Unit 1 Test Review — Study Guide for Tonight

Tomorrow's test covers everything from Topics 1.1 through 1.5 — vectors, velocity and acceleration, reading motion graphs, reference frames, and 2D projectile motion. This post is a complete review of all five topics, with worked practice problems (different numbers than anything on the actual test) so you can check your understanding before walking in tomorrow. You'll have a formula sheet on the test itself, so you don't need to memorize equations tonight — focus on knowing which equation to reach for and why.

Topic 1.1: Scalars, Vectors, and Sign Conventions

What to remember:

Displacement is a vector — it's the straight-line change in position, not the total distance you walked.
In one dimension, direction is shown with a plus or minus sign. Pick a positive direction first and stay consistent.
$\Delta v = v_{final} - v_{initial}$. If an object reverses direction, one of those velocities is negative — don't lose that sign.

Worked Example: A hiker walks 12 m east, then turns around and walks 5 m west. What is the magnitude and direction of her total displacement? What is her total distance traveled?

Solution: Taking east as positive: displacement $= (+12) + (-5) = +7\text{ m}$, so 7 m east. Distance traveled is NOT a vector sum — it's just how much ground she covered: $12+5=17\text{ m}$. Notice these two answers are different on purpose — that's the whole point of the displacement-vs-distance distinction.

Topic 1.2: Displacement, Velocity, and Acceleration

What to remember:

The four kinematic equations only work during a stretch of constant acceleration.
5 variables ($v_i$, $v$, $a$, $\Delta x$, $t$), 4 equations, each equation leaves exactly one variable out — find what you're missing, then pick the equation that also leaves that one out.
"Starts from rest" means $v_i=0$. "Comes to rest" means the final $v=0$.

Worked Example: A car speeds up from 15 m/s to 35 m/s over 4 seconds. (a) What is its acceleration? (b) How far did it travel during those 4 seconds?

Solution: (a) $a=\dfrac{\Delta v}{\Delta t}=\dfrac{35-15}{4}=5\text{ m/s}^2$. (b) Since we know $v_i$, $v$, and $t$ but not $\Delta x$, use $\Delta x = \tfrac{1}{2}(v_i+v)t = \tfrac{1}{2}(15+35)(4) = 100\text{ m}$.

Topic 1.3: Representing Motion (Graphs and Free Fall)

What to remember:

On a position-time graph, the slope is the velocity. On a velocity-time graph, the area under the graph is the displacement, and the slope is the acceleration.
Free fall is just the kinematic equations with $a=g\approx9.8\text{ m/s}^2$ downward — gravity never turns off, even for an instant at the top of a throw.
An object's speed changes as it falls, so it covers more distance in each successive second — equal time intervals do NOT mean equal distances during free fall.

Worked Example: A rock is dropped from a cliff (starts at rest). How far does it fall during just the 3rd second of its fall (that is, between $t=2\text{ s}$ and $t=3\text{ s}$)?

Solution: Use $\Delta x = \tfrac{1}{2}gt^2$ (since $v_i=0$) to find the total distance fallen at each time, then subtract. At $t=2\text{ s}$: $\tfrac{1}{2}(9.8)(2)^2=19.6\text{ m}$. At $t=3\text{ s}$: $\tfrac{1}{2}(9.8)(3)^2=44.1\text{ m}$. Distance during the 3rd second $=44.1-19.6=24.5\text{ m}$ — notice this is more than the 19.6 m it fell during the entire first 2 seconds combined, because it's speeding up the whole time.

Topic 1.4: Reference Frames and Relative Motion

What to remember:

$v_{A-C}=v_{A-B}+v_{B-C}$ — velocities relative to a chain of frames add together.
Pick ONE positive direction and apply it to every velocity in the problem, even ones given relative to a different, moving frame.
With three or more frames (person → vehicle → ground), work one layer at a time.

Worked Example: A person walks toward the front of a train at 1.5 m/s relative to the train. The train moves at 20 m/s relative to the ground, in the same direction the person is walking. What is the person's velocity relative to the ground?

Solution: $v_{person-ground}=v_{person-train}+v_{train-ground}=1.5+20=21.5\text{ m/s}$, same direction as the train's motion.

Try this one yourself (answer at the bottom): Same train, still moving at 20 m/s, but now the person walks toward the BACK of the train at 1.5 m/s instead. What is the person's velocity relative to the ground?

Topic 1.5: Vectors and Motion in Two Dimensions (Projectiles)

What to remember:

Horizontal and vertical motion are totally independent. Solve vertical first to get the time in the air, then use that same time horizontally.
Changing the horizontal launch speed changes how far a projectile goes, never how long it's in the air.
Mass never shows up in any projectile equation — it doesn't affect height, range, or time of flight.
Horizontal launch ($\theta=0$): $v_{0y}=0$, so just solve $h=\tfrac12gt^2$ for time.
Launched at an angle: $v_x=v_0\cos\theta$, $v_{0y}=v_0\sin\theta$, and (same launch/landing height) total flight time $=\dfrac{2v_0\sin\theta}{g}$.

Worked Example 1 (horizontal launch): A ball rolls off a table 1.2 m high with a horizontal speed of 2.0 m/s. How long does it take to hit the floor, and how far from the table does it land?

Solution: Vertical first: $h=\tfrac12gt^2 \Rightarrow t=\sqrt{\dfrac{2h}{g}}=\sqrt{\dfrac{2(1.2)}{9.8}}\approx0.50\text{ s}$. Horizontal: $\Delta x = v_0t = (2.0)(0.50)\approx1.0\text{ m}$.

Worked Example 2 (angled launch): A ball is kicked from level ground at 12 m/s, at an angle of 40° above horizontal. Find the total time in the air and the horizontal range.

Solution: $v_{0y}=v_0\sin\theta=12\sin(40°)\approx7.7\text{ m/s}$. Total time: $t=\dfrac{2v_{0y}}{g}=\dfrac{2(7.7)}{9.8}\approx1.6\text{ s}$. Horizontal speed: $v_x=v_0\cos\theta=12\cos(40°)\approx9.2\text{ m/s}$. Range $=v_x t \approx (9.2)(1.6)\approx14.5\text{ m}$.

Test-Day Reminders

You'll get a formula sheet — you don't need equations memorized, but you DO need to know which one to use and why.
Calculator is allowed the whole test, matching the real AP exam policy.
Set up every equation symbolically first (with letters), then substitute numbers — it's much easier to catch a wrong equation choice before you've done any arithmetic.
If you get stuck on a multiple-choice question, don't spend more than a minute or two — flag it, keep moving, and come back if you have time at the end.

Missed a day this unit or want a deeper refresher on one topic? Rewatch the video linked in that day's lesson page — this review is meant to jog your memory, not replace what we covered in class.

Answer to the "try this yourself" problem: Walking toward the back means the person moves opposite the train's direction. Taking the train's direction as positive: $v_{person-ground}=(-1.5)+20=18.5\text{ m/s}$, same direction as the train, just slower than the train itself.