Solution Set 1

PHYSICS
SOLUTION SET 1

1. The movie Spiderman brought in $114,000,000 during its opening weekend. Express this amount in scientific notation.
Solution:
I need to pick a number between 1 and 10 from the large number above, so I pick “1.14.” So, What would I have to multiply 1.14 by to get 114,000,000? I think 10^8. So the answer is 1.14 x 10^8.
Solution:
2. The speed of light is 299,792,458 m/s. Express that in scientific notation, and round to 3 significant figures.
Solution:
If I merely lopped off this number with the first three numbers, I'd have 299,000,000. But since the numbers after “299” are greater than 50% of the next number up, I have to round up to 299,800,000. But that's 4 significant figures. So, it must be 300,000,000. And in scientific notation, that is 3.00 x 10^8 m/s.

3. 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”

4. The irrational number p = 3.14159265358979…..Round this to seven significant figures
Solution:
To lop off the first seven digits would make it 3.141592, but since the next number is past 5, then we have to round off to 3.141593.

5. The largest blue whale observed was 108 feet long. Find that in meters.
Solution:
Since 3 feet = 1.0 yard, and 1 yard almost equals 1.0 meter, then a good answer would be 108 divided by 3, or, 36 (approx)

6. Woody the Woodpecker can accelerate its beak to 98 m/s2. Express that in feet per square second, ft/s2.
Solution:
The acceleration of gravity is 9.8 m/sec^2, or, 32 ft/sec^2. This number is ten times that, so the answer is 320 ft/sec^2.

7. 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. A quarter has a mass of about 5.6 grams. $12 million in quarters is 48,000,000 quarters and if we multiply that by 5.6 grams, we get (48 million x 5.6 grams) = 1,537,708,800 grams = 1.538 x 10^6 kg.
b. A dollar bill is about 1.0 gram, so, 12 million grams, or, 12,000 kg.

8. Take a look at the diagram (not) below. 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.
a. How far did you walk, i.e., total distanced from your house to the Library to the Park?
b. 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…
a. 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.
b. You started at your house and ended at the park, which is 0.75 mi.

9. The golfer, not seen below, 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
a. The ball traveled 10 + 2.5 + 2.5 = 15 meters.
b. The ball started at the golfer and ended up 10 meters away.

10. 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.

11. 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.

12. A dog named Fido runs back and forth between Jack and Jill, as not seen in the diagram below. 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.

13. 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.
a. 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).
b. Okay, I did. Thanks.
c. Ditto.

14. 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.
a. t = (12 m/s)/(156 m/s2) = 0.076923076 seconds, or = 0.077 sec.
b. 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.

15. 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

a. 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.

b. 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.
c. 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.

16. 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) 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

a. 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:

(i) (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.
b. 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.

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

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

17. 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!
PHYSICS
SOLUTION SET 1

1. The movie Spiderman brought in $114,000,000 during its opening weekend. Express this amount in scientific notation.
Solution:
I need to pick a number between 1 and 10 from the large number above, so I pick “1.14.” So, What would I have to multiply 1.14 by to get 114,000,000? I think 10^8. So the answer is 1.14 x 10^8.
Solution:
2. The speed of light is 299,792,458 m/s. Express that in scientific notation, and round to 3 significant figures.
Solution:
If I merely lopped off this number with the first three numbers, I'd have 299,000,000. But since the numbers after “299” are greater than 50% of the next number up, I have to round up to 299,800,000. But that's 4 significant figures. So, it must be 300,000,000. And in scientific notation, that is 3.00 x 10^8 m/s.

3. 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”

4. The irrational number p = 3.14159265358979…..Round this to seven significant figures
Solution:
To lop off the first seven digits would make it 3.141592, but since the next number is past 5, then we have to round off to 3.141593.

5. The largest blue whale observed was 108 feet long. Find that in meters.
Solution:
Since 3 feet = 1.0 yard, and 1 yard almost equals 1.0 meter, then a good answer would be 108 divided by 3, or, 36 (approx)

6. Woody the Woodpecker can accelerate its beak to 98 m/s2. Express that in feet per square second, ft/s2.
Solution:
The acceleration of gravity is 9.8 m/sec^2, or, 32 ft/sec^2. This number is ten times that, so the answer is 320 ft/sec^2.

7. 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. A quarter has a mass of about 5.6 grams. $12 million in quarters is 48,000,000 quarters and if we multiply that by 5.6 grams, we get (48 million x 5.6 grams) = 1,537,708,800 grams = 1.538 x 10^6 kg.
b. A dollar bill is about 1.0 gram, so, 12 million grams, or, 12,000 kg.

8. Take a look at the diagram (not) below. 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.
a. How far did you walk, i.e., total distanced from your house to the Library to the Park?
b. 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…
a. 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.
b. You started at your house and ended at the park, which is 0.75 mi.

9. The golfer, not seen below, 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
a. The ball traveled 10 + 2.5 + 2.5 = 15 meters.
b. The ball started at the golfer and ended up 10 meters away.

10. 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.

11. 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.

12. A dog named Fido runs back and forth between Jack and Jill, as not seen in the diagram below. 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.

13. 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.
a. 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).
b. Okay, I did. Thanks.
c. Ditto.

14. 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.
a. t = (12 m/s)/(156 m/s2) = 0.076923076 seconds, or = 0.077 sec.
b. 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.

15. 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

a. 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.

b. 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.
c. 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.

16. 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) 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

a. 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:

(i) (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.
b. 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.

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

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

17. 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!