1.5 Vectors and Motion in Two Dimensions

Learning objectives

1.5A—Describe the perpendicular components of a vectorDescribe the perpendicular components of a vector.• Vectors can be mathematically modeled as the resultant of two perpendicular components.• Vectors can be resolved into components using a chosen coordinate system.• Vectors can be resolved into perpendicular components using trigonometric functions and relationships. Relevant equations: sin cos tan += ab c22 2 TOPIC 1.5 Vectors and Motion in Two Dimensions 1.5B—Describe the motion of an object moving in two dimensionsDescribe the motion of an object moving in two dimensions.• Motion in two dimensions can be analyzed using one-dimensional kinematic relationships if the motion is separated into components.• Projectile motion is a special case of twodimensional motion that has zero acceleration in one dimension and constant, nonzero acceleration in the second dimension. AP Physics 1: Algebra-Based Course and Exam Description AP PHYSICS 1 UNIT Force and Translational Dynamics 2 18–23% AP EXAM WEIGHTING ~22–27 CLASS PERIODS 3535 | AP Physics 1: Algebra-Based Course and Exam Description Remember to go to AP Classroom to assign students the online Progress Check for this unit. Whether assigned as homework or completed in class, the Progress Check provides each student with immediate feedback related to this unit’s topics and science practices. Progress Check 2 Multiple-choice: ~30 questions Free-response: 4 questions

Choose perpendicular axes

Choose perpendicular $x$ and $y$ axes. A vector $\vec A$ can be replaced by components $A_x$ and $A_y$ whose vector sum is the original vector.

Resolve and reconstruct

If $\theta$ is measured from the +x-axis	Relationship
Adjacent component	$A_x=A\cos\theta$
Opposite component	$A_y=A\sin\theta$
Reconstruct magnitude	$A=\sqrt{A_x^2+A_y^2}$
Reconstruct direction	$\tan\theta=A_y/A_x$

Calculate both components

Hypothetical example: a $10\,\mathrm{N}$ vector points $30^\circ$ above +x. Then $A_x=(10)\cos30^\circ=8.66\,\mathrm{N}$ and $A_y=(10)\sin30^\circ=5.00\,\mathrm{N}$ . Check: $\sqrt{8.66^2+5.00^2}=10.0\,\mathrm{N}$ .

Keep coordinate signs

The trigonometric magnitudes do not replace the coordinate signs. Determine each sign from the vector’s direction relative to the chosen axes. Components are simultaneous parts of one vector—not two vectors applied one after the other.

Separate the components

Resolve position, velocity and acceleration into perpendicular components. Analyze the $x$ and $y$ motions separately with one-dimensional kinematics, then combine them at the same time $t$ .

Compare horizontal and vertical motion

Projectile component	Acceleration	Consequence
Horizontal $x$	$a_x=0$	$v_x$ is constant and $\Delta x=v_xt$
Vertical $y$	$a_y=-g$ if up is positive	$v_y$ changes at a constant rate

Use one shared time

Hypothetical example: a projectile launches horizontally at $v_x=6\,\mathrm{m\,s^{-1}}$ from $20\,\mathrm{m}$ above the ground. With up positive and $g=10\,\mathrm{m\,s^{-2}}$ , $-20=\tfrac12(-10)t^2$ , so $t=2\,\mathrm{s}$ . During that same time, $\Delta x=(6)(2)=12\,\mathrm{m}$ .

Recombine the story

Zero horizontal acceleration does not mean zero horizontal velocity. The components evolve independently under their own accelerations, but they are synchronized by the same time interval and together form one curved trajectory.

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