Topic 1.5: Vectors and Motion in Two Dimensions

What Topic 1.5 requires

The CED gives Topic 1.5 two learning objectives, and every idea on this page sits under one of them.

1.5.A: describe the perpendicular components of a vector. A vector can be modeled as the resultant of two perpendicular components (EK 1.5.A.1); vectors are resolved into components using a coordinate system you choose (1.5.A.2); and that resolution runs on trigonometric relationships (1.5.A.3), for which the CED prints $\sin\theta = a/c$, $\cos\theta = b/c$, $\tan\theta = a/b$, and $a^2 + b^2 = c^2$.

1.5.B: describe 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 (1.5.B.1), and projectile motion is a special case that has zero acceleration in one dimension and constant, nonzero acceleration in the second (1.5.B.2).

Topic 1.5 closes Unit 1, worth 10 to 15 percent of the multiple-choice section, and it is the bridge into Unit 2: the same decomposition move that splits a launch velocity also splits weight on a ramp.

Resolving a vector into perpendicular components

A vector carries a magnitude and a direction. Components trade that pair for two numbers along axes you control, and the CED calls for perpendicular components in EK 1.5.A.1 and 1.5.A.3, because perpendicular axes do not overlap: each one carries information the other cannot.

Put the tail at the origin, magnitude $M$, angle $\theta$ measured counterclockwise from the $+x$ axis:

The vector is now the hypotenuse of a right triangle whose legs are its own components, which is why the AP right-triangle relations do all the work. The equation sheet prints them in its Geometry and Trigonometry box: $\sin\theta = a/c$, $\cos\theta = b/c$, $\tan\theta = a/b$, and $a^2 + b^2 = c^2$, with $c$ the hypotenuse, $b$ the side adjacent to $\theta$, and $a$ the side opposite.

Store the pairing as a fact about the triangle, not about the letters $x$ and $y$: the component adjacent to the angle takes cosine, the component opposite takes sine. Cosine only lands on $x$ when the angle is measured from the $x$ axis. Measure the same angle from the vertical and the pairing flips.

The coordinate system is your choice

EK 1.5.A.2 says vectors are resolved into components using a chosen coordinate system, and the word chosen is load-bearing. The physics does not hand you axes. You pick them, and a good pick makes the algebra short.

• Projectiles: horizontal $x$, vertical $y$. Gravity then sits entirely in the vertical component, making the vertical axis EK 1.5.B.2's constant, nonzero acceleration dimension and the horizontal axis its zero-acceleration dimension.
• Inclines: tilt the axes so $x$ runs along the slope and $y$ runs perpendicular to it. The acceleration now lives on one axis and only the weight needs decomposing, which is the whole method behind inclined plane problems.
• Circular motion: aim one axis at the center of the circle. That resolves the forces, but neither component acceleration is constant there, so the constant-acceleration equations below do not carry over.

Two rules survive whatever you choose. The axes must be perpendicular, and the choice must hold from the first line of the solution to the last. Rotating the axes changes the components but never the vector itself: a 50 N force is 50 N however you slice it.

From components back to magnitude and direction

Decomposition runs both ways, and AP questions ask for both. Given perpendicular components $A_x$ and $A_y$:

The first is the Pythagorean theorem straight off the AP sheet. The second is $\tan\theta = a/b$, opposite over adjacent.

Two cautions. Magnitudes never add directly: 3 m/s east combined with 4 m/s north is 5 m/s, not 7 m/s, because perpendicular pieces combine through the square root. And an inverse tangent on a calculator returns an angle between $-90^\circ$ and $90^\circ$, so it cannot tell up-and-left from down-and-right. Sketch the two components as arrows before you trust the sign it gives you.

State every direction with a reference attached: 37 degrees above the horizontal, or 20 degrees north of east. A bare angle is ambiguous, and free-response rubrics do not award ambiguous answers.

Two dimensions become two one-dimensional problems

EK 1.5.B.1 is the operating instruction for the rest of the course: motion in two dimensions can be analyzed using one-dimensional kinematic relationships if the motion is separated into components. Split the motion and every tool from Topics 1.2 and 1.3 keeps working, one axis at a time.

