HOMEWORK SOLUTION SET 1

Chapter 1: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39 on pp 15-16.

The movie Spiderman brought in $114,000,000 during its opening weekend. Express this amount in (a) gigadollars, (b) teradollars
Solution: 114,000,000 = 1.14 x 10^8, so
1.0 gigadollar = 1.0 x 10^9, so (1.14 x 10^8)/(1.0 x 10^9) = 0.114 gigadollar.
(remember that 10^9 means “ten to the 9th power)”
1.0 teradollar = 1.0 x 10^12, so (1.14 x 10^8)/(1.0 x 10^12) = 1.14 x 10^(-4) teradollar.

The speed of light is about 0.3 Giga-m/s. Express that in meters/sec.
Solution: 0.3 Giga-meters is 0.3 x 10^9 meters = 3 x 10^8 meters, so 0.3 Giga-m/s = 3 x 10^8 meters/sec.

9. If acceleration is expressed as, a = 2xtp, then find out what the number “p” is (the exponent of t). Here, x is distance, t is time.
Solution: a = 2xtp = (2)(m)(s-2),
Because
(a) x is in units of meters (m)
(b)t is in units of seconds (s)
(c)p is the exponent (a number), and has to be “-2”,

Why? Because (s-2) ≡ 1/s2 (in this case, the symbol “≡” means “defined as” or “is the same as”

13. The irrational number p = 3.14159265358979…..
a. Round this to three significant figures
b. Round it to five sig figs
c. Do it again to 7 s.f.
Solution:
a. Round this to three significant figures p = 3.14 ;
b. Round it to five sig figs, p = 3.1416 ;
c. Do it again to 7 s.f., p = 3.141593 .

19. Peacock Mantis Shrimp can crush the shell of a snail by slamming it at 23 m/s. Find how fast this is in (a) feet per second; (b) miles/hour.
Solution: 1.0 m = 39.37 inches = 3 feet 3.37 inches = 3.281 feet
23 m = 23(3.281) = 75.459 ft, so, 23 m/s = 75.459 ft /s. Or, sig. figs, 75 ft/s.
23 m/s = 75 ft/s = (3600 s)(75 ft/s) = 271,653 ft/hour =(271,653 ft/h) /(5280 ft/mi) = 51.449 mi/h, or, 51 mph.

25. The largest blue whale observed was 108 feet long. Find that in meters.
Solution: 1.0 m = 3.281 ft, so (108 ft)/(3.281 ft/m) = 32.91679366 m = 32.9 m.

29. Woody the Woodpecker can accelerate its beak to 98 m/s2. Express that in feet per square second, ft/s2.
Solution: Dr Dave already knows that 9.8 m/s2 = 32 ft/s2. 320 ft/s2 ? KISMIE?
Or 1.0 m = 3.281 feet, so 98 m = (98)(3.281) = 321.538 ft/s2 = 320 ft/s2.
3.280833333

31. a. A Mutchkin is a Scottish unit of liquid measure, about 0.42 liter. This is not a munchkin, which is a fictional small resident of the Land of Oz.) How many mutchkins will fill a cube that measures 1.0 foot on a side? (Convert this to metric first).
b. A noggin is 0.28 mutchkin. Give the conversion factor between noggins and gallons. (Convert gallons to metric first).
Solution: 1.0 ft3 = 0.028 m3. [This is because 1.0 ft = 12 inches = (12)(2.54) = 30.48 cm = 0.3048 m; so 1.0 ft3 = (0.3048 m)3 = 0.028344726 m3.]
1.0 mutch = 420 cm3 = (0.0748887 m)3. = 0.00042 m3. So, (0.028344726)/(0.00042) = 67.4875074 = 67.
1.0 noggin = 0.28 mutchkin = (0.28)(0.42) = 0.1176 liters; but 3.78 liters = 1.0 gal, so 0.1176 liters = (0.1176 liters)/(3.78 liters/gal) = 0.031 gallons = 1.0 noggin, or, 32.14 noggins per gallon = 32.

39. Antonio just won a $12 million pay out from the local state lottery.
a. If he took all $12 million in quarters, how much mass is that?
b. If he took it all in $1 bills, how heavy, in mass, is that?
Solution: A quarter is 5.67 grams*; a dollar bill is 1.0 g (Wiki answers).
*per http://www.webs1.uidaho.edu/dl2/on_target/mass_of_quarter.htm
a. $12 million = 48 million quarters = (48 x 10^6)(5.67 x 10^-3 kg) = 272.17 x 10^3 = 2.7217 x 10^5 = 2.7E5 kg = 5.98774E5 = 6.0 E5 lbs.

