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I HAVE A UNIT 1 PHYSICS AP TEST , and i know nothing i have 2 hours and i need you to give me notes that can help me learn everthing and how to use the first 4 formulas on the formula sheet

YES. You have 2 hours, so we are NOT trying to learn every tiny detail. We’re going to focus on what AP Physics 1 Unit 1 actually tests, the first 4 equations on the AP formula sheet, and—most importantly—how to recognize which equation to use.

🚨 AP PHYSICS 1 UNIT 1 CRASH COURSE

⏰ Your 2-Hour Plan

1. THE BIG IDEA OF UNIT 1

Unit 1 is basically KINEMATICS.

Kinematics = describing how objects move.

You will mostly deal with:

• Position

• Displacement

• Velocity

• Acceleration

• Time

• Motion graphs

• Free fall

The basic story:

Position → Velocity → Acceleration

Think of it like:

Where am I? → How fast/direction am I moving? → How is my velocity changing?

2. POSITION

Position tells you where an object is relative to a reference point.

Usually we call position:

xx

Example:

You are 10 m to the right of a tree.

x=+10mx=+10m

If you are 5 m to the left:

x=−5mx=-5m

IMPORTANT

Positive and negative don't automatically mean "good" or "bad."

They represent direction.

You have to choose a positive direction.

For example:

➡ right = positive

⬅ left = negative

3. DISPLACEMENT

Displacement is change in position.

Δx=xf−xi\boxed{\Delta x=x_f-x_i}

where:

• xfx_f = final position

• xix_i = initial position

Example

You start at:

xi=2mx_i=2m

and finish at:

xf=10mx_f=10m

Then:

Δx=10−2\Delta x=10-2 Δx=8m\boxed{\Delta x=8m}

⚠ DISTANCE vs DISPLACEMENT

This is VERY important.

Distance

How much ground you traveled.

Displacement

How far you ended up from where you started with direction.

Example:

You walk:

➡ 10 m

then:

⬅ 4 m

Distance:

10+4=14m10+4=14m

Displacement:

10−4=6m10-4=6m

So:

Distance = 14 m

Displacement = 6 m right

4. VELOCITY

Velocity tells you how quickly position changes.

Average velocity:

vavg=ΔxΔt\boxed{v_{avg}=\frac{\Delta x}{\Delta t}}

Units:

m/sm/s

Example

You move 20 m in 4 seconds.

v=204v=\frac{20}{4} v=5m/s\boxed{v=5m/s}

🚨 SPEED vs VELOCITY

Speed

How fast.

Velocity

How fast AND in what direction.

So:

20m/s20m/s

is speed.

But:

20m/s east20m/s\text{ east}

is velocity.

5. ACCELERATION

Acceleration = change in velocity over time.

a=ΔvΔt\boxed{a=\frac{\Delta v}{\Delta t}}

or

a=vf−viΔt\boxed{a=\frac{v_f-v_i}{\Delta t}}

Units:

m/s2m/s^2

Example

A car goes from:

10m/s10m/s

to

30m/s30m/s

in 5 seconds.

a=30−105a=\frac{30-10}{5} a=4m/s2\boxed{a=4m/s^2}

🚨 IMPORTANT: ACCELERATION DOES NOT MEAN "SPEEDING UP"

This is a huge AP Physics concept.

An object accelerates whenever its velocity changes.

That can mean:

• speeding up

• slowing down

• changing direction

6. THE FIRST 4 FORMULAS ON YOUR FORMULA SHEET

This is what you REALLY need to know.

The first four kinematics equations are:

FORMULA 1

v=v0+at\boxed{v=v_0+at}

FORMULA 2

Δx=v0t+12at2\boxed{\Delta x=v_0t+\frac12at^2}

FORMULA 3

v2=v02+2aΔx\boxed{v^2=v_0^2+2a\Delta x}

FORMULA 4

Δx=12(v0+v)t\boxed{\Delta x=\frac12(v_0+v)t}

🧠 WHAT ALL THE LETTERS MEAN

Memorize this:

0 = starting

no 0 = ending

So:

v0=starting velocityv_0 = \text{starting velocity} v=ending velocityv = \text{ending velocity}

🔥 HOW DO I KNOW WHICH FORMULA TO USE?

