A whole new class of problems raises its ugly head when we consider two simultaneous motions of a particle. We have had some contact with this already, but it is most graphically illustrated by problems of boats traveling across streams. Before we launch into this, however, we need to return for a moment to the contemplation of vectors.

In physics a vector is any quantity described by both a magnitude and a direction, while a scalar needs only magnitude. Thus quantities like velocity, displacement, electric field, and momentum are vectors, quantities like pressure, speed, energy, and time are scalars.

Vectors add in a peculiar but logical way. Can you think of how a person could walk 3.0 miles, stop for lunch, walk 4.0 more miles and still be only 5.0 miles from her starting point? Here's how it's done:

Travel straight for 3.0 miles, make a turn to the left, and walk for 4 more miles.

You may prefer to think of it this way.

We will deal almost exclusively with vectors that are at right angles to each other, but it is not difficult to see how more general addition can be accomplished. Just end up with little parallelograms instead of little rectangles. To subtract vectors, simply add the negative of the subtracted vector.

23. Note the direction and magnitude of each of the vectors A, B, C, and D.

a)

b)

c)

d)

24. Kent Dewitt possesses a small motorized craft that can travel at 2.1 m/sec in still water. He heads his boat directly away from shore on a stream traveling at 1.3 m/sec, and is therefore swept downstream. At what angle to his intended direction does he in fact travel?

tanØ = opp/adj tanØ = (1.3 m/s) / (2.1 m/s) tanØ = .619 Ø = 31.8° To find "R", his reluctant speed, simply use: R2 = 1.32 + 2.12 R = 2.47 m/sec

25. This time Kent uses his boat in a stream whose velocity is 0.92 m/sec. What is the angle of his motion, and how fast is he moving from his point of origin?

tanØ = .92 / 2.1 Ø = 23.7° You may choose a different angle, but indicate it in your diagram. R = (2.12 + .922)1/2 R = 2.29 m/sec

26. A plane, traveling at 450 m/sec fires a bullet perpendicularly to its flight path. If the muzzle velocity of the bullet is 522 m/sec, what is the direction and combined velocity of the bullet?

tanØ = 522 m/s / 450 m/sec tanØ = 1.16 Ø = 49.2° R = (4502 + 5222)1/2 R = 689 m/sec

27. A ship cruises at 4.8 m/sec. On deck a sailor walks cross-ways to the motion of the ship at 1.7 m/sec. What is his resultant speed and direction?

tanØ = 1.7 m/sec / (4.8 m/sec) Ø = 19.5° R = (1.72 + 4.82)1/2 R = 5.1 m/sec

28. A boat, whose speed in still water is 3.4 m/sec travels across a current of unknown speed. If the boat is forced 22° from its intended direction, what is the speed of the current?

tan22° = vc / (3.4 m/sec) vc = 3.4tan22 vc = 1.37 m/sec

29.* Miles Tugo wants to get across a stream without drifting downstream. The stream flows t 3.1 m/sec while the boat travels at 7.3 m/sec. In what direction should he aim the craft to achieve his end?

We need a slightly different diagram. Miles must aim upstream enough to balance the 3.1 m/sec downstream current. sinØ = opp/hyp sinØ = 3.1 / 7.3 sinØ = .42 Ø = 25.1°

30. An airplane must travel due east, but a wind of 40 knots blows from the north. In what direction must the pilot aim if his air speed is 224 knots? (Air speed indicates speed relative to still air.)

Using the reasoning from prblem 27: sinØ = 40 knot / 224 knot sinØ = .18 Ø = 10.3° The captain must aim the plane 10.3° n. of his intended direction, so it will be blown back on course. Notice that his final speed is less than 224 knots.

31. A man finds he must aim his boat 18° upstream to go straight across. If his speed relative to the water is 16.3 mi/hr, what is the speed of the stream?

vy = stream velocity vx = final velocity of boat relative to shore sin18° = opp/hyp sin18° = vy/ 16.3 vy = 16.3sin18 vy = 5.0 mi/hr

We return, now, to motion in one dimension. We will remove the restriction that the velocity be uniform, and consider cases of acceleration, that is, changing velocity.

Although they may seem a bit awkward at the moment, the following exercises will be most helpful later ( and sooner) in the course. We will visualize motion not simply as a particle traveling in a line, but as its various graphical representations. When doing these graphs, plot distance on the vertical axis and time on the horizontal axis. Except for unusual circumstances, time is routinely plotted on the horizontal. This convention aids quick comprehension of the meaning of a graph.

32. Graph all the following before checking the solution:
a) A particle, starting at the origin, travels at 5 m/sec.
b) A particle, starting at the origin, travels at 7 m/sec.
c) A particle, starting at the origin, remains at rest.
d) A particle starts 3 meters from the origin, at 5 m/sec.

33. Yes, as in problem 32, you should decide for yourself what scale to use for each axis. Graph on.
a) Starting at the origin, a particle travels at 4 m/sec.
b) Starting 6 m behind the origin, it travels at 7 m/sec.
c) From 4 m beyond the origin, it moves backwards at 6 m/sec.

34. One last review:
a) Start 8 m beyond the origin and remain at rest.
b) Start 4 m behind the origin and go backwards at 2 m/sec.
c) Start 4 m beyond the origin and go forward at 4 m/sec.

