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Agenda / August 25, 2008
• Problem of the Day
• Chapter 3
– 3.1 & 3.2 Two dimensional kinematics
• Practice Problems 6, 9, 12, 14, 15
Problem of the Day
• A football player starts from rest 10 meters
from the goal line and accelerates away
from the goal line at 5 m/s2
. How far from
the goal line is the player after 4 seconds?
Kinematics in Two Dimensions
Chapter 3
3.2 Equations of Kinematics in Two Dimensions
Equations of Kinematics
( ) tvvx o += 2
1
2
2
1
attvx o +=
atvv o +=
axvv o 222
+=
3.2 Equations of Kinematics in Two Dimensions
tavv xoxx += ( ) tvvx xox += 2
1
xavv xoxx 222
+=2
2
1
tatvx xox +=
3.2 Equations of Kinematics in Two Dimensions
tavv yoyy +=
2
2
1
tatvy yoy +=
( )tvvy yoy += 2
1
yavv yoyy 222
+=
3.2 Equations of Kinematics in Two Dimensions
The x part of the motion occurs exactly as it would if the
y part did not occur at all, and vice versa.
3.2 Equations of Kinematics in Two Dimensions
Example 1 A Moving Spacecraft
In the x direction, the spacecraft has an initial velocity component
of +22 m/s and an acceleration of +24 m/s2
. In the y direction, the
analogous quantities are +14 m/s and an acceleration of +12 m/s2
.
Find (a) x and vx, (b) y and vy, and (c) the final velocity of the
spacecraft at time 7.0 s.
3.2 Equations of Kinematics in Two Dimensions
Reasoning Strategy
1. Make a drawing.
2. Decide which directions are to be called positive (+) and
negative (-).
3. Write down the values that are given for any of the five
kinematic variables associated with each direction.
4. Verify that the information contains values for at least three
of the kinematic variables. Do this for x and y. Select the
appropriate equation.
5. When the motion is divided into segments, remember that
the final velocity of one segment is the initial velocity for the next.
6. Keep in mind that there may be two possible answers to a
kinematics problem.
3.2 Equations of Kinematics in Two Dimensions
Example 1 A Moving Spacecraft
In the x direction, the spacecraft has an initial velocity component
of +22 m/s and an acceleration of +24 m/s2
. In the y direction, the
analogous quantities are +14 m/s and an acceleration of +12 m/s2
.
Find (a) x and vx, (b) y and vy, and (c) the final velocity of the
spacecraft at time 7.0 s.
x ax vx vox t
? +24.0 m/s2
? +22 m/s 7.0 s
y ay vy voy t
? +12.0 m/s2
? +14 m/s 7.0 s
3.2 Equations of Kinematics in Two Dimensions
x ax vx vox t
? +24.0 m/s2
? +22 m/s 7.0 s
( )( ) ( )( ) m740s0.7sm24s0.7sm22
22
2
1
2
2
1
+=+=
+= tatvx xox
( ) ( )( ) sm190s0.7sm24sm22 2
+=+=
+= tavv xoxx
3.2 Equations of Kinematics in Two Dimensions
y ay vy voy t
? +12.0 m/s2
? +14 m/s 7.0 s
( )( ) ( )( ) m390s0.7sm12s0.7sm14
22
2
1
2
2
1
+=+=
+= tatvy yoy
( ) ( )( ) sm98s0.7sm12sm14 2
+=+=
+= tavv yoyy
3.2 Equations of Kinematics in Two Dimensions
v
sm98=yv
sm190=xv
( ) ( ) sm210sm98sm190
22
=+=v
θ
( ) 
2719098tan 1
== −
θ
3.2 Equations of Kinematics in Two Dimensions
3.3 Projectile Motion
Under the influence of gravity alone, an object near the
surface of the Earth will accelerate downwards at 9.80m/s2
.
2
sm80.9−=ya 0=xa
constant== oxx vv
3.3 Projectile Motion
Example 3 A Falling Care Package
The airplane is moving horizontally with a constant velocity of
+115 m/s at an altitude of 1050m. Determine the time required
for the care package to hit the ground.