THIS is the trick.

Before touching a formula, write down what you know.

For example:

A car starts at 5 m/s and accelerates at 2 m/s² for 4 seconds. What is its final velocity?

Known:

v0=5v_0=5 a=2a=2 t=4t=4

Looking for:

v=?v=?

Which formula contains those things?

v=v0+at\boxed{v=v_0+at}

DONE.

FORMULA 1

v=v0+at\boxed{v=v_0+at}

Use it when you know:

• initial velocity

• acceleration

• time

and want:

• final velocity

Example

A car starts at 10 m/s and accelerates at 3 m/s² for 4 seconds.

v=10+(3)(4)v=10+(3)(4) v=22m/s\boxed{v=22m/s}

FORMULA 2

Δx=v0t+12at2\boxed{\Delta x=v_0t+\frac12at^2}

Use it when you know:

• initial velocity

• acceleration

• time

and want:

• displacement

Example

A car starts from rest and accelerates at:

4m/s24m/s^2

for:

3s3s

"Starts from rest" means:

v0=0v_0=0

So:

Δx=(0)(3)+12(4)(32)\Delta x=(0)(3)+\frac12(4)(3^2) Δx=2(9)\Delta x=2(9) Δx=18m\boxed{\Delta x=18m}

FORMULA 3

v2=v02+2aΔx\boxed{v^2=v_0^2+2a\Delta x}

USE THIS ONE WHEN THERE IS NO TIME.

🚨 This is the easiest way to recognize it.

If the problem gives you:

• v0v_0

• vv

• aa

• Δx\Delta x

but NOT time

think:

FORMULA 3

Example

A car starts at 5 m/s and accelerates at 2 m/s² for a distance of 20 m.

Find final velocity.

We know:

v0=5v_0=5 a=2a=2 Δx=20\Delta x=20

No time.

So:

v2=52+2(2)(20)v^2=5^2+2(2)(20) v2=25+80v^2=25+80 v2=105v^2=105

Take the square root:

v≈10.25m/s\boxed{v\approx10.25m/s}

FORMULA 4

Δx=12(v0+v)t\boxed{\Delta x=\frac12(v_0+v)t}

This is useful when you know:

• initial velocity

• final velocity

• time

and want:

• displacement

Example

A car travels for 5 seconds.

Initial velocity:

4m/s4m/s

Final velocity:

10m/s10m/s

Then:

Δx=12(4+10)(5)\Delta x=\frac12(4+10)(5) Δx=12(14)(5)\Delta x=\frac12(14)(5) Δx=35m\boxed{\Delta x=35m}

🚨 THE BIG FORMULA CHEAT SHEET

Memorize this table:

EVEN BETTER:

Ask:

Does the problem give me time?

If NO:

v2=v02+2aΔx\boxed{v^2=v_0^2+2a\Delta x}

If YES, look at what you're trying to find.

7. "STARTS FROM REST"

You WILL see this.

"Starts from rest" means:

v0=0\boxed{v_0=0}

Example:

A ball starts from rest and accelerates at 5 m/s².

That means:

v0=0v_0=0

NOT:

v=0v=0

The initial velocity is zero.

8. FREE FALL

This is extremely important.

Near Earth's surface:

g≈9.8m/s2\boxed{g\approx9.8m/s^2}

Sometimes AP problems use:

g=10m/s2g=10m/s^2

depending on the problem.

Gravity always accelerates objects downward.

If UP is positive:

a=−9.8m/s2\boxed{a=-9.8m/s^2}

If DOWN is positive:

a=+9.8m/s2\boxed{a=+9.8m/s^2}

You choose the coordinate system.

