Homework

**100 points**

**Instructions: **

·
**Bold** **Blue** numbers are ones I will do
on the board ** after** you have
attempted them.

·
**Bold Blue **numbers are also the ones I am looking
for when I give you a stamp. Don’t
expect me to give you a stamp if you haven’t at least **attempted** these particular problems. The stamp is worth 20 out of the 100 points.

· Try the problems on scratch paper and then transfer your final solution to your homework paper. That will keep everything neat and organized. Remember that a drawing is worth a thousand words/formulas. Also 25 points out of the 100 points for homework is for neatness/organization/showing-your-work.

· Keep page 57 handy. It has all of the formulas you need to answer these questions

· Answers to odd numbered problems begin on A-15 in the back of the book. I wrote the even numbered answers on this worksheet

·
Try the problems **BEFORE**
looking at my helpful tips!

**2.1 and 2.2 Distance
and speed, Displacement and Velocity**

__P. 57-58:__
1,3,4,5,7,8,9,11,12, **13**, 14, **15**, 19, 20, 26

__Helpful Tips:__

12 - see p. 37 example 2.2

**2.3 Acceleration:**

__P. 59:__ 28,
29, 30, 31, 32, 34, 37, **38**

**2.4 Kinematic
Equations with Constant Acceleration**

__P.60-61:__ 42,
43, **45**,
46, **48**,
**49, 51**,
52, 54, **55,**
56, 64

__Helpful Tips:__

45: recall that the
Average Velocity of the challenger car = Dv_{challenger}/2

Since v_{ochallenger} = 0 m/s.

48: Convert Km/h into m/s first.

49: Convert Km/h into m/s to do the problem. Convert back to Km/h to find the final answer. Also, notice that the acceleration will be negative.

51: You had better at least **attempt**
this problem! I know it’s hard, but you
won’t understand my answer in class if you haven’t at least struggled with it a
bit.

Step 1) Convert V_{ox}
into m/s.

Step 2) V_{ox}
is in the +x direction.

Step 3) acceleration is in the –x direction.

Step 4) Realize that at the time the negative acceleration is applied, the car will continue going in the +x direction for a while until it stops and goes backwards (-x). Therefore, 175m from the point at which the –a was applied will be in the +x direction.

Step 5) Use X=V_{ox}t +1/2at^{2} where X= 175 m and use the quadratic formula
to solve for two solutions to time (t).
Remember that a is negative and will change +1/2at^{2} to –1/2at^{2}.

Step 6) What do __each__
of the answers you get for time tell you about this problem?

52: For part b, can
you tell the __actual__ speeds of the cars a and b? ANSWER:
Car A will be 32.5 m/s and Car B will be 35 m/s. Do you see how I did that?

54 - They are asking for the magnitude of the de-acceleration. Also, remember to convert cm into meters

55 - convert cm to meters. Also, you will need to use 2 equations to solve this. Use the first equation to solve for a. Use the second to solve for t. You actually use the same equation both times, but for different unknowns.

56 - convert cm to meters

**Graphing and finding
the area under the curve**:

__P. 60-61:__ **60**, 61,
**62**__
When you are done with problem #62 on graph paper, turn on G.A. and plot figure
2.24 and use the “integrate” function to determine the displacement.__

__Helpful Tips:__

60: Use the formula
x = v_{o}t + 1/2at^{2} to help you see how this works.

Assume that x = area under the curve = distance traveled. Notice that depending upon what is zero initially, you will only use certain parts of that equation.

62 - (Re-draw figure 2.24 on GRAPH paper and
start shading in the shapes and figure out their areas. Here’s how to do it: The baseline is 0 m/sec. Realize that you will have several different
“shapes” to deal with; you will need to find the **area** of these shapes).
Realize that 0-1 seconds is a line that __never__ leaves the baseline
and has an area of 0!

**2.5 Free Fall: **

**Helpful tips for this section: ****FOR CRYING OUT
LOUD, READ THESE WILL YOU! THEY JUST
MIGHT SAVE YOUR LIFE/GRADE!**

**1) It is important to remember that g = the
acceleration due to gravity near the surface of the earth. g = 9.8 m/s ^{2}. g is the same thing as a in the
equations. We just use g when talking
about gravity.**

** **

2) g can be + or – depending upon the vantage point. For example: If you throw a rock up into the air, it experiences –g upon it. If you throw the rock down (off of a cliff for example) it experiences +g upon it.