3.3 Projectile Motion
y ay vy voy t
-1050 m -9.80 m/s2
0 m/s ?
3.3 Projectile Motion
y ay vy voy t
-1050 m -9.80 m/s2
0 m/s ?
2
2
1
tatvy yoy += 2
2
1
tay y=
( ) s6.14
sm9.80
m105022
2
=
−
−
==
ya
y
t
3.3 Projectile Motion
Example 4 The Velocity of the Care Package
What are the magnitude and direction of the final velocity of
the care package?
3.3 Projectile Motion
y ay vy voy t
-1050 m -9.80 m/s2
? 0 m/s 14.6 s
3.3 Projectile Motion
y ay vy voy t
-1050 m -9.80 m/s2
? 0 m/s 14.6 s
( )( )
sm143
s6.14sm80.90 2
−=
−+=+= tavv yoyy
3.3 Projectile Motion
Conceptual Example 5 I Shot a Bullet into the Air...
Suppose you are driving a convertible with the top down.
The car is moving to the right at constant velocity. You point
a rifle straight up into the air and fire it. In the absence of air
resistance, where would the bullet land – behind you, ahead
of you, or in the barrel of the rifle?
3.3 Projectile Motion
Example 6 The Height of a Kickoff
A placekicker kicks a football at and angle of 40.0 degrees and
the initial speed of the ball is 22 m/s. Ignoring air resistance,
determine the maximum height that the ball attains.
3.3 Projectile Motion
ov
oxv
oyv
θ
( ) sm1440sinsm22sin === 
θooy vv
( ) sm1740cossm22cos === 
θoox vv
3.3 Projectile Motion
y ay vy voy t
? -9.80 m/s2
0 14 m/s
3.3 Projectile Motion
y ay vy voy t
? -9.80 m/s2
0 14 m/s
yavv yoyy 222
+=
y
oyy
a
vv
y
2
22
−
=
( )
( ) m10
sm8.92
sm140
2
2
+=
−
−
=y
3.3 Projectile Motion
Example 7 The Time of Flight of a Kickoff
What is the time of flight between kickoff and landing?
3.3 Projectile Motion
y ay vy voy t
0 -9.80 m/s2
14 m/s ?
3.3 Projectile Motion
y ay vy voy t
0 -9.80 m/s2
14 m/s ?
2
2
1
tatvy yoy +=
( ) ( ) 22
2
1
sm80.9sm140 tt −+=
( ) ( )t2
sm80.9sm1420 −+=
s9.2=t
3.3 Projectile Motion
Example 8 The Range of a Kickoff
Calculate the range R of the projectile.
( )( ) m49s9.2sm17
2
2
1
+==
=+= tvtatvx oxxox
3.3 Projectile Motion
Conceptual Example 10 Two Ways to Throw a Stone
From the top of a cliff, a person throws two stones. The stones
have identical initial speeds, but stone 1 is thrown downward
at some angle above the horizontal and stone 2 is thrown at
the same angle below the horizontal. Neglecting air resistance,
which stone, if either, strikes the water with greater velocity?
3.4 Relative Velocity
TGPTPG vvv

+=
3.4 Relative Velocity
Example 11 Crossing a River
The engine of a boat drives it across a river that is 1800m wide.
The velocity of the boat relative to the water is 4.0m/s directed
perpendicular to the current. The velocity of the water relative
to the shore is 2.0m/s.
(a) What is the velocity of the
boat relative to the shore?
(b) How long does it take for
the boat to cross the river?
3.4 Relative Velocity
( ) ( )
sm5.4
sm0.2sm0.4
2222
=
+=+= WSBWBS vvv
WSBWBS vvv

+=

63
0.2
0.4
tan 1
=





= −
θ
3.4 Relative Velocity
s450
sm4.0
m1800
==t

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Cutnellpptch03 1219726892727682-8

  • 1. Agenda / August 25, 2008 • Problem of the Day • Chapter 3 – 3.1 & 3.2 Two dimensional kinematics • Practice Problems 6, 9, 12, 14, 15
  • 2. Problem of the Day • A football player starts from rest 10 meters from the goal line and accelerates away from the goal line at 5 m/s2 . How far from the goal line is the player after 4 seconds?
