A Relative Velocity Question

Johnny and Mary, living in sunny south-east Asia, are assigned the following question by their Canadian-born teacher:

An ice-boat is sailing over a frozen lake with a velocity of 6 metres per second, due East. The captain of the ice-boat perceives the wind to be blowing from the South-West, although the wind is actually blowing from a different direction at a constant 11 metres per second. Use your knowledge of vectors and relative velocity to find the actual wind direction.

http://minnesota.publicradio.org/display/web/2009/01/29/ice_sailing/

Mary decides to tackle the problem algebraically, while Johnny decides to model it using Geogebra.

Mary’s reasoning

Mary knows the apparent or relative wind velocity, V_R, is the actual wind velocity, V_A, minus the velocity of the iceboat, V_R = V_A - V_B.

Therefore the actual wind velocity is V_A = V_R + V_B. She knows V_B. She could easily calculate the direction of V_A if she knew the magnitude of V_R. As she is missing this apparently vital information she begins a seemingly idle sketch.

 

Recalling the triangle law of vector addition, Mary starts to sketch.

 

She labels the known facts.

Mary’s final sketch prompts her to recall the sine law.

 

sine 135°/11 = sine y°/6

sine y° = (6/11) × sine 135° = 0.3856 …

y = sin^-1 (.3856 …) = 22.69° (to two decimal places)

x = 180° – 135° – 22.69° = 22.31° (to two decimal places)

z = 90° - 22.31° = 67.69°

The bearing of the wind is from 180° + 67.69° = 247.7° (to one decimal place)

Johnny’s Reasoning

No.	Geogebra	Reasoning
1	O = (0, 0)	Origin
2	P = (11, 0)	Needed to define a circle
3	Circle[O, P]	The actual wind speed is 11 km/h, with an unknown direction. The tip of the velocity vector is somewhere on a circle of radius 11.
4	i = (1, 0)	Unit vector, East
5	j = (0, 1)	Unit vector, North
6	Number k is created from the slider pull-down menu	A slider to change the magnitude of the relative wind velocity
7	u = k i + k j	Wind from the south west is going north east. The direction is somewhat special; consequently the vector components are equal. Status: a hidden vector in quadrant I.
8	Q = O + -u	A hidden point; the tail of the relative wind velocity.
9	relwα = Vector[Q, O]	Shows the wind in quadrant III, from the south west.
10	iceboat = 6 i	The actual velocity of the iceboat
11	A = (6, 0)	The tip of the iceboat velocity vector.
	relwβ = A + relwα	A translation (or copy) of the relative wind vector is created and added to the iceboat velocity vector.
12	R = A + u	The point at the tip of the iceboat velocity vector.
13	actualw = Vector[O, R]	Move point k on the slider until the actual wind velocity has a magnitude of 11 (i.e., just touching the circle).
14	α = Angle[P,A,R]
15	β= Angle[R,O,A]

insert Geogebra file

Moving the slider creates directly proportional changes in the magnitude of the relative wind velocity, which in turn affects the actual wind velocity. The slider can be adjusted so that the actual wind velocity touches the circle representing its magnitude. A careful comparison reveals Johnny and Mary have reached the same result by different methods.

Kinematics

A tiny speck moves along a number line. The velocity of the speck is given by a polynomial function; for example v : t à 3t² - 8t + 4. The function is a gift that allows us quite arbitrarily to set the position of the speck on the number line at zero (or perhaps some other number) for the very instant that we start the timer.

What follows is inevitably a request that we calculate the distance of the speck from the origin for a particular time, the time for which the speck returns to the origin, the acceleration of the speck, the time for which the instantaneous velocity is zero, or the distance travelled by the particle between two instants of time.

