER 9,  FOR EXPOSURE TO NON UNIFORM CIRCULAR MOTION. WE WILL  BREAK THIS PROBLEM DOWN TO CHAPTER 3 ESSENTIALS.
2. READ EXAMPLE 3.1.  See section 3.1. Use the definition: Average velocity vector = change in r-vector/change in time. See eqn 3.2 . 

(a) Remember each component of the average velocity vector is simply given by the x or y displacement over change in time: Vx average = (xb - xa)/(tb - ta) and 
Vy average = (yb - ya)/(tb - ta). For the case at hand: Vx average = (xb - xa)/(tb - ta)  = -3.8 m/s, where you are given( tb - ta)  = 12.0 seconds. Find (xb - xa) and follow the same procedure for the y component. Find (yb - ya) in the same way.

(b) Compute the magnitude of the displacement vector whose x and y components are  given by
 (xb - xa) and  (yb - ya).  Use the Pythagorean Theorem. Note the displacement vector = avearge 

(xb - xa)*i + (yb - ya)*j  , where i and j in bold represent  our standard unit vectors. Review quiz 1.
3.  READ EXAMPLE 3.1 . Great problem for you future or current web designers out there using the same procedures as previous problem. The position vector = 2.5t2*i + 5.0*t*j = x(t)*i + y(t)* j, unit vectors in bold. 
(a) Remember each component of the average velocity vector is simply given by the x or y displacement over change in time: Vx average = (xb - xa)/(tb - ta) and 
Vy average = (yb - ya)/(tb - ta).    Here is an example for the x-component of the problem:
x((t)  = 2.5t2.  We evaluate x(2) and x(0) by plugging in the time values into the formula for x(t). Compute x(2) - x(0), then compute [x(2) - x(0)]/(2 - 0).  Also compute:  [y(2) - y(0)]/(2 - 0). Once you get each component of the average velocity, find the magnitude using the Pythagorean Theorem. Find direction by determining the angle the vector makes with the x axis and the quadrant it points.  See Ch. 1 and 4 problems and icq's/ quiz 1.
(b) For each formula x(t) and y(t), compute the derivative: for example dx/dt = 5.0*t. Differentiate and plug in the numbers for each component at the 3 different times.
(c) Sketch qualitatively the curve of motion on an x-y axes like we did with  projectile motion and other examples. Hint: The curve is a parabola since x is proportional to the square of y.  
8. Great intuitive qualitative  problem to close out the vector section of this quiz. See my notes and figures 3.10 and 3.12. See section 3.2 and read example 3.3. See also section 3.4. 

(a) If the particle moves with constant speed along the curved path, the acceleration is zero on the straight parts and not zero on the curved parts. Let's look at the curved parts. which may be considered to be arcs of a circle. In that approximation, we see  the acceleration points to the center of the circle defined by the arced segment of the path. Thus between A and B, the acceleration  vector points rightward and  somewhat below the horizontal toward the center of the circle defined by arc A-B . Use the same logic for arc D-E. 

(b) See figures 3.10 and 3.12 carefully. But first look at the straight section of the path. On the straight parts of the curve, the acceleration vector  clearly points in the same direction as the velocity. See the scantron portion of test #1. Let's look at curved section A-B, which corresponds to figure 3.12 (b). The acceleration vector has a component tangent to the circular arc and also a component pointing to the center of the circle defined by the arc. Thus the acceleration does not point to the center of the circle as in part (a) for the same arced section. It points slightly above the center of the circle as shown in figure 3.12 (b) . Apply this same logic to the other arced section, D-E. 