  • 3. Kinematics in Two Dimensions Chapter 3
  • 4. 3.2 Equations of Kinematics in Two Dimensions Equations of Kinematics ( ) tvvx o += 2 1 2 2 1 attvx o += atvv o += axvv o 222 +=
  • 5. 3.2 Equations of Kinematics in Two Dimensions tavv xoxx += ( ) tvvx xox += 2 1 xavv xoxx 222 +=2 2 1 tatvx xox +=
  • 6. 3.2 Equations of Kinematics in Two Dimensions tavv yoyy += 2 2 1 tatvy yoy += ( )tvvy yoy += 2 1 yavv yoyy 222 +=
  • 7. 3.2 Equations of Kinematics in Two Dimensions The x part of the motion occurs exactly as it would if the y part did not occur at all, and vice versa.
  • 8. 3.2 Equations of Kinematics in Two Dimensions Example 1 A Moving Spacecraft In the x direction, the spacecraft has an initial velocity component of +22 m/s and an acceleration of +24 m/s2 . In the y direction, the analogous quantities are +14 m/s and an acceleration of +12 m/s2 . Find (a) x and vx, (b) y and vy, and (c) the final velocity of the spacecraft at time 7.0 s.
  • 9. 3.2 Equations of Kinematics in Two Dimensions Reasoning Strategy 1. Make a drawing. 2. Decide which directions are to be called positive (+) and negative (-). 3. Write down the values that are given for any of the five kinematic variables associated with each direction. 4. Verify that the information contains values for at least three of the kinematic variables. Do this for x and y. Select the appropriate equation. 5. When the motion is divided into segments, remember that the final velocity of one segment is the initial velocity for the next. 6. Keep in mind that there may be two possible answers to a kinematics problem.
  • 10. 3.2 Equations of Kinematics in Two Dimensions Example 1 A Moving Spacecraft In the x direction, the spacecraft has an initial velocity component of +22 m/s and an acceleration of +24 m/s2 . In the y direction, the analogous quantities are +14 m/s and an acceleration of +12 m/s2 . Find (a) x and vx, (b) y and vy, and (c) the final velocity of the spacecraft at time 7.0 s. x ax vx vox t ? +24.0 m/s2 ? +22 m/s 7.0 s y ay vy voy t ? +12.0 m/s2 ? +14 m/s 7.0 s
  • 11. 3.2 Equations of Kinematics in Two Dimensions x ax vx vox t ? +24.0 m/s2 ? +22 m/s 7.0 s ( )( ) ( )( ) m740s0.7sm24s0.7sm22 22 2 1 2 2 1 +=+= += tatvx xox ( ) ( )( ) sm190s0.7sm24sm22 2 +=+= += tavv xoxx
  • 12. 3.2 Equations of Kinematics in Two Dimensions y ay vy voy t ? +12.0 m/s2 ? +14 m/s 7.0 s ( )( ) ( )( ) m390s0.7sm12s0.7sm14 22 2 1 2 2 1 +=+= += tatvy yoy ( ) ( )( ) sm98s0.7sm12sm14 2 +=+= += tavv yoyy
  • 13. 3.2 Equations of Kinematics in Two Dimensions v sm98=yv sm190=xv ( ) ( ) sm210sm98sm190 22 =+=v θ ( )  2719098tan 1 == − θ
  • 14. 3.2 Equations of Kinematics in Two Dimensions
  • 15. 3.3 Projectile Motion Under the influence of gravity alone, an object near the surface of the Earth will accelerate downwards at 9.80m/s2 . 2 sm80.9−=ya 0=xa constant== oxx vv
  • 16. 3.3 Projectile Motion Example 3 A Falling Care Package The airplane is moving horizontally with a constant velocity of +115 m/s at an altitude of 1050m. Determine the time required for the care package to hit the ground.