Chapter 2: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and maybe 53, 63, 65, and 71 on pp 48-54.
Take a look at the diagram in the book. Imagine leaving your house and walking east to the Library. When you are finished, you turn around and walk west to the local Park. The distance from your house to the park is 0.75 miles and the Library is another 0.6 miles east of that.
How far did you walk, i.e., total distance from your house to the Library to the Park?
What was your displacement, i.e., net distance from your house to the Park?
Solution: Distance is total amount of length traveled, while displacement is the difference between your starting point and your ending point, so…
From the house to the library = 0.75 + 0.60 = 1.35, plus, from the library to the park = 0.6, for a total of 1.35 + 0.6 = 1.95 mi.
You started at your house and ended at the park, which is 0.75 mi.

3. The golfer, seen in the book, stands 10 meters to the west of the hole, and sinks the ball in two putts. His first putt misses the hole and travels 2.5 meters further east of the hole. On his second putt, the ball travels 2.5 meters west and falls in.
a. How far did the ball travel overall?
b. What was the total displacement of the ball?
Solution: The ball travels past the hole and has to come back, retracing its steps, so to speak
The ball traveled 10 + 2.5 + 2.5 = 15 meters.
The ball started at the golfer and ended up 10 meters away.

9. The Olympic record for the 200 meter dash was 19.75 seconds in 1988. How fast is that in meters per second? Miles per hour?
Solution: Some runner traveled 200 meters in 19.75 seconds. 200 meters is 0.2 kilometers. 1.0 kilometer = 0.6214 mile*.
*http://en.wikipedia.org/wiki/Kilometer
a. (200 m)/(19.75 s) = 10.12658228 m/s, or 10.1 m/s.
b. 10.12658228 m/s = (10.12658228 m/sec)(3600 sec/hour) = 36,455.69621 m/hour = 36.45569621 km/hr, and since 1.0 kilometer = 0.6214 mile, then 36.45569621 km = 22.65356962 mi, so, the answer is 22.65356962mi/hour = 22.7 mi/hr.

13. Radio waves are light waves and thus travel at the speed of light, about 186,000 miles per second. How much time would it take for a radio wave to travel from Earth to the Moon and back? (The Moon is, on average, about 240,000 miles from Earth).
Solution: The distance from Earth to the Moon, and back, is about 480,000 miles. So, the light travels 480,000 miles at 186,000 mi/hr; thus, divide 480,000 by 186,000: (480,000)/(186,000) = (480)/(186) = 2.58 sec.

19. A dog named Fido runs back and forth between Jack and Jill, as seen in the book. However, Jack is walking towards Jill at 1.3 m/s while Jill is walking towards Jack at 1.3 m/s. If Fido begins to run when Jack and Jill are 10.0 meters apart, and if he travels at 3.0 m/s, how far will Fido travel when Jack and Jill crash into each other?
Solution: In order to KISMIE, forget about the dog until later. Then, pretend either Jack or Jill is standing still while the other is moving. Relative to each other, it’s the same as if Jack is moving at 2.6 m/s towards a non-moving Jill. How long would it take Jack to travel 10.0 meters at 2.6 m/s? Divide 10 by 2.6 to get approximately 4 seconds: (10.0)/(2.6) = 3.846153846 sec. Now, let’s go back to the dog. How far does Fido travel in if he travels at a constant speed of 3.0 m/s? x = (3.846153846 sec)(3.0 m/s) = 11.53846154 m. or 12 m.

25. The position of a particle as a function of time is given by this formula: x = (6 m/s)t + (- 2 m/s2)t2.
a. Plot a graph of x vs. t for the time from 0.0 s to 2.0 s.
b. Find the average velocity of the particle from t = 0.0 sec to 1.0 sec.
c. Find the average speed during that same time period.
Solution: In this problem since no direction is given, speed = velocity
See graph somewhere
Find the velocity at 0.0 sec and at 1.0 sec, then add them and divide by 2.0:
v = x/t, so v = (6 m/s) + (-2 m/s2)t ; at t = 0, v = 6 m/s; at t = 1.0 sec, v = 6 m/s + (-2 m/s) = 4 m/s. Average is 6 + 4 divided by 2, or, 10/2 = 5 m/s.

