1. Problems – 1
A roller coaster moves 85 m horizontally, then travels 45 m at an angle of 30.0° above the horizontal. What is its displacement from its starting point? Use graphical techniques (i.e. draw a scale diagram).
10°
30°

2. Problems – 2
A bird dog on the trail of a grouse walks 65 m along a heading of 75°, then turns to walk 48 m along a heading of 27°. Use graphical techniques to find the resultant displacement of the dog.
Dx1 = 65 75o (15° North of East)
Dx2 = 48 27o (27° East of North)
Dxr = ?
Scale: 1 cm = 10 m
Dx2 (4.8 cm) = 48 27°
90o
270o 0o
180o
Dxr (11 cm) 0o
90o 0o
35°
Dx1 (6.5 cm) = 65 75°
Dxr = 110 m
@ 35o North of East

3. Problems – 3
A novice pilot sets a plane’s controls, thinking the plane will fly at 250 km/h to the north. If the wind blows at 75 km/h toward the southeast what is the plane’s resultant velocity? Use graphical techniques.
45°
15°

4. Problems – 4
While flying over the Grand Canyon, the pilot slows the plane down to one-half the velocity in item 3. If the wind’s velocity is still 75 km/h toward the southeast, what will the plane’s new resultant velocity be?
45°
36°

5. Problems – 5
A dog searching for a bone walks 3.50 m south, then 8.20 m at an angle of 30.0 north of east, and finally 15.0 m west. Use graphical techniques to find the dog’s resultant displacement vector.
4.3°
30°

6. Problems – 6
While following the directions on a treasure map, a pirate walks 45.0 m north and then turns and walks 7.5 m east. What single straight-line displacement could the pirate have taken to reach the treasure? Use graphical techniques.
10°

7. Problems – 7
While following the directions on a treasure map, a pirate walks 45.0 m north and then turns and walks 7.5 m east. What single straight-line displacement could the pirate have taken to reach the treasure? q

9. Problems – 9

10. Problems - 10

11. Problems - 11

12. Problems - 12

13. Problems – 14

14. Problems – 17
A roller coaster moves 85 m horizontally, then travels 45 m at an angle of 30.0° above the horizontal. What is its displacement from its starting point? Use algebraic methods.
30°

15. Problems – 18
A novice pilot sets a plane’s controls, thinking the plane will fly at 250 km/h to the north. If the wind blows at 75 km/h toward the southeast what is the plane’s resultant velocity? Use algebraic methods.
45°

16. Problems – 19
A football player runs directly down the field for 35 m before turning to the right at an angle of 25° from his original direction and running an additional 15 m before getting tackled. Use algebraic methods to find the magnitude and direction of the runner’s total displacement.
25°

17. Projectiles Launched Horizontally
The initial velocity is the initial horizontal velocity.
The initial vertical velocity is 0 (vy,i = 0).
Horizontal Component
(Constant Velocity)
vx = vx,i = constant
Dx = vx · Dt
Vertical Component (Constant Acceleration)
ay = -g = m/s2
vy,f = ay(Dt)
Dy = ½ay(Dt)2
vy,f2 = 2ay(Dy)
There is only 1 equation necessary to address the horizontal component of motion because there is no acceleration in that direction.
The vertical component is addressed by the uniform acceleration equations derived in the last chapter, revised for the case where ay = -g = m/s2 and vy,i = 0.
A system of equations is created using the horizontal equation in combination with one of the vertical equations. Solve the system for the unknown and substitute values only at the end.

18. Problems - 22

19. Problems - 23

20. Problems - 24

21. Problems - 26

22. Problems - 27

23. Problems – 28

24. Problems – 29

25. Problems – 30

26. Projectiles Launched at an Angle
Resolve the initial velocity into x and y components.
The initial horizontal velocity is the x-component.
The initial vertical velocity is the y- component.
Horizontal (Constant Velocity)
vx = vx,i = vicosq = constant
Dx = vicosq · Dt
Vertical (Constant Acceleration)
ay = -g = m/s2
vy,f = visinq + ay(Dt)
Dy = (visinq)Dt +½ay(Dt)2
vy,f2 = vi2(sinq)2 + 2ay(Dy)
Dy = ½(visinq + vy,f )Dt (not included – not necessary)
These equations are on the bottom of page 99.

27. Problems – 31

28. Problems – 32

29. Problems – 33
Dy = 0.55 m
Dx = 2.0 m
vi = ? q

30. Problems – 34
The author left this equation out of the list. The problem can be done with the others, but it is much easier with this one.

31. Problems – 35 1.
Yay !! He Makes it!!

32. Problems – 36
Yeah! He makes it by 4m-3.05m, or 1m!

33. Practice Problems
35.
Since Vy,f is negative, ball is on its way down.