🚨 THROWING A BALL UP

Suppose you throw a ball upward at:

20m/s20m/s

Choose up as positive.

Then:

v0=20m/sv_0=20m/s a=−9.8m/s2a=-9.8m/s^2

At the highest point:

v=0\boxed{v=0}

This is VERY important.

The ball isn't accelerating at zero at the top.

Instead:

v=0v=0

but:

a=−9.8m/s2a=-9.8m/s^2

9. MOTION GRAPHS

You NEED these.

There are three main graphs:

Position vs Time

x−tx-t

Velocity vs Time

v−tv-t

Acceleration vs Time

a−ta-t

POSITION-TIME GRAPH

The slope tells you velocity.

slope of x−t=v\boxed{\text{slope of }x-t=v}

Straight horizontal line:

Object isn't moving.

v=0v=0

Positive slope:

Positive velocity.

Negative slope:

Negative velocity.

VELOCITY-TIME GRAPH

The slope tells you acceleration.

slope of v−t=a\boxed{\text{slope of }v-t=a}

Positive slope:

a>0a>0

Negative slope:

a<0a<0

Horizontal line:

a=0a=0

🚨 AREA UNDER VELOCITY GRAPH

This is HUGE.

The area under a velocity-time graph gives:

Δx\boxed{\Delta x}

So:

slope of velocity graph → acceleration

area under velocity graph → displacement

ACCELERATION-TIME GRAPH

The area under an acceleration-time graph gives:

Δv\boxed{\Delta v}

So:

10. CONSTANT VELOCITY

If:

a=0a=0

then velocity doesn't change.

Example:

v=10m/sv=10m/s

for 5 seconds.

Then:

Δx=vt\Delta x=vt Δx=(10)(5)\Delta x=(10)(5) 50m\boxed{50m}

11. VECTORS

AP Physics loves directions.

A vector has:

magnitude + direction

Examples:

20m/s east20m/s\text{ east} 5m north5m\text{ north}

Velocity and displacement are vectors.

Speed and distance are scalars.

12. NEGATIVE VELOCITY DOES NOT NECESSARILY MEAN SLOWING DOWN

This is another BIG AP trap.

Imagine:

v=−20m/sv=-20m/s

and:

a=−5m/s2a=-5m/s^2

Velocity and acceleration have the same sign.

The object is speeding up in the negative direction.

But:

v=−20v=-20 a=+5a=+5

They have opposite signs.

The object is slowing down.

Memorize:

Same direction → speeding up

Opposite directions → slowing down

13. HOW TO SOLVE ANY KINEMATICS PROBLEM

Use this EVERY TIME.

STEP 1 — Pick positive direction

Example:

Up = +

STEP 2 — Write what you know

v0=?v_0=? v=?v=? a=?a=? t=?t=? Δx=?\Delta x=?

STEP 3 — Identify what you're looking for.

STEP 4 — Pick an equation.

STEP 5 — Substitute.

STEP 6 — Check units and direction.

🧠 EXAMPLE

A car starts from rest and accelerates at 3m/s23m/s^2 for 6 seconds. Find its final velocity.

Step 1:

Starts from rest:

v0=0v_0=0

Step 2:

a=3a=3 t=6t=6

Step 3:

Want:

v=?v=?

Step 4:

Use:

v=v0+atv=v_0+at

Step 5:

v=0+(3)(6)v=0+(3)(6)

Answer:

18m/s\boxed{18m/s}

🔥 NOW A HARDER ONE

A car is traveling at 8 m/s and accelerates at 2 m/s² over a distance of 30 m. Find its final velocity.

Known:

v0=8v_0=8 a=2a=2 Δx=30\Delta x=30

Want:

vv

Do we know time?

❌ NO.

