Unit 1 Student Formula Sheet

You may use this sheet on the Unit 1 Test. It gives you the equations — it does NOT tell you which one to use or do the thinking for you. Read each problem carefully, decide what you know and what you're looking for, and then pick the matching tool from below.

1. Vectors and Signs (Topic 1.1)

Variable definitions

$\Delta x$ = displacement (change in position) — a vector, not just a distance
$v$ = velocity — a vector; speed is just its magnitude (size), with no direction attached
$\Delta v$ = change in velocity = $v_{final} - v_{initial}$ (always final minus initial, never the other way around)

Tips

In one dimension, direction is shown by a plus or minus sign — pick which direction is positive at the start of the problem and stay consistent for the whole problem.
Displacement is a vector sum, not a distance sum. Moving +4 units then −3 units gives a total displacement of $+4+(-3)=+1$, NOT $4+3=7$.
$\Delta v = v_{final}-v_{initial}$ is a subtraction of two signed numbers. If the object reverses direction, $v_{initial}$ and $v_{final}$ have opposite signs — don't drop the negative sign of whichever one is negative.
Speed can never be negative (it's just a magnitude). Velocity, acceleration, and displacement CAN be negative — that's how they show direction in 1D.

2. The Kinematic Equations (Constant Acceleration Only) — Topics 1.2 & 1.3

Variable definitions

$v_i$ = initial velocity
$v$ = final velocity
$a$ = acceleration (constant)
$\Delta x$ = displacement
$t$ = time elapsed

#	Equation	Leaves Out
1	$v = v_i + at$	$\Delta x$
2	$\Delta x = v_it + \tfrac{1}{2}at^2$	$v$
3	$v^2 = v_i^2 + 2a\Delta x$	$t$
4	$\Delta x = \tfrac{1}{2}(v_i+v)t$	$a$

Tips

There are 5 variables total ($v_i$, $v$, $a$, $\Delta x$, $t$) and 4 equations — each equation uses 4 of the 5 and leaves exactly one out.
How to choose: list what you're given, list what you're asked to find, then identify the ONE variable that's neither given nor asked for. Use the equation that also leaves that variable out.
Set up the equation symbolically first (with letters), THEN substitute numbers. This makes it much easier to catch a wrong equation choice before you've done any arithmetic.
"Starts from rest" means $v_i = 0$. "Comes to rest" or "stops" means the final $v = 0$. Watch for these phrases — they hand you a variable for free.
A cart or object that reverses direction, slows down, or changes its motion mid-problem may need to be split into two separate intervals, each with its own constant acceleration — the equations above only work within ONE interval of constant $a$.

3. Free Fall — Special Case of Section 2

Variable definitions

$g$ = acceleration due to gravity $\approx 9.8\text{ m/s}^2$, always directed downward

Tips

Free fall is just the same 4 kinematic equations above, with $a = g$.
Pick a sign convention before you start and hold it for the whole problem. Most common: up = positive, so $a = -9.8\text{ m/s}^2$.
Thrown upward and caught at the top: $v = 0$ at that instant, but $a$ is STILL $-9.8\text{ m/s}^2$ the whole time — gravity never turns off, even when the object is momentarily motionless.
A trip that goes up and comes back down to the same height is symmetric: time going up = time coming down, and the speed returning to the starting height equals the initial launch speed.

4. Projectile Motion (2D) — Topic 1.5

Variable definitions

$v_0$ = initial (launch) speed
$\theta$ = launch angle, measured above the horizontal
$v_x$ = horizontal component of velocity — constant the entire flight (no horizontal acceleration)
$v_{0y}$ = initial vertical component of velocity
$h$ or $H$ = height (of a cliff/table, or maximum height reached)

Components of the launch velocity: $$v_x = v_0\cos\theta \qquad\qquad v_{0y}=v_0\sin\theta$$

Key results (launched and landing at the SAME height): $$t_{up} = \frac{v_{0y}}{g} \qquad t_{total}=\frac{2v_{0y}}{g}=\frac{2v_0\sin\theta}{g} \qquad h_{max}=\frac{v_{0y}^2}{2g} \qquad \text{range} = v_x \times t_{total}$$

Launched horizontally (θ = 0°, e.g., off a table or cliff): $v_{0y}=0$, so solve $h=\tfrac{1}{2}gt^2$ for time first, then horizontal distance $=v_x \times t$.

Tips

Horizontal and vertical motion are completely independent. Solve the vertical direction first to find the time in the air — then use that SAME time in the horizontal direction.
Changing the horizontal launch speed changes how FAR a projectile travels, but never changes how LONG it's in the air — time of flight depends only on the vertical motion (initial vertical velocity and $g$).
Mass never appears in any of these equations — it does not affect the height, range, or time of flight of a projectile.
Instantaneous speed at any point is the Pythagorean combination of the horizontal and vertical velocity components at that instant: $v=\sqrt{v_x^2+v_y^2}$.

5. Reference Frames and Relative Velocity (1D) — Topic 1.4

Variable definitions

$v_{A-B}$ = velocity of object A relative to object B (read the subscripts right to left: "A relative to B")

The rule: $$v_{A-C} = v_{A-B} + v_{B-C}$$

Velocities relative to a chain of frames add together, as long as the "inner" subscripts match up (the $B$ in $v_{A-B}$ cancels with the $B$ in $v_{B-C}$).

Tips

Pick one positive direction and use it for every velocity in the problem — this is the single most common place to lose points. If a velocity is given as moving in the negative direction, its value goes into the equation as a negative number.
Work one layer at a time when there are three or more reference frames (e.g., passenger → boat → shore): find the passenger's velocity relative to the boat first, then combine that with the boat's velocity relative to the shore.
"Relative to the ground/shore/road" almost always means the stationary, outside observer's frame — that's usually the frame the question is actually asking about, even if the given information is in a different, moving frame.

Keep this sheet out during the Unit 1 Test — it will not be collected, and you won't need to memorize any of these equations before test day.