29. The position of a particle as a function of time is given by this formula: x = (2 m/s)t + (- 3 m/s3)t3.
a. Plot a graph of x vs. t for the time from 0.0 s to 1.0 s.
b. Find the average velocity of the particle from t = 0.35 sec to 0.45 sec.
c. Find the average velocity of the particle from t = 0.39 sec to 0.41 sec.
d. Do you expect the instantaneous velocity at t = 0.40 s to be closer to 0.54 m/s, 0.56 m/s, or 0.58 m/s? Explain.
Solution: v = (2 m/s) + (-3m/s2)t2 so…
See the graph somewhere
v(0.35) = (2 m/s) + (-3m/s2)(0.35)2 = (2 m/s) + (-3m/s2)(0.1225) = (2 m/s) +
(-0.3675) = 1.6325; and v(0.45) = (2 m/s) + (-3m/s2)(0.45)2 = (2 m/s) +
(-3m/s2)(0.2025) = (2 m/s) + (-0.6075) = 1.3925. The average is then [½(1.6325 + 1.3925)] = ½ [3.025] = 1.5125 m/s, or 1.5 m/s.
v(0.39) = (2 m/s) + (-3m/s2)(0.39)2 = (2 m/s) + (-3m/s2)(0.1521) = (2 m/s) +
(-0.4563) = 1.5437; and v(0.41) = (2 m/s) + (-3m/s2)(0.41)2 = (2 m/s) +
(-3m/s2)(0.1681) = (2 m/s) + (-0.5043) = 1.4957. The average is then [½(1.5437 + 1.4957)] = ½ [3.0394] = 1.5197 m/s, or 1.5 m/s.
d. TBD

31. Two bows shoot identical arrows with the same launch speed. The accomplish this, the string in bow 1 must be pulled back farther when shooting its arrow than the string in bow 2.
a. Is the acceleration of the arrow shot by bow 1 greater than, less than, or equal to the acceleration of the arrow shot by bow 2?
b. Choose the best explanation from among the following:
I. The arrow in bow 2 accelerates for a great time.
II. Both arrows start from rest.
III. The arrow in bow 1 accelerates for a greater time.
Solution: Both arrows are identical, same size, shape, mass, density. So, no tricks there.
The same
None. The bows are different in tensile strength.

39. Assume that the brakes in your car create a constant deceleration of 4.2 m/s2 regardless of how fast you are driving. If you double your driving speed from 16 m/s to 32 m/s
a. does the time required to stop increase by a factor of two or a factor of four? Explain.
b. Verify your answer by calculating the stopping times for the initial speeds of 16 m/s
c. Verify your answer by calculating the stopping times for the initial speeds of 32 m/s.
Solution: Realize that acceleration equals velocity divided by time, or, a = v/t.
Since a = v/t, then t = v/a, which means that time and velocity are directly related. Thus, double the velocity and you double the time (factor of two).
Okay, I did. Thanks.
Ditto.

53. Approximate 0.1% of the bacteria in the intestine are E coli. These bacteria have been observed to move with speeds of up to 15 m/s (microns per second) and max accelerations of 166 m/s2. Suppose an E coli bacterium in your intestine starts at rest and accelerates at 156 m/s2.
a. How much time is required for the bacterium to reach a speed of 12 m/s?
b. How much distance is required for the bacterium to reach a speed of 12 m/s?
Solution: From problem 39, we know that a = v/t, or, t = v/a.
t = (12 m/s)/(156 m/s2) = 0.076923076 seconds, or = 0.077 sec.
The standard distance equation, when starting from zero (0.0 m) and with an initial velocity of zero (0.0 m/s) is: x = ½ a t2. Since a = (156 m/s2) and the time we just found to be t = 0.076923076 seconds, then x = ½ a t2 =
½ (156 m/s2)(0.076923076 s)2 = ½ (156 m/s2)(0.005917159621 s2) = 0.46153845 m. Or, just 0.46 m.