  • 17. 3.3 Projectile Motion y ay vy voy t -1050 m -9.80 m/s2 0 m/s ?
  • 18. 3.3 Projectile Motion y ay vy voy t -1050 m -9.80 m/s2 0 m/s ? 2 2 1 tatvy yoy += 2 2 1 tay y= ( ) s6.14 sm9.80 m105022 2 = − − == ya y t
  • 19. 3.3 Projectile Motion Example 4 The Velocity of the Care Package What are the magnitude and direction of the final velocity of the care package?
  • 20. 3.3 Projectile Motion y ay vy voy t -1050 m -9.80 m/s2 ? 0 m/s 14.6 s
  • 21. 3.3 Projectile Motion y ay vy voy t -1050 m -9.80 m/s2 ? 0 m/s 14.6 s ( )( ) sm143 s6.14sm80.90 2 −= −+=+= tavv yoyy
  • 22. 3.3 Projectile Motion Conceptual Example 5 I Shot a Bullet into the Air... Suppose you are driving a convertible with the top down. The car is moving to the right at constant velocity. You point a rifle straight up into the air and fire it. In the absence of air resistance, where would the bullet land – behind you, ahead of you, or in the barrel of the rifle?
  • 23. 3.3 Projectile Motion Example 6 The Height of a Kickoff A placekicker kicks a football at and angle of 40.0 degrees and the initial speed of the ball is 22 m/s. Ignoring air resistance, determine the maximum height that the ball attains.
  • 24. 3.3 Projectile Motion ov oxv oyv θ ( ) sm1440sinsm22sin ===  θooy vv ( ) sm1740cossm22cos ===  θoox vv
  • 25. 3.3 Projectile Motion y ay vy voy t ? -9.80 m/s2 0 14 m/s
  • 26. 3.3 Projectile Motion y ay vy voy t ? -9.80 m/s2 0 14 m/s yavv yoyy 222 += y oyy a vv y 2 22 − = ( ) ( ) m10 sm8.92 sm140 2 2 += − − =y
  • 27. 3.3 Projectile Motion Example 7 The Time of Flight of a Kickoff What is the time of flight between kickoff and landing?
  • 28. 3.3 Projectile Motion y ay vy voy t 0 -9.80 m/s2 14 m/s ?
  • 29. 3.3 Projectile Motion y ay vy voy t 0 -9.80 m/s2 14 m/s ? 2 2 1 tatvy yoy += ( ) ( ) 22 2 1 sm80.9sm140 tt −+= ( ) ( )t2 sm80.9sm1420 −+= s9.2=t
  • 30. 3.3 Projectile Motion Example 8 The Range of a Kickoff Calculate the range R of the projectile. ( )( ) m49s9.2sm17 2 2 1 +== =+= tvtatvx oxxox
  • 31. 3.3 Projectile Motion Conceptual Example 10 Two Ways to Throw a Stone From the top of a cliff, a person throws two stones. The stones have identical initial speeds, but stone 1 is thrown downward at some angle above the horizontal and stone 2 is thrown at the same angle below the horizontal. Neglecting air resistance, which stone, if either, strikes the water with greater velocity?
  • 32. 3.4 Relative Velocity TGPTPG vvv  +=
  • 33. 3.4 Relative Velocity Example 11 Crossing a River The engine of a boat drives it across a river that is 1800m wide. The velocity of the boat relative to the water is 4.0m/s directed perpendicular to the current. The velocity of the water relative to the shore is 2.0m/s. (a) What is the velocity of the boat relative to the shore? (b) How long does it take for the boat to cross the river?
  • 34. 3.4 Relative Velocity ( ) ( ) sm5.4 sm0.2sm0.4 2222 = +=+= WSBWBS vvv WSBWBS vvv  +=  63 0.2 0.4 tan 1 =      = − θ