Motion in Two DimensionsIn two dimensions, it is necessary to use vector notation to describe physical quantities withboth magnitude and direction. In this lecture, we recap the definitions of displacement, velocityand acceleration as vectors in two dimensions. We also discuss the solution of projectile motionproblems in two dimensions.

A recap of VectorsVectors give us a simple way of handling motion when it is not in a straight line. Vectors arespecified by a magnitude and a direction. For example, if you drive down a road to the 47o

north east direction with a speed of 30 km/h; your velocity vector when will be specified by amagnitude of 30 km/h (your speed) and a direction of 47o north of east.

Fig. 3.1 A velocity vector

Symbolically, the easiest way to represent a vector is by an arrow, where the length of the arrowis proportional to the magnitude of the vector (say, a length of 3 cm represents a velocity of30 km/h as in the example of Figure 3.1 above), and the direction of the arrow represents thedirection of the vector.In this course we will represent a vector quantity by an algebraic symbolfor the vector’s magnitude with a small arrow above it (pointing to the right)as shown with thevelocity vector in Figure 3.1 above.

Using vectors to represent motion in 2-D

Two methods to combine vectors in order to represent motion in 2-d include: the triangle ruleand the parallelogram rule. the Triangle rule is applied whenever one vectors acts followedby another. The parallelogram rule is applied whenever vectors act at the same time from thesame point. For an example of the triangle rule; suppose you travel 30 m due south and then 40m due east; what is your displacement from the original position? (recall how we illustratedthis on the board in class). We find the displacement by drawing the vectors involved.

• Draw an arrow 3 cm from a starting point to show a displacement of 30 m south.

• Draw an arrow 4 cm from the head of the first arrow and at right angles to the first arrowto represent a displacement of 40 m east.

• Join the starting point of the first arrow with the end of the second arrow. This is yourdisplacement

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You can then measure the distance and direction of the displacement from your scale diagram.Or more mathematically, a method you will use in this course, you can use trigonometry.

Vector components

Recall from our first lecture that the components of a vector completely define the vector.Where the components of the vector are the projections of a vector along the axes of arectangular co-ordinate system.

Ay

Ax

~A

For an example of the parallelogram rule, take the velocity vector in Figure 3.1 above. Now,imagine that a strong light is shining down on the velocity vector in the diagram from above;this light would cast a shadow on the east axis. If we were to call the east direction the positivex-direction, then we would say that the x-component of the velocity vector, denoted vx, is justthe length of this shadow along the x-axis. Likewise, if a strong light shines from the right ontothe vector, then it would cast a shadow on the north axis, which we can call the positive y-axis;this would give us the y-component of the velocity vector, denoted vy. These two componentsare shown in the revised figure below.

Fig. 3.4 The parallelogram

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Note that we have also drawn in the x- and y-components of the velocity vector above and tothe right of the velocity arrow, thus forming a rectangle whose sides are the two components ofthe vector, and whose diagonal is the vector itself. The vector therefore splits the rectangle intotwo right triangles. It then follows that we may apply the definitions of trigonometry to eitherof these two triangles, whichever is more convenient for you. (We usually know an angle fromthe vector to either the x- or y-axis, since this gives us the direction of the vector. We wouldthen simply pick the triangle which contains the given angle!)

To see how trigonometry fits into this, consider, for example, the x-component of the velocityvector~v. This component is the side of the right triangle adjacent to the angle θ = 47o.

Fig. 3.5 Components of the velocity vector

From the definition of the cosine function, cosθ = ad j/hyp, we get that ad j = (hyp)cosθ .But, for our triangle, adj is just vx, and hyp is just v (the magnitude of the velocity vector; noarrow above the symbol!), so that vx = vcosθ .

Likewise, from the definition of the sine function, we get that opp = (hyp)sinθ , so that, sinceopp = vy, and hyp = v, we have that vy = vsinθ . Using the numbers from above (v = 30 km/hand θ = 47o), you should be able to use your calculator to show that

vx = (30km/h)cos(47o) = 36.1km/h and vy = (30km/h)sin(47o) = 38.8km/h.

(Ask your group partners or your instructor if you don’t know how to use your calculator toevaluate trigonometric functions - this will be very important in this course!)

The next two examples will give you more practice with using the trigonometric functions inconjunction with vectors.

A recap of the Kinematic quantities and definitionsDisplacement

In Figure 3.2, an object is at position~ri at time ti (point P). Some time later, t f , the object is atposition~r f (point Q). The displacement vector of the object is given by:

∆~r =~r f −~ri (1)

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Fig. 3.2 Displacement

Average Velocity

Definition:~̄v =

∆~r∆t

=~r f −~ri

t f − ti(2)

As with the 1-D definition, ~̄v is independent of the path between the end points.

Instantaneous Velocity

Definition:~v = lim

∆t→0

∆~r∆t

(3)

Fig. 3.3: Instantaneous velocity in 2-D

Interpretation: When ∆t → 0, the point Q in the figure gets closer and closer to the point Pand the direction of ∆~r approaches the direction of a tangent to the curve at point P. Thus theinstantaneous velocity~v is parallel to the tangent and in the same direction as the motion.

Average Acceleration

Definition:~̄a≡ ∆~v

∆t(4)

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Instantaneous Acceleration

Definition:~a≡ lim

∆t→0

∆~v∆t

(5)

Note: a particle can accelerate in different ways:

1. The magnitude of~v can change in time, while the direction of motion stays the same.

2. The magnitude of~v , |~v| , can stay constant, while the direction of motion changes. Thisonly happens in more than one dimension. We will discuss this further in Unit 4..

3. Both |~v| and the direction of~v can change.

Projectile MotionProjectile motion is a particular kind of 2 dimensional motion. Take an example of fee fall; wemake the following assumptions:

• The only force present is the force due to gravity.

• The magnitude of the acceleration due to gravity is |~a| = g = 9.8 m/s 2 We choose acoordinate system in which the positive y-axis points up perpendicular to the earth’ssurface. This definition gives, ay =−g and ax = 0.

• The rotation of the earth does not affect the motion.

Initial Conditions:We choose the coordinate system so that the particle leaves the origin ( x0 = 0, y0 = 0 ) at timet0 = 0 with an initial velocity of~v0 .

Procedure for Solving Projectile Motion Problems

1. Separate the motion into the x (horizontal) part and y (vertical) part.

2. Consider each part separately using the appropriate equations. The equations of motionbecome,

• x motion (ax = 0):vx = vx0 = constantx = vx0t

• y motion ( ay =−g ):vy = vy0−gty = vy0t− 1

2gt2

v2y = v2

y0−2gy.

3. Solve the resulting system of equations for the unknown quantities.

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