** **

**3) Equation x = v _{o}t + 1/2at^{2}
can also be written **

**Equation v ^{2}
= v_{o}^{2} +2ax be written as: v^{2} = v_{o}^{2}
+2a**

** **

**4) In the equations mentioned in hint 3) above,
please realize that you can use g instead of a. Also please realize that g might be – or + depending upon the
particular situation (see hint 2) **

__P. 61-62:__ 65,
66, 67, 68, 69, **70,
71**, 72, 73, 74, 75, **76**, **77**, 78, 79, 80, 81, **82, 83, 84**, **85, 86**

__Helpful Tips:__

69 – Assume it is dropped from v_{o} = 0 m/sec. You will use the same equation twice. Each time you are solving for a different
variable

70 – Be sure to sketch in the correct DIRECTION. Think about it – is the velocity in a + or – direction? Is the y in a + or – direction?). y is the height.

74 – At the top of the flight, v = 0 m/sec, v_{o} is given in the problem

76 – Ask yourself: does the mass of the bullet have anything
to do with this problem? Answer: it doesn’t make a lick of difference how
heavy the bullet is! Also, please
realize that g = -9.8m/s^{2} since gravity is pulling down on the
bullet. PART B IS A QUADRATIC
EQUATION!!!!!!!!

77 – Remember that v = 0 m/sec at the top of the flight.

79 - Do you add g or subtract g when an object goes down?

80 – Note that g = +9.8m/s^{2} because the
gravitational acceleration is in the same direction as the throw.

81 – No math needed to answer this question.

82 - For b you’ll need to use the equation v^{2}
= v_{o}^{2} + 2gy to find y_{max}. Realize that

g = -9.8m/s^{2} and v = 0 m/s. How do you figure out t at y_{max}? I think you can figure that one out for
yourself.

83 - Remember at the top of the rebound, v = 0 m/sec.

84 - I think I see a
QUADRATIC EQUATION HERE! x = v_{o}t
+ 1/2at^{2} is your equation of choice. v_{o} = +10m/s for ball A and –10m/s for ball B. a = -9.8m/s^{2} in both cases.

85 - You should be able to figure out that the balloon misses her. But, in order for you to figure out how far in front of her it hits the ground, you need to realize that the balloon will fall the entire height of the building (18.0 m) instead of the 16.3 m it would have to fall to hit her on the head. Therefore, to figure out how far in front of her it lands, you need to calculate the time for the entire drop.

86 – The camera’s v_{o} = +12.5 m/sec. Realize the camera will go up first to a v =
0 m/sec and then it will drop.
Therefore, it will fall a greater distance than 60.0 m.

For part a: y = v_{o}t-1/2gt^{2} and
then use quadratic to solve for t. (y =
60.0 m, v_{o} = 12.5).

For part b: v = v_{o }+ at since you now know
time from part a.

Realize that a = g = -9.8m/s^{2}

4) Yes. Reverse not possible

8) 300 m

12) a. 60 km/h, 75 km/h b. 67 km/h

14) a. 63 m/s b. zero

20) a. AB = 0, BC = 3 CD = 1.3 EF = -1.7…

b. you can do this for yourself

c. 0 m/s

26) 12.5 sec; 56.3 m (relative to the left runner)

28) d

30) b

32) you can do this for yourself or ask me in class.

34) 6.9 m/s^{2}

38) any time after 7.0 s

42) c

46) a. 10m/s b. 25 m

48) -3.1 m/s^{2}

52) a. 25 m b. Car B

54) 1.8 x 10^{5}
m/s^{2}

56) 2.0 x 10^{2}
m/s^{2}

64) 11 m/s^{2}

60) a. Area = v_{o}t b.
A = 1/2at^{2} c. A = v_{o}t + 1/2at^{2}

62) a. –12 m/s ; -4.0 m/s b. –18m c. 50 m

66) c

68) yes; y up and down

70) a. linear, slope –g b. parabola – you should draw it!

72) a. 27 m/s b. 38 m

74) 11 m

76) a. 19 m
b. t_{up} = 0.68 s; t_{down} = 3.6 s

78) a. Dt = 0.407 sec b. 9.51 sec

80) a. 23.1 m b. 21.6 m/s downward

82) Too much to put it all here. You should draw it. Ask me in class if you don’t get it.

84) a. Ball B b. 2.04 s c. no

86) a. 5.00 sec b. 36.5 m/s