Projections

About two thousand five hundred years ago, Socrates explained that the ‘reality’ we believe in is just illusion. Socrates told Plato's brother, Glaucon, to think of a prisoner who has been immobilised in a home cinema since birth. His body is chained and his head is fixed in position so he can see nothing but the screen in front of him. Obviously the prisoner will take the movies he sees on the screen for reality. (Please don’t worry about how the prisoner is fed or other details like how Socrates subscribed to the movie channel thousands of years ago. The story is an allegory. You can read more about it here. http://www.wisegeek.com/what-is-the-allegory-of-the-cave.htm)

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Figure 1. Socrates’ allegory (The clipart images of the face and the torch (a.k.a. flashlight) are from the royalty fre website http://www.clker.com/ )

The essential point for kinematics is that the speck-like particle need not move along a number line at all. Only its projection does so. I prefer to think that the particle moves 1 unit of distance in the x-direction for every 1 unit of time. How do I know this is true? Actually I don’t – I’m a prisoner in the cinema. It’s just a convenient fiction. However, motion in the y-direction will be controlled by the given function, and it is perfectly reasonable to believe that motion in the x-direction is independent of motion in the y-direction. Problem-solvers are often told to strip everything down to the essentials, but I always prefer a back-story. Why is the particle behaving this way? Who does it call, “Mom”? (Or how did it come to life?)

Figure 2. The image on the left represents either the path of a ball tossed straight up and falling back down or the projection (i.e., shadow) of the motion of a ball tossed upwards to the right and falling back down.

Integration and differentiation

Johnny knows that fundamentally, kinematics is simply differentiation and integration applied to moving objects. If he knows the object’s position function with respect to time he can differentiate once to get its velocity function. Then he differentiates the velocity function to get the acceleration function.

Or, if he is given the velocity function he integrates to find the position (also called displacement) function and differentiates to find the acceleration function.

Or, if he is given the acceleration function, he integrates firstly to get the velocity function and integrates again to get the position function.

Suppose Johnny and Mary are given, v : t à 3t² - 8t + 4, 0 ≤ t ≤ 3

Johnny writes p: t à ∫ v(t) dt = t³ – 4t² + 4t + C. Mary reminds him C = 0 as the timer starts at the instant the particle is at the origin.

Mart differentiates the velocity function to get the acceleration, a: t à dv/dt = 6t – 8

a) How far is the particle from the origin when t = 1.5?

Johnny and Mary will simply substitute t = 1.5 into their position function.

b) When is the velocity instantaneously zero?

Johnny and Mary will simply set the (given) velocity function to zero and solve for t.

c) What is the acceleration when t = 2?

Johnny and Mary will simply substitute the value t = 2 into their acceleration function.

d) What is the total distance travelled by the function in the time from t = 0 to t = 3? Mary remembers that if the particle reverses direction, its total distance travelled will not be the same as its final position. She needs to identify the turning points of the function. How? She recalls that, at the turning points of polynomial functions the first derivative must be zero. (Be careful, the first derivative could also equal zero for some inflection points).

Figure 3. Geogebra of the kinematics problem. Move the slider to see point Q move along the position function. Point P is the projection of Q onto the y-axis.

Mary creates a graph of the position function and projects a point from the function onto the y-axis. The process is quite simple. Click on view … Construction protocol for details. From the graph the turning points of the position function occur at x = 0.7 and x = 2. Alternatively, factorise the velocity function: v = 0 = 3t² - 8t + 4 = (3t – 2) (t – 2) ßà t = 2/3 or t = 2. Factorising gives the exact values of t; although Mary notes 2/3 = 0.7 when rounded to one decimal place. From the graph (but Johnny uses the the second derivative) the first value of t produces a local maximum while the second produces a local minimum. Because the particle P rises and falls back to the origin in the first two seconds, its distance travelled in that interval is double of the local maximum. Also, in this case, the distance the particle travels in the third second (2 ≤ t ≤ 3) is just its position at t = 3.

Thus the total distance travelled is 2 ((⅔)³ – 4(⅔)² + 4(⅔)) + (3³ – 4(3)² + 4(3)) ≈ 5.37.

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