Kinematics in Two Directions

Analysis of Motion

Resolve vectors into x and y components (table 3.1, page 61)

Solve for x and y directions

Combine x and y components for resultant

Variables Involved

X direction: x, a_x, v_x, v_ox, t

Y direction: y, a_y, v_y, v_oy, t

Kinematic Example

Rocket has horizontal and vertical engines

Engines operate independent of each other

Case 1

t = 0

Neither engine on

R = 0

Case 2

x engine on, y engine off

a_x positive

v_x positive

a_y = 0

v_y = 0

R = x

Case 3

y engine on, x engine off

a_x = 0

v_x = 0

a_y positive

v_y positive

R = y

Case 4

x and y engines on

a_x positive

v_x positive

a_y positive

v_y positive

R = x + y

Independence of Motion

Motion in one direction is independent of motion in another

Observed motion is a vector sum of the independent motions

Problem Solving

Read carefully and identify the question being asked

Make a drawing and establish a frame of reference

ID variables and select appropriate formulas

t will be the same for both directions

Solve separately for x and y directions

Combine results to determine final answer

Projectile Motion

Object having constant horizontal motion is accelerated by gravity

Analyze motion by separating into x and y components

Analysis of Projectile Motion

No acceleration in X direction: x = v_xt

g accelerates in Y direction

y = v_yt+ ½gt²

g = 9.80 m/s²

v_x remains constant

v_y changes

Maximum at ground level

Minimum at top of trajectory

Always the same at the same height

Modeling

Use parametric mode

T represents the variable that changes

X_1T and Y_1T represent the horizontal and vertical axes

Modeling Motion

T = time

X_1T = x displacement

Y_1T = Y displacement

Modeling Motion

General Equations

x = vCosqt

y = h + vSinqt + ½gt²

Situation 1

Object projected horizontally from height (h)

h > 0

v > 0

q = 0

x_1T = vt

Cos(q) = Cos(0) = 1

y_1T = h + ½gt²

Sin(q) = Sin(0) = 0

Situation 2

Object projected at angle from ground level

h = 0

v > 0

q > 0

x_1T=vCosqt

y_1T=vSinqt + ½ gt²

Situation 3

Projected at angle from a height

h > 0

v > 0

q > 0

x = vCos(q)t

y = h + vSin(q)t + ½gt²