63. A model rocket rises with constant acceleration to a height of 3.2 m, at which point its speed is 26.0 m/s.
a. How much time does it take for the rocket to reach this height?
b. What was the magnitude of the rocket’s acceleration?
c. Find the height and speed of the rocket 0.10 s after launch.
Solution: The standard height equations, when starting from zero (0.0 m) and with an initial velocity that is non-zero (0.0 m/s) are:

(i) y = y0 + v0t + ½ a t2. And

(ii) v = v0 + a t

We assume the rocket starts from ground level, i.e., y0 = 0.0 meters, and that it starts from rest, i.e., v0 = 0.0 m/s. So…we can rewrite equation (i) above as:

y = y0 + v0t + ½ a t2. Or, y = 0 + 0 + ½ a t2. Which leaves only y = ½ a t2.

We know “y”, but don’t know “a” even though we know that it is a constant acceleration, so we can’t find “t” just from equation (i). So, let’s use equation (ii) to see if it helps:

v = v0 + a t, but since v0 = 0.0 m/s, that leaves us with v = a t. Again, here we know “v” but, again, don’t know “a” even though we know that it is a constant acceleration.

Let’s use both equations.
First, we found that y = ½ a t2. Which we re-write as t = √ [(2y)/a].
Second, we found that v = a t. Which we re-write as t = v/a.

Since they both equal “t”, we can set them equal to find “a” then go back to get “t.”

Or, √ [(2y)/a] = v/a. Now, square both sides to get rid of the square root:

[(2y)/a] = (v/a)2. = (v2)/(a2). Now, multiply both sides by a2 to get

2ya = v2. Now, divide both sides by “2y” to get: a = (v) 2 / (2y) = (26.0 m/s)2/[(2)(3.2 m)] = (676)/(6.4) = 105.625 m/s2.

Now we use v = a t and rewrite it as t = v/a = (26.0)/(105.625) = 0.246153846 seconds, or 0.25 sec.

We found this while doing part (a.), so a =105.625 m/s2 or 106 m/s2. Or 1.06 x 10^2 m/s2.
To find the height, we use equation (i): y = ½ a t2 or, y(0.1) = ½ (105.625)(0.1)2 = ½ (105.625)(0.01) = 0.528125 m, or 0.528 m.
To find its speed, we use equation (ii): v = a t = (105.625)(0.10s) = 10.5625 m/s or 10.6 m/s.

65. A bicyclist named Bob is finishing his repair of a flat tire when a friend named Bill rides by with a constant speed of 3.5 m/s. Two seconds (2.0 s) later Bob hops on his bike and accelerates at 2.4 m/s2 until he catches up with Bill.
a. How much time does it take for Bob to catch up with Bill?
b. How far has Bob traveled in this time?
c.What is Bob’s speed when catches up with Bill?
Solution: The standard distance equations, when starting from zero (0.0 m) and with an initial velocity that is non-zero (0.0 m/s) are:

(i) x = x0 + v0t + ½ a t2. And

(ii) v = v0 + a t

When Bob starts, his initial distance (x0) and velocity (v0) are zero. Thus, these two equations for Bob now are modified to be:

(i) x = ½ a t2. And we know “a” but don’t know “t”, so we can’t find “x” yet, plus

(ii) v = a t. Here, we also know “a” but not “t” so we can’t find “v” yet.

When Bob starts (t0 = 0.0 s), Bill is at a distance of x = v t = (3.5 m/s)(2.0 s) = 7.0 meters from Bob, and at the end, “t,” Bill is at a distance of zero (0.0 m) from Bob. This means that Bob has to travel some distance, “x” to reach Bill, who is also at the same place, “x.”

For Bill, the two equations are a little different:

(x = x0 + v0t + ½ a t2) becomes (x = 0 + v0t + 0 = v0t) because Bill’s initial distance, x0, is the same as Bob’s, or, zero (0.0 m) and Bill’s initial speed is not zero (0.0 m/s) but 3.5 m/s. Plus, since we are told that Bill has a constant velocity, there is NO acceleration, or, for Bill, acceleration is zero (0.0 m/s2). Thus, for Bill:
x = v0t = (3.5 m/s)t

(ii) v = v0 + a t. Which becomes v = v0 since a = 0.0 m/s2.

So, Bob’s “x” and Bill’s “x” are the same:

½ a t2. = v0t, or ½ a t = v0. Thus, t = (2 v0)/a = (2)(3.5)/(2.4) = 7.0/2.4 = 2.917 s or 2.9 sec.
How far has Bob traveled? We use equation (i):

x = x0 + v0t + ½ a t2. = 0.0 m + (0.0 m/s)(2.9 sec) + ½ (2.4)(2.9)2 = 0 + 0 +
½ (2.4)(8.5) = 10.2 meters.