So immediately:

v2=v02+2aΔx\boxed{v^2=v_0^2+2a\Delta x}

Substitute:

v2=82+2(2)(30)v^2=8^2+2(2)(30) v2=64+120v^2=64+120 v2=184v^2=184 v=184v=\sqrt{184} v≈13.6m/s\boxed{v\approx13.6m/s}

🚨 AP PHYSICS WORDS YOU NEED TO KNOW

"Starts from rest"

v0=0v_0=0

"Comes to rest"

v=0v=0

"Constant velocity"

a=0a=0

"Constant acceleration"

Acceleration doesn't change.

"Drops"

Usually:

v0=0v_0=0

unless the problem says it was thrown.

"Thrown upward"

Initial velocity is positive if you choose up as positive.

"Highest point"

v=0v=0

"Returns to starting point"

Δx=0\Delta x=0

⚠ This does NOT mean distance = 0.

🧪 MINI PRACTICE TEST

Don't look at the answers immediately.

1.

A car starts from rest and accelerates at 4m/s24m/s^2 for 5 seconds.

What is its final velocity?

2.

A car starts from rest and accelerates at 2m/s22m/s^2 for 6 seconds.

What is its displacement?

3.

A car travels 50 m while accelerating from 5 m/s at 3m/s23m/s^2.

What is its final velocity?

4.

A ball is thrown upward at 20 m/s.

What is its velocity at the highest point?

5.

If a velocity-time graph has a slope of −4m/s2-4m/s^2, what is the acceleration?

6.

What does the area under a velocity-time graph represent?

✅ ANSWERS

1.

v=v0+atv=v_0+at v=0+(4)(5)v=0+(4)(5) 20m/s\boxed{20m/s}

2.

Δx=v0t+12at2\Delta x=v_0t+\frac12at^2 Δx=0+12(2)(62)\Delta x=0+\frac12(2)(6^2) 36m\boxed{36m}

3.

NO TIME.

So:

v2=v02+2aΔxv^2=v_0^2+2a\Delta x v2=52+2(3)(50)v^2=5^2+2(3)(50) v2=325v^2=325 v≈18.0m/s\boxed{v\approx18.0m/s}

4.

0m/s\boxed{0m/s}

5.

−4m/s2\boxed{-4m/s^2}

6.

Displacement\boxed{\text{Displacement}}

🚨 YOUR 5-MINUTE MEMORIZATION SHEET

Before your test, know THESE:

DEFINITIONS

Δx=xf−xi\boxed{\Delta x=x_f-x_i} vavg=ΔxΔt\boxed{v_{avg}=\frac{\Delta x}{\Delta t}} a=ΔvΔt\boxed{a=\frac{\Delta v}{\Delta t}}

THE 4 EQUATIONS

v=v0+at\boxed{v=v_0+at}

Final velocity

Δx=v0t+12at2\boxed{\Delta x=v_0t+\frac12at^2}

Displacement

v2=v02+2aΔx\boxed{v^2=v_0^2+2a\Delta x}

NO TIME

Δx=12(v0+v)t\boxed{\Delta x=\frac12(v_0+v)t}

Displacement using initial + final velocity

GRAPHS

Position-time slope = velocity

Velocity-time slope = acceleration

Velocity-time area = displacement

Acceleration-time area = change in velocity

GRAVITY

g=9.8m/s2\boxed{g=9.8m/s^2}

Downward.

If up is positive:

a=−9.8m/s2\boxed{a=-9.8m/s^2}

MOST IMPORTANT TEST TRICKS

Starts from rest → v0=0v_0=0

Comes to rest → v=0v=0

Top of trajectory → v=0v=0

Constant velocity → a=0a=0

No time given → Formula 3

Velocity and acceleration same direction → speeding up

Velocity and acceleration opposite directions → slowing down

🧠 THE ONE THING I WANT YOU TO DO RIGHT NOW

Don't just read these notes again.

Take the 6 practice questions above, cover the answers, and solve them yourself. Then send me a picture of your work/answers (even if you think they're completely wrong).

I can then do a rapid-fire AP Physics 1 Unit 1 review with you, question by question, and focus ONLY on the things you're getting wrong.