Conceptual Physics Review (Chapters 2 & 3) Solutions Sample Calculations 1. My friend and I decide to race down a straight stretch of road. We both get in our cars and start from rest. I hold the steering wheel steady, slam on the gas, and my speedometer shows an increase in speed of 10 meters per second every second. a) What is my acceleration?

!

a ="v"t

=v final # vinitial

t=10m /s# 0m /s

1s=10m /s2

b) If I maintain that constant acceleration, how fast will I be going at the end of 10 seconds?

!

v f = vi + at

v f = 0 + (10m /s2)(10s) =100m /s

c) How far have I traveled in those 10 seconds?

!

d = vavgt, and vavg =vi + v f

2=

0 +100m /s2

= 50m /s

d = (50m /s)(10s) = 500mOR

d = vit +12at 2 = 0 +

12

10m /s2( ) 10s( )2= 500m

2. Grace is driving her sports car at 30 m/s when a ball rolls out into the street in front of her. She slams on the brakes and comes to a stop in 3.0 s. What was the acceleration of Grace’s car? (Use a positive sign if she is speeding up, and a negative sign if she is slowing down)

!

a ="vt

=v f # vit

a =0 # 30m /s

3s= #10m /s2

3. A torpedo fired from a submerged submarine is propelled through the water with a constant speed of 20.00 m/s and explodes upon impact with a target 2000.0 m away. a) How long does it take the torpedo to reach its target?

!

v =dt

, so t =dv

=2000m

20.0m /s=100s

b) If the sound of the impact is heard by the people inside the submarine 101.4 s after the torpedo was fired, what is the speed of sound in water? The torpedo takes 100 seconds to reach its target. Only then does it hit the target, making a sound when it explodes. So, the sound itself travels all the way back to the submarine in the remaining time of 1.4 seconds:

!

tsound = ttotal " ttorpedo =101.4 "100 =1.4s

v =dt

=2000m1.4s

=1429m /s

4. Bobby, a lab, rat has 2.0 minutes to go through a maze to get the yummy cheese. He starts off by meandering 100 cm down a path at a rate of 2.0 cm/s before stopping, sniffing and deciding to go left or right. It takes him 10.0 seconds to decide to turn right. He then scurries down a 50 cm path at a rate of 2.5 cm/s before making a left, speeding up to 3.0 cm/s and racing to the cheese, which is 90 cm away. Does he make it before a trap door crashes down, blocking the cheese? a) Does Bobby get to eat the yummy cheese? (Show your work)

!

speed =distance

time,so t =

distancespeed

tsegment"1 =100cm

2.0cm /s= 50s & tsegment"3 =

50cm2.5cm /s

= 20s

tsegment"4 =90cm

3.0cm /s= 30s

ttotal = 50s+10s+ 20s+ 30s =110s100s

1#

1min60s

=1.8minutes

...so, yes, he gets to eat!

b) Draw a distance vs. time graph of the situation. Label your axes with words, units, and a numbered scale.

5. King Kong carries Fay Wray up the Umpire State Building in New Fork City. At the top of the skyscraper, Fay Wray’s shoe falls from her foot. It hits the ground 15 seconds later. For this problem, ignore air resistance. a) How fast is the shoe going the instant before it hits the ground?

!

v f = vi + at

v f = 0 + ("10m /s2)(15s) = "150m /s

The negative sign indicates that the object is falling down, and is a result of using -10 m/s2 for the acceleration. After all, gravity causes objects to accelerate downward, in the negativedirection.

b) How tall is the Umpire State Building? Well, how far did the shoe fall during the 15s?

d = 12 at

2 = 12 (!10m / s2 )(15s)2 = !1125m

or

vavg =v i+vf

2=

0+!150m / s2

= !75m / s

d = vavgt = (!75m / s)(15s) = !1125mA negative distance is actually called a displacement,which is a vector quantity. This quantity is negative because the object ends up at a position below where it started. Since the question asks how tall the building is, it is appropriate to give your answer as a positive number, 1125 meters.

6. The Steamboat Geyser in Jellystone National Park is capable of shooting its hot water up from the ground with a speed of 50 m/s. How high can the geyser shoot? Ignore air resistance.

We can figure this out using

!

d = 12 at

2 if we can figure out for what length of time the water is in the air. Conceptually, you should be able to see that water with an initial velocity of 50 m/s will slow to 0m/s in 5s, because the acceleration is -10m/s2 when it’s shooting through the air. Or:

!

a =v f " vit

,so t =v f " via

=0 " 50m /s"10m /s2 = 5s

Now, how far does the water travel during those 5s?

!

d = 12 at

2 = 12 ("10m /s2)(5s)2 =125m, or

d = vavgt = (25m /s)(5s) =125m

7. At Six Flags Great Adventure Amusement Park in New Jersey, a popular ride known as “Free Fall” carries passengers up to a height of 33.5 m and drops them to the ground inside a small cage. How fast are the passengers going at the bottom of this exhilarating journey? Ignore air resistance. To get velocity, we’ll need to know for how long the cage falls, so we have to get time first:

!

d = 12 at

2, which rearranges to become t =2da

t =2("33.5m)"10m /s2 = 2.58s

Now:

!

v f = vi + at = 0 + ("10m /s2)(2.58s) = "25.8m /s 8. A pop fly is hit straight up at an initial speed of 40 m/s. Ignore air resistance. a) How long is it in the air if it is caught at the same height from which it was hit?

There are lots of different strategies for solving this. Conceptually, we know that it will take 4 seconds to slow down from 40m/s to 0m/s, at -10m/s2, and the same amount of time to come down, so that gives us 8 seconds. You can calculate this using a formula as well:

!

t =v f " via

=0 " 40m /s"10m /s2 = 4s (for going up)# 2 = 8s

b) How high is it 2 seconds after it is hit? Velocity after 2 seconds of going up is 20m/s, so...