What is Bob’s speed at that time? Use equation (ii):

v = a t = (2.4)(2.9) = 7.0 m/s.


71. Review the image of the car falling off a cliff. The driver says, “Let’s see if we can go from zero to sixty in three seconds.” Prove the driver right or wrong.
Solution: The standard height equations, when starting from zero (0.0 m) and with an initial velocity that is non-zero (0.0 m/s) are:

(i) y = y0 + v0t + ½ a t2. And

(ii) v = v0 + a t

In this question, y0 = 0.0 meters, or, the edge of the cliff; v0=0.0 m/s as it is falling down not zooming down (it is moving to the left, but that’s the x-axis and not relevant). And we assume that “zero to sixty” means from v0= 0.0 m/s (and 0.0 miles per hours) to v = 60.0 miles/hour = 96.6 km/hour = 96,600 meters/hour = 26.8 m/s. We also know that the acceleration here is ONLY gravity, or, a = - 9.8 m/s2. Also, we don’t seem to need equation (i), so we will use only equation (ii). So,

v = v0 + a t = 0 + (g)(t) = (- 9.8 /s2)(3.0 s) = - 29.4 m/s. (it is negative, as it is falling down, not up). So, the statement is false. It cannot go from 0.0 m/s to 60 miles per hour (- 26.8 m/s) in 3.0 seconds of falling. But it comes close!

Chapter 3: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 76-80.

1.Suppose that each component of a certain vector is doubled.
By what multiplicative factor does the magnitude of the vector change?
By what multiplicative factor does the direction angle of the vector change?
Solution:
Double
No change

3.Rank the vectors in figurer 3-31 on page 76 in order of increasing value of their x-component.
Solution: C, B, A, D. Although D will be in the negative direction.

9. A baseball diamond is a square with sides 90 feet (27.4 meters) in length. If the positive x-axis is home plate to first base, and if the positive y-axis points from home plate to third base , find the displacement vector of a base runner who has just hit a
a. double (safely hit the ball and made it to second base)
b. triple (safely hit the ball and made it to third base)
c. home run (safely hit the ball and made it all the way back to home plate)
Solution:
(27.4 m)√2 = 38.7 meters, or (90ft)√2 = 127ft.
27.4 meters = 90 feet
Zero (0.0 m = 0.0 feet)

13. In 1755, John Harrison completed his 4th precision chronometer (clock), known as the “H4”. When the minute hand of the H4 indicated 10 minutes past the hour, it extended 3.0 cm in the horizontal direction.
a. How long was the H4’s minute hand?
b. At 10 minutes past the hour was the extension of the minute hand in the vertical direction more than, less than, or equal to 3.0 cm? Explain.
c. Calculate the vertical extension of the minute hand at 10 minutes past the hour.
Solution: 10 minutes is 1/6 of an hour, or 1/6 of 360°, or 60°.
The x-component is 3.0 cm, we are told, and since in this case, x = h sinθ, where h = the length of the minute hand (the hypotenuse). So, h = (x)/(sinθ) = (3.0 cm)/(sin60°) = (3.0)/(0.866) = 3.46 cm.
Longer.
It will be (3.0 cm)/(cos60) = (3.0)/(0.5) = 6.0 cm.

19. Refer to figure 3-36 on page 78. A refers to a vector.
a. Is the magnitude of A + D greater than, less than, or equal to the magnitude of A + E?
b. Is the magnitude of A + E greater than, less than, or equal to the magnitude of A + F?
Solution: D is in a negative direction, while A, E, and F are all in the positive direction.
the magnitude of A + D is less than, or equal to the magnitude of A + E
the magnitude of A + E equal to the magnitude of A + F? Since E and F are equal in the y-direction.

25. Vector A points in the negative y-direction and has a magnitude of 5 units. Vector B has twice the magnitude and points in the positive x-direction. Find the direction and magnitude of
a. A + B
b. A – B
c. B – A
Solution: A = - 5 ŷ and B = + 10 x
a. A + B will create a vector that will be in Quadrant IV, with a magnitude of √(125) = 11.2 and an angle of 333°.
b. A – B will create a vector that will be in Quadrant III, with a magnitude of √(125) = 11.2 and an angle of 206.8°.
c. B – A will create a vector that will be in Quadrant I with a magnitude of √(125) = 11.2 and an angle of 26.8°.