(c) For  the curved parts, use the same logic discussed in part (b) but  use figure 3.12 (c) instead. For the straight part what is the direction of the acceleration compared to the velocity's  as the particle slows down? 
MORE DISCUSSIONS TO FOLLOW; WITH  RESPECT TO PROJECTILE  MOTION, PROBLEMS 12, 14, 18, 20, 21 CAN BE CONSIDERED "WARM UP "SINCE THEY ARE EASIER BUT ARE POWERFUL TEACHING AND LEARNING TOOLS.
12. Think two dimensionally. Review the logic beyond the take home virtual lab: click here.   Read book examples. In this case  it is only matter of finding the time from  y motion equations and plugging that time into an x-motion equation. 
14.  See previous discussion comment; Given the cliff height 2.75 m, for what range of speeds vo will the x-displacement be between 1.50 m and 2.00 m,  assuming of course all launches are purely horizontal? Remember your constraint is the cliff height which determines the time of flight.   
18. SEE class notes on this symmetric path with vertex at the highest point. Remember  you are responsible for deriving the range formula of example 3.8.  Do not just plug in numbers without understanding how to derive it.
20. SEE class notes and textbook examples on this asymmetric path with vertex at the highest point. Contrast this with previous exercise,  where the particle returns to the same vertical level. In this problem, the particle lands at a lower level. General comments: Remember Vy = Voy - gt, where Voy = Vo*sin 51.0. You are given the time of flight. Vx is constant. The acceleration is constant and points downward. 
21. General comments. There is a nice derivation of the parabolic path on page 81.  See equation 3.27. You could use this expression to find the height but please understand how to derive it. Of course you must use other means to find  the y-component of velocity since  equation 3.27 does not give that.
33. SEE  LECTURE  CLASS EXAMPLES WHERE I SHOWED A PARTICLE IN UNIFORM CIRCULAR MOTION AT 4 OR SO POINTS ALONG THE PATH.  SEE ALSO EXAMPLES 3.11 AND 3.12. Remember, the centripetal acceleration  vector always points to the center of the circle whereas the velocity vector is always tangent to path.
For  relative motion problems, covered in the next 4 exercises, always remember  
VPAx = VPBx + VBAx 
VPAy = VPBy + VBAy, 
where  we have written the x and y components of the following equation
vector-VPA = vector-VPB + vector-VBA . This simply states  that the velocity of a particle P relative to frame A equals the velocity of the particle relative to frame B plus the velocity of frame B relative to frame A. Remember the BART train examples. Let Frame B be the train. Let Frame A be the station, and let P be the person on a skateboard coasting along  the floor inside the train.   Recall in one example I said if VPBx = 10 mph and VBAx = 80 mph (Bart train's top speed), then VPAx = 90 mph as expected. 
36. You are to find VPBx  in three cases when VBAx =13.0 m/s . Discussion of (a) opens the door to the other parts: 
(a) VPAx = VPBx + VBAx  translates to 10.0 m/s = VPBx  + 13.0 m/s. Solve for VPBx . 
(b) Same equation but now VPAx= -3.0 m/s.
(c) Use the same method to reach an  obvious conclusion.
39.  USE 
VPAx = VPBx + VBAx 
VPAy = VPBy + VBAy,  .

A = EARTH FRAME

B = RIVER FRAME

P = BOAT

ON YOUR PAPER , LET UP BE NORTH (N), DOWN BE SOUTH (S),  RIGHT BE EAST (E) AND LEFT BE WEST (W). RIGHT IS THE POSITIVE X DIRECTION AND UP IS THE POSITIVE Y DIRECTION ON YOUR AXES.

VPAx =  +0.40*cos45 (m/s)
VPAy = -0.40*sin45   (m/s)

Note: SOUTHEAST means 45 degrees south of east, in the 4th quadrant; NORTHWEST means 45 degrees north of west, in the second quadrant, etc. 

VBAx = 0.50 m/s and VBAy = 0.

USING THIS INFO, FIND THE MAGNITUDE OF VECTOR-VPB  BY FINDING BOTH OF ITS COMPONENTS AND APPLYING THE PYTHAGOREAN THEOREM TO THEM.
Find the direction by accounting for the signs of the components which will give you the quadrant in which VECTOR-VPB  points.  Find the related angle theta thIS vector makes with the x-axis by evaluating tan theta = |VPBy | / | VPBx  | . 
For problems 41 and 42, see in-class  lecture notes and  book examples 3.14 and 3.15 for reference. With regard to #41, see nearly identical  lecture example I did with a river flowing to the right and a boat velocity relative to the river directed upward  on the paper.  Thus the boat velocity  relative to the shore  was not straight upward on the paper.   
MORE DISCUSSIONS  SHOULD   FOLLOW BUT TRY THE REMAINING  PROBLEMS USING THE ABOVE  CONCEPTS. THEY REPRESENT AN IMPORTANT FOUNDATION, LIKE "WARM UPS" FOR THE MAIN RACE.   FREE TO EMAIL ME  !
54. This problem is like the egg and the professor, #76, Ch. 2. The only difference is the non-vertical launch direction toward the front of the incoming boat.  The horizontal displacement  of the equipment from the  launch point is x = Vo*cosao*t. Time t  of course is obtained using Ch. 2 methods on the  y- motion equations for projectile motion.  Where V is the boat's speed, add V*t to the value of x to get the initial distance D of the boat from  shore, 
58. 

(a) Go to page 87, figure 3.26 to see why you must always aim above a target to hit it   for any launch  angle. If you aim at or below the target you will never hit target  because of the downward pull of the gravitational force also known as the weight. Note: The difference between the straight line and the actual path is (1/2)*g*t2, simply seen by evaluating  

 Vo*sinao*t - (Vo*sinao*t - (1/2)*gt2) , 

the vertical difference  between the straight line (no gravity path) and actual trajectory. 

If you want , rotate the figure shown and pretend you have a horizontal launch to better see  you must always aim above target.

In any event find the angle if you aimed directly at the target. To experiment with this idea find the angle then plug in an  angle less than this value into  equation 3.27. For the given y and x, there is no value of  Vo   satisfying the equation. 
(b) Use equation 3.27 but know how to derive it. Note 45 degrees is larger than the angle of part (a). x = Vo*sinao*t