!

d = vavgt

d =vi + v f2

"

# $

%

& ' t =

40 + 202

"

# $

%

& ' 2 = 60m

c) How high does it go?

!

d = vavgt

d =vi + v f2

"

# $

%

& ' t =

40 + 02

"

# $

%

& ' 4 = 80m

d) What is its velocity 6 seconds after it was hit?

Conceptually, we know that the ball is falling back down from its maximum height at time t=4 s. Two seconds later, it has been accelerating down at 10m/s2, so it must be going -20 m/s. Mathematically:

!

v f = vi + at = 0 + ("10m /s2)(2s) = "20m /s e) How high above its starting point is it 6 seconds after it was hit?

Well, how far has it fallen from its maximum height of 80m during those 2 seconds?

!

d = 12 at

2 = 12 ("10m /s2)(2s)2 = "20m

80m " 20m = 60m above the ground

9. Write a description of an object whose motion is represented by the following graph. Be sure to give a chronological description, starting with what is happening at time = 0. Use appropriate physics vocabulary such as velocity and acceleration.

The object begins at a location 0.5m and starts traveling in the positive direction (to the right). From 0 to 1.5 seconds, the object is speeding up (increasing velocity, or accelerating positively here) as it moves in the positive direction, and then from 1.5 to 2.5 seconds it appears to be traveling at a relatively constant velocity. Then, from 2.5 to 4 seconds it begins to slow down (acceleration in the negative direction), still traveling in the positive direction but with a lesser speed until it comes to a halt at location 3.7 meters.

10. Write a description of an object whose motion is represented by the following graph. Be sure to give a chronological description, starting with what is happening at time = 0. Use appropriate physics vocabulary such as velocity and acceleration.

The object here begins at some location with a positive velocity, but that velocity is decreasing over time (=acceleration, negative in this case). The object continues to lose velocity until, at time t=5 seconds, it comes to halt for an instant, and then begins moving again with a negative velocity, traveling back the direction that it originally came from. The object has a constant acceleration, which is to say its velocity is constantly changing. Although the acceleration of the object in this graph is -

5.0m/s2, if it were -10 m/s2, the shape of this graph would be consistent with a ball that was thrown up into the air with an initial velocity of 20m/s. That ball would decrease its velocity until it reached an instantaneous velocity of 0 at the top of its path, before starting to fall down with increasing speed (=increasingly negative velocity). 11. Rochelle is flying to New York for her big Broadway debut. If the plane heads out of Los Angeles with a velocity of 220.0 m/s in a northeast direction, relative to the ground, and encounters a wind blowing head-on at 45 m/s, what is the resultant velocity of the plane, relative to the ground? Draw a picture of the situation. 220.0m/s – 45m/s = 175m/s northeast

0

0.5

1

1.5

2

2.5

3

3.5

4

0 1 2 3 4 5

time t (seconds)

po

sit

ion

x (

mete

rs)

velocity vs. time

-40

-30

-20

-10

0

10

20

30

0 2 4 6 8 10 12

Time (seconds)

Velo

cit

y (

mete

rs/

seco

nd

)

12. In her physics lab, Melanie rolls a 10-g marble down a ramp and off the table with a horizontal velocity of 1.2 m/s. The marble falls into a cup placed on the floor 0.51 m from the bottom of the table. How high is the table? Draw a picture of the situation. Horizontal d = vt

!

t =dv

=0.51m1.2m /s

= 0.43s

Vertical d = 1/2at2 = ½ (9.8m/s2) (0.43s)2 = 0.91 m 13. Bert is standing on a ladder picking apples in his grandfather’s orchard. As he pulls each apple off the tree, he tosses it into a basket that sits on the ground 3.0 m below at a horizontal distance of 2.0 m from Bert. How fast must Bert throw the apples (horizontally) in order for them to land in the basket? Draw a picture of the situation. Vertical: d = 1/2at2 3.0m = ½(9.8m/s2) t2

0.61s2 = t2

Don’t forget to take the square root of both sides!!! 0.78s = t Horizontal: d = vt

!

v =dt

=2.0m0.78sec

= 2.56m /sec

14. A cannon is fired up from the ground at an angle of 53° from the horizontal, with a velocity of 100 m/s. Draw a picture of the cannon and label the angle of launch and draw the velocity vector, in the direction the cannonball will be fired. a. On your diagram above, draw and label the component vectors (horizontal and vertical) for the initial velocity of a cannonball to be fired from the cannon. Hint: Based on the angle of launch, you can infer that the right triangle formed by the two component vectors and the resultant is a 3-4-5 right triangle. This is a 3,4,5 right triangle, so vx=60m/s and vy =80m/s b. How long will it take the cannonball, after it is fired, to reach its highest point?

vf = vi + gt 0 = 80m/s + (-10m/s2) t t = 8 sec

c. How long after being fired will the cannonball hit the ground?

In the absence of air resistance, it takes an equal amount of time to go up and come down, so total time in the air = 2(8 sec) = 16 sec.

d. How high above the ground is the cannonball at its highest point? Analyzing from the top down: d =1/2 gt2 d = ½(-10m/s2) (8sec)2 d = -320 m Or, the ball fell 320 m, so the highest point must be 320 meters above the ground. e. How far away from the cannon does the cannonball land (horizontal distance)? dx = vxt = (60m/s) (16 sec) = 960 m f. In the space below, draw the cannonball at two-second intervals and draw the horizontal and vertical components of its velocity (vectors) at each two-second interval.