29. Vector A has a length of 6.1 meters and points in the negative x-direction.
a. Find the x-component of – 3.7 A.
b. Find the magnitude of the vector – 3.7 A.
Solution: Since vector is in the negative x-direction, multiplying it by a negative would make a vector in the positive direction
x-component is (+3.7)(6.1) = 22.57, or 23 meters along the x-axis.
Same

31. Find the direction and magnitude of the vectors. x refers to the unit vector in the x-direction, and ŷ refers to the unit vector in the y-direction.
a. A = (5.0 m)x + (-2.0 m)ŷ
b. B = (-2.0 m)x + (5.0 m)ŷ
c. A + B.
Solution: a. A = (5.0 m)x + (-2.0 m)ŷ will create a vector that will be in Quadrant IV, with a magnitude of √(29) = 5.4 and an angle of 338°.
b. B = (-2.0 m)x + (5.0 m)ŷ will create a vector that will be in Quadrant II, with a magnitude of √(29) = 5.4 and an angle of 112°.
c. A + B. will create a vector, C = (3.0 m)x + (3.0 m)ŷ that will be in Quadrant I, with a magnitude of (3.0) √(2) = 4.2 and an angle of 45°.

39. A neighborhood cat takes 45 minutes to travel 120 meters due north (+ y-direction) and then takes 17 minutes to travel 72 meters due west (- x-direction). Find the magnitude and direction of its average velocity over its 62 minute (3720 sec) jaunt.
Solution: The feline’s foray sketches a right triangle with one side (120 meters)ŷ and the other side (– 72 meters)x. The cat ends up in Quadrant II, thus, it will be more than 90 degrees (North) but less than 180 degrees (West). So, it’s average velocity (or speed) is distance over time, or, (120 + 72)/(3720 sec) = (192 m)/(3720 sec) = 0.0516 m/s. About 5 cm/sec is not too fast. The cat’s displacement will be the hypotenuse of the right triangle made, or, letting the hypotenuse = h, then x2 + y2 = h2. Or, (72)2 + (120)2 = (5184) + (14400) = 19,584 = h2. So, h = √(19,584) = 139.9428455 meters. The cat traveled 192 meters, but its displacement was only 140 meters. And what direction? y = h cosθ here, so cos θ = y/h = (120 m)/(140 m) = 0.857142857, thus the angle, θ, equals the arc-cosine of (0.857142857), or
cos-1(0.857142857) = 31°. So the hypotenuse is 31° to the west of north, but the proper way would be to say how far it is away from the + x-axis, thus we need to add 90° to get 31° + 90° = 121°. The question did NOT ask for the vector, for if it did, then the answer would be 140 m @ 121°.

53. An airplane pilot wants to fly due north but a 65 km/hour wind blowing eastward makes him make changes.
a. In what direction should the pilot head if its relative air speed is 340 km/h?
b. Draw a vector diagram of this
c. If the pilot decreases the plane’s air speed but still wants to fly due north, should the angle found in (a) be increased or decreased?
Solution: If the pilot is going 340 km/h north and there is a west wind (wind from the west, blowing him eastward) of 65 km/h, then he is really traveling √[(340)2 + (65)2] = √[(115600) + (4225)] = √(119825) = 346 km/h at an angle of cos-1(340/346) =
cos-1(0.982658959) = 10.7° east of north (or, 346 km/hr @ 79.3°).
To compensate, he will have to travel 10.7° west of north, or, 346 km/h @ 100.7°.
See a drawing somewhere
Increased to compensate

Chapter 4: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 103-107.

Walking down the street, you toss a ball up into the sky.
If you want the ball to land in your hand when it comes back down, should you toss the ball straight up, forward, or backwards to your movement?
Choose the best explanation
I If the ball is thrown straight up you will leave it behind and it will fall in back of you
II You have to throw the ball in the direction that you are walking
III The ball moves in the forward direction with your walking speed at all times.
Solution:
If you want the ball to land in your hand when it comes back down, you should you toss the ball straight up to your movement.
Choose the best explanation - III The ball moves in the forward direction with your walking speed at all times.