The split is legal because acceleration is a vector too. The $x$ component of the acceleration changes only the $x$ component of the velocity; nothing inside $a_x$ can reach $v_y$. So the two axes run as two separate problems, side by side. The one-dimensional kinematic relations then hold on an axis only while that axis's acceleration stays constant, which is why projectiles split so cleanly and why circular motion does not.

They share exactly one quantity: time. One clock runs for both axes, which makes $t$ the only bridge between them. That is why nearly every two-dimensional problem is solved by finding $t$ on whichever axis you know most about, then carrying it across.

In practice, draw a two-column table with $x$ on the left and $y$ on the right, and file each known under one column only. An $x$ quantity that wanders into a $y$ equation quietly voids everything after it. The kinematic equations guide covers choosing the right equation once the split is made.

Projectile motion, the CED's special case

EK 1.5.B.2 defines a projectile with unusual precision: two-dimensional motion with zero acceleration in one dimension and constant, nonzero acceleration in the second. With the standard axes, that reads:

The conventions printed with the exam's table of information assume air resistance is negligible unless a problem says otherwise, and that assumption is what holds $a_x$ at zero.

The value of $g$ itself carries a caveat worth knowing. A boundary statement under Topic 1.3 says that for all situations in which a numerical quantity is required for $g$, the value $g \approx 10 \text{ m/s}^2$ will be used, and that students will not be penalized for correctly using the more precise commonly accepted values $9.81 \text{ m/s}^2$ or $9.8 \text{ m/s}^2$. This site works in $9.8 \text{ m/s}^2$, the figure printed in the table of information, so expect small last-digit gaps against a key built on 10.

Two consequences students underuse. Horizontal velocity at impact equals horizontal velocity at launch, always. And flight time is set entirely by the vertical column, so a ball dropped and a ball fired sideways from the same height hit the floor together. For the full solving method see how to solve projectile motion problems, check numbers against the projectile motion calculator, and watch the split live in the projectile launcher.

The trig values the AP sheet hands you

The AP Physics 1 table of information prints sine, cosine, and tangent for seven angles: 0, 30, 37, 45, 53, 60, and 90 degrees. Knowing that table is worth more than it looks. The five middle angles are the ones that do real work, so they are the ones reproduced here.

The 37 and 53 degree entries are the 3-4-5 right triangle, good to about two significant figures: the true $\sin 37^\circ$ is $0.6018$ and the sheet gives $3/5$. Problem writers choose those angles deliberately, so a 25 m/s launch at 37 degrees splits into a clean 20 m/s and 15 m/s. Take $\cos 37^\circ$ and $\sin 37^\circ$ off a calculator instead and the same launch gives 19.97 m/s and 15.05 m/s. Both earn credit, and the tools on this site use the calculator values, so expect a gap that size when you cross-check. Seeing a 3-4-5 fall out of your components is a signal that your setup matches the one the question was built around.

Notice the mirror pairs, 30 with 60 and 37 with 53: the sine of one is the cosine of the other, because the two acute angles of a right triangle add to 90 degrees. Both the trig relations and the three kinematic equations behind this topic are printed on the AP Physics 1 formula sheet.

Common mistakes, and how the exam tests this

Errors that turn up often on Topic 1.5 work:

• Swapping sine and cosine. Locate the angle in the triangle first. Adjacent takes cosine, opposite takes sine, whatever the axes are named.
• Feeding the full speed into a vertical equation. Only $v_{y0} = v_0 \sin\theta$ belongs there.
• **Setting $v_y = 0$ at landing.** Vertical velocity reaches zero at the peak of the arc, not at the ground.
• Adding magnitudes. Perpendicular pieces combine as $\sqrt{A_x^2 + A_y^2}$, never by plain addition.
• Flipping the positive direction midway. Fix the sign convention once and hold it.

The CED lists five suggested skills for this topic: 1.B (create quantitative graphs), 2.A (derive a symbolic expression), 2.D (predict new values using functional dependence between variables), 3.A (create experimental procedures), and 3.C (justify a claim using evidence). One of them, 3.A, is experimental design, and the CED's sample free-response set aligns Question 3, the Experimental Design and Analysis question, to LO 1.5.B with skills 1.B, 2.B, 2.D, and 3.A. That is why launcher labs keep appearing: the CED's own sample activities for 1.5 have students launch a ball from an upper story and graph the components, then predict where a spring-loaded launcher will land.