35. This time we consider a single particle doing a sequence of motions. Draw them as one continuous graph of the particle's motion.
a) Start 8 m behind the origin, go forward at 3 m/sec for 4 sec.
b) Continue at 6 m/sec for 2 sec.
c) Stop for 3 seconds, and
d) finally go backward at 8 m/sec for 3 seconds.

By now you should be striving to ";read"; the graphs in a direct way. That is, steep slopes should indicate to you rapid motion, negative slopes a backward motion, a horizontal line no motion.

36. Try another:
a) Start 10 m behind the origin, go forward at 5 m/sec for 3 seconds, b) then go forward at 2 m/sec for 5 sec.
c) Go backward at 8 m/sec for 4 seconds, and finally
d) go backward at 2 m/sec for 5 seconds.

37. This time plot velocity on the vertical and time on the horizontal. Repeat the directions of problem 32, but using the new variable.

Think about this so it makes sense - any constant velocity will be represented by a horizontal line.

38. Repeat problem 37, but this time with the motions from problem 33.

39. You guessed it, do problem 34 with a velocity vs. time graph.

Notice that the starting position of the particle is irrelevant to this graph - only the velocity matters.

40. Now repeat problem 35 with the new axes (velocity vs. time).

41. And finally, repeat problem 36.

Look back at what you have just done. In plotting velocity vs. time, the area under the curve will be given by the height times the length, or in this case, velocity times time. But velocity time time is just the distance traveled. Hence on this type of graph, the area under the curve represents the distance traveled. This looks supremely unprofound at the moment, but the concept is crucial to important insights in the future.

42. Determine the distance traveled in each graph.

a)	b)
c)	d)

a) Area = l x w Distance = (8 m/sec)(5 sec) Distance = 40.0 m	b) Distance = (7 m/sec)(3 sec) + (3 m/sec)(6 sec) Distance = 39.0 m
c) Distance = (-4 m/sec)(6 sec) Distance = -24.0 m	d) Distance = (-2 m/sec)(4 sec) + (4 m/sec)(2 sec) Distance = 0.0 m note: the particle returned to the starting point

43. Again, find the distance traveled.

a)	b)
a) s = [(3 m/sec) (2 sec)] + [(5 m/sec) (3 sec)] + [(-4 m/sec) (2 sec)] s = 6 m + 15 m - 8 m s = 13m	b) s = 1/2 [(5 m/sec) (2 sec)] + [(5 m/sec) (3 sec)] + 1/2 [(-6 m/sec) (3 sec)] s = 5 m + 15 m - 9 m s = 11 m It is important to recognize that "distance" in these equations means distance from the origin, not total distance covered.

44. Determine the distance on these goodies:

a	b	c
d	e	f
a) s = area s = (4 m/sec) (8 sec) s = 32 m	b) s = (1/2)v (t) s = (1/2 * 4 m/sec) (8 sec) s = 16 m	c) s = (1/2 * 6 m/sec) (10 sec) s = 30 m
d) s = (5 m/sec) (4 sec) + (.5 * 5 m/sec)(6 sec) s = 20 + 15 s = 35 m	e) s = 1/2[(3.14)r2] s = (1/2)(3.14) (4 m/sec) (4 sec) s = 25.1 m	f) Counting squares: s = 31 m

45. A car starts at rest and steadily accelerates to a speed of 15 ft/sec in 10 seconds. How far will it have traveled in this time?

The area under the triangle is: s = 1/2 (15 m/sec) (10 sec) s = 75 ft

46. A rock is thrown down with an initial velocity of 4.0 m/sec. In 3 seconds it is traveling at 33.4 m/sec. How far has it traveled in this time?

The area is a trapezoid, whose area is: A = ((a + b)/2)h Where a and b are the parallel sides. so: s = ((v0 + v) / 2)t s = ((-4.0 - 33.4)/2) (3) s = -56.1 m

47. A particle starts at rest and accelerates uniformly to a speed of 7.5 m/sec in 6.0 seconds. It then slows to a stop in 9.0 seconds. How far has it traveled in this time?

area = 1/2 (weight * base) s = (1/2) (7.5 m/sec) (9 sec) s = 33.8 m

48*. Cora Napple has a boat that travels 2.4 m/sec in still water. She wishes to go upstream 500 m and then return to her origin. If the stream flows at 1.1 m/sec, what time will be required for the trip?

s = vt t = s/v For her trip up: v = 2.4 - 1.1 v = 1.3 m/sec For her trip down: v = 2.4 + 1.1 v = 3.5 m/sec t = tup + tdown t = [500 m / (1.3 m/sec)] + [(500 m / (3.5 m/sec)] t = 527 sec

49**. A bird flies in still air at 2.8 m/sec. One morning the wind is blowing from the north at an unknown speed, but the bird notes that it takes her 60 seconds to visit her friend 76 m due east. What, then, is the speed of the wind?

The actual velocity of the bird si given by: v = s/t v = 76 m / 60 sec v = 1.27 m/sec We see that: vw = (2.82 - v2)1/2 vw = (2.82 - 1.272)1/2 vw = 2.5 m/sec