3.While walking to class at a constant speed of 1.75 m/s, you are moving in a direction that is 18.0° north of east. How much time does it take to change your displacement by
a. 20.0 meters east
b. 30.0 meters north
Solution: Essentially, what is the x-component of this velocity vector, and, what is the y-component, then use the distances and speed to find time. The x-component of the speed, vx = (1.75)Cos 18.0° = (1.75)(0.95) = 1.66 m/s; and vy = (1.75) Sin 18.0° = (1.75)(0.31) = 0.54 m/s
vx = x/tx so, tx = x/vx = (20.0 meters)/(1.66 m/s) = 12 sec.
vy = y/ty so, tx = y/vy = (30.0 meters)/(0.54 m/s) = 55 sec.

9. Two kids jump/drop off a cliff overhang into a lake. Johnny drops straight down, while Jimmy runs off the cliff with an initial horizontal speed of v0. (Assume no air friction)
a. Is Jimmy’s splashdown speed larger, the same, or less than Johnny’s splashdown speed? (splashdown speed is the vertical speed that the boys hit the water).
b. Choose the best explanation
I. Both kids are in free fall and hence they will have the same splashdown speed.
II. The kids have the same vertical speed but Jimmy has more horizontal speed
III. Johnny hits faster since he goes straight down.
Solution:
a. Jimmy’s splashdown is the same as Johnny’s splashdown speed.
b. Choose the best explanation - II. The kids have the same vertical speed but Jimmy has more horizontal speed

13. An astronaut exploring the asteroid Ceres tosses a rock horizontally at 6.95 m/s. The rock falls vertically 1.4 meters and lands a horizontal distance of 8.75 meters from the astronaut. Find the acceleration of gravity on Ceres.
Solution: we will use the relationships of v = x/t, and y = ½ g t2. In this case, we are looking for gCeres. First, how much time does it take to go 8.75 meters if the rock travels at 6.95 m/s? It will be x/v = (8.75 m)/(6.95 m/s) = 1.26 sec. What is the g for this asteroid? It will be (2y)/t2 = (2)(1.4)/(1.26)2 = (2.8)/(1.58) = 1.77 m/s2.

19. There is a tradition in Denver, Colorado, in which children bring their old pumpkins from Halloween and toss them off a 9.0-meter high tower, while trying to hit a target that is 3.5 meters from the base of the tower (See figure 4-15 on page 105). Find the horizontal speed needed to hit the target.
Solution: Well, the pumpkin must travel 3.5 meters in the x-direction in the same amount of time it takes to fall 9.0 meters down = y. Denver is on Earth, so we need to use the acceleration to be 9.8 m/s2. But we won’t use a negative, as we are not looking for a vector, and we can’t find a time that is “imaginary” if we took the square root of a negative number. So, we need to find the time, t, and divide x by it to get vx. How long does it take for an object to fall 9.0 meters? We use y = ½ g t2 and re-write it to be t = √(2y/g) = √[(2)(9)/(9.8)] = √(18/9.8) = √(1.8367) = 1.36 sec. So, the pumpkin needs to move 3.5 meters in 1.36 seconds (in the x-direction) to hit the spot, so the horizontal speed is vx = (3.5)/(1.36) = 2.58 m/s.

25. A ball rolls off a table at Sea Level on Earth, and falls 0.75 meters to the floor, landing with a speed of 4.0 m/s. Ignore air friction.
a. What is the acceleration of the ball just before it strikes the ground?
b. What was the initial speed of the ball?
c. What initial speed must the ball have it if is to land with a speed of 5.0 m/s?
Solution: First of all, it is the feet of the table’s legs that are at sea level, so the ball is 0.75 meters above sea level, but the difference in gravitational acceleration due to that is negligible.
The acceleration of any object on Earth, g, near the surface, is 9.8 m/s2.
The initial vertical speed of the ball was zero, vy = 0.0 m/s if it merely rolls off the table. However, for a ball to land at 4.0 m/s after traveling 0.75 meters, it means that we have to use a special equation: vf2 = vi2 + 2 gy, or to find vi, we re-write it to be vi2 = vf2 - 2 gy = [(4)2 – (2)(9.8)(0.75)] = (16 – 14.7) = 1.3. So, if vi2 = 1.3, then vi = √(1.3) = 1.14 m/s. We must conclude that a “fairy” or some other mythical creature propelled it downward the moment it rolled off the table.
If it lands at 5.0 m/s instead of 4.0 m/s, we use the same equations and substitute “5” for “4”: vi2 = vf2 - 2 gy = [(5)2 – (2)(9.8)(0.75)] = (25 – 14.7) = 10.3. So, if vi2 = 10.3, then vi = √(10.3) = 3.2 m/s. Probably from another more powerful fairy.

29. In a baseball game, the second baseman picks up the baseball and tosses it to the first baseman, who catches it at the same level from which is it was thrown. The throw is made with an initial speed of 18.0 m/s at an angle 37.5° from the horizontal.
a. What is the horizontal component of the ball’s velocity just before it’s caught?
b. How long is the ball “in the air”?
Solution: Ignoring air friction, there are no other forces along the x-axis. And, of course, the ball takes as long to reach its maximum height as it does to come back to the original level.
vx is constant, so it’s always (18.0 m/s)Cos(37.5°) = (18.0)(0.793) = 14.3 m/s.
vy is, initially, (18.0)Sin(37.5°) = (18.0)(0.608) = 10.96 m/s, but as it goes up, it slows to, eventually, zero (0.0 m/s) before falling back down. We use the relationship of vf2 = vi2 + 2 gy, where vf = 0.0 m/s at the top. So, now we have the relationship 0 = vi2 + 2 gy, and we can now find “y” the distance it went up before stopping: vi2 = - 2 gy, or y = - (vi2)/(2g) = - [(10.96)2] /[(2)(-9.8)] = - (120)/(- 19.6) = + 6.13 meters. So, the ball goes up 6.13 meters, stops, and goes down 6.13 meters. Since vf = vi + gt we can find how much time it took to get up. We know that vf = 0.0 m/s. So, 0 = vi + gt, and vi = - gt, and t = -(vi/g) = - (10.96)/(-9.8) = + 1.12 sec. Obviously, it takes another 1.12 seconds to come down, or 2.4 sec.

31. A cork shoots out of a champagne bottle at an angle of 35.0° from the horizontal. If the cork travels a horizontal distance 1.30 meters in 1.25 seconds, what was its initial speed?
Solution: If it travels 1.30 meters per 1.25 sec, that means that the x-component of the velocity is vx = (1.30)/(1.25) = 1.04 m/s. Since vx = v0Cos(35.0°), then
v0 = (vx)/[ Cos(35.0°)] = (1.04 m/s)/(0.8192) = 1.27 m/s.

39. The “hang time” of a punt (the kick of the football in American Football) is 4.50 seconds. If the ball were kicked at an angle of 63.0° to the horizontal and was caught at the same level from which it was kicked, what was its initial speed?
Solution: This means it took t = 2.25 seconds to reach its maximum height, y. We are looking for v0. (It then takes another 2.25 seconds to come back down). We know that vx = v0 Cos(63.0°) and vy = v0 Sin (63.0°). So, if we find vy, then we will be able to get v0. We learned from an earlier problem that vy = - gt = - (-9.8)(2.25) = + 22.05 m/s. So, v0 = (vy)/[Sin (63.0°)] = (22.05 m/s)/(0.89) = 24.7 m/s.

53. A soccer ball is kicked with an initial speed 10.2 m/s in a direction 25.0° above the horizontal. Find the magnitude and direction of its velocity
a. 0.250 seconds after being kicked
b. 0.500 seconds after being kicked
c. Is the ball at its greatest height before or after 0.500 second? Explain.
Solution: Using all the same stuff that we already had, we can find out how much time it stays above ground. We have to assume that the ball was kicked from the ground by a midget about 1 micron in size. We learned from before that vy = v0 Sin (25.0°) = (10.2 m/s) Sin (25.0°) = (10.2 m/s)(0.4226) = 4.31 m/s. And remember, since vy = - gt, we find that t = - (vy)/(g) = - (4.31 m/s)/(- 9.8 m/s2) = + 0.44 sec.
Okay, v(0.250sec) = - (g)(0.250 sec) @ 25.0°. = - (- 9.8 m/s2)(0.250 sec) = + 2.45 m/s.
At 0.500 sec, we know that it stopped at 0.44 seconds when it reached the top. So this is like having the ball in free fall for a period of (0.500 – 0.44) sec = 0.06 seconds. And thus, v = gt = (- 9.8)(0.06 sec) = - 0.588 m/s. It’s negative, since it is going down.
We did this already. It reaches its greatest height at 0.44 seconds, or 0.06 seconds before 0.500 sec.

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