AP Physics 1 Kinematics Study Guide: Equations & Practice Problems
Cheatsheet and worksheet downloads are found at bottom of the page. This Kinematics guide covers the definitions, the four constant-acceleration equations, motion graphs, free fall and projectiles, then gives you 24 practice problems with a full answer key and a projectile simulator you can experiment with.
Written by Berke SahbazogluOwner and tutor, Burke Tutoring In Fremont · BA Biochemistry, Washington University in St. Louis · 10+ years tutoring math and science, including AP Physics 1 · Full bio Published September 9, 2026 · Last updated September 9, 2026 · Free to print and use in class
What kinematics is. Kinematics describes how objects move using position, velocity, and acceleration. It does not explain the forces that cause the motion. That comes in Unit 2. Your first job in any kinematics problem is to choose a positive direction, list what you know with signs, and pick a representation that makes the situation clear.
01 / BUILD THE FOUNDATION
Position, velocity, and acceleration
A scalar has magnitude only. A vector has magnitude and direction. In one dimension you do not need arrows or components. A positive or negative sign carries the direction, relative to the axis you chose.
Quantity
Meaning
SI unit
Position, x
Location relative to an origin. A coordinate can be negative.
m
Displacement, Δx
xf − xi. The change in position, with direction.
m
Distance
Total path length traveled. A nonnegative scalar.
m
Average velocity
Δx / Δt. Displacement per elapsed time.
m/s
Average speed
Total distance / Δt. Not the same as average velocity.
m/s
Average acceleration
Δv / Δt. Change in velocity per elapsed time.
m/s2
Worth doing in your head: walk 3 m east, then 1 m west, in 4 s. With east positive, displacement is +2 m and distance is 4 m. Average velocity is +0.5 m/s and average speed is 1 m/s. Two different questions, two different answers, same trip.
Instantaneous versus average
Instantaneous velocity tells you how fast and in which direction an object is moving at one moment. On a position-time graph it is the slope of the tangent at that moment. Average velocity is the slope of the straight line between two points.
Negative acceleration does not mean slowing down. An object speeds up when velocity and acceleration share a sign, and slows down when their signs are opposite. A car moving in the negative direction with negative acceleration is speeding up.
02 / CHOOSE YOUR TOOL
The constant-acceleration equations
These four relationships apply over an interval only when acceleration is constant through that interval. Here v0 is initial velocity, v is final velocity, Δx is displacement, a is acceleration, and t is elapsed time. Each equation leaves out one variable, which is how you choose between them.
v = v0 + at
Leaves out displacement.
Δx = v0t + ½at2
Leaves out final velocity.
v2 = v02 + 2aΔx
Leaves out time.
Δx = ½(v0 + v)t
Leaves out acceleration.
A method you can repeat under time pressure
Sketch the motion and mark a positive direction on the sketch.
List every known value with its sign and unit, and circle the one you want.
Pick the equation that contains your target and leaves out the variable you neither know nor need.
Solve symbolically first, then substitute. Check the sign and the units at the end.
If the acceleration changes partway through, split the motion into stages and solve each one separately. The final position and velocity of one stage become the initial conditions of the next. When you take a square root to find a velocity, the algebra gives you two roots and the physical situation tells you which sign to keep.
03 / READ THE PICTURE
Motion graphs: slope and area
Graph
Slope gives you
Signed area gives you
Position versus time
Velocity
No standard quantity used in this unit
Velocity versus time
Acceleration
Displacement
Acceleration versus time
Rate of change of acceleration (not needed here)
Change in velocity
On a velocity-time graph, area above the time axis counts as positive displacement and area below counts as negative. Add the signed areas to get displacement. Add the magnitudes to get distance.
Key
Blue line Velocity, +2.0 to +8.0 m/s
Lower region Rectangle, area = 6.0 m
Upper region Triangle, area = 9.0 m
Slope Acceleration, +2.0 m/s2
Total area Displacement, +15 m
The slope of this line is the acceleration. The area beneath it is the displacement. Check whether the question asks for acceleration or displacement before choosing slope or area.
With constant positive acceleration, a position-time graph curves upward, a velocity-time graph is a straight line with positive slope, and an acceleration-time graph is a horizontal line above zero. A horizontal position graph means the object is at rest. A horizontal velocity graph means constant velocity, which is not the same thing.
Motion diagrams and measured data
Dots drawn at equal time intervals are position snapshots. Growing gaps mean increasing speed, equal gaps mean constant speed, shrinking gaps mean slowing down. For measured data, calculate the slope over each interval or fit a line to the graph that should be straight. Starting from rest at constant acceleration, Δx = ½at2, so a plot of Δx against t2 is a straight line with slope a/2. The slope of this linearized graph can be used to determine acceleration from measured data.
04 / EXTEND TO TWO DIMENSIONS
Free fall and projectile motion
In free fall near Earth's surface, with air resistance neglected, acceleration is constant and directed downward. This guide uses g = 9.8 m/s2. If a question specifies a different value, use theirs.
Taking upward as positive, ay = −g. An object thrown straight up has positive vertical velocity going up, zero vertical velocity at the top, and negative vertical velocity coming down. Its acceleration is −g the entire time, including at the top.
Two directions, one clock
A projectile has zero horizontal acceleration and constant downward vertical acceleration. Solve the two directions separately, but the elapsed time is shared. That shared time is what connects them. With launch speed v0 at angle θ above the horizontal:
v0x = v0 cos θ
Δx = v0xt
Horizontal velocity never changes.
v0y = v0 sin θ
Δy = v0yt − ½gt2
Upward is positive.
For a horizontal launch, v0y = 0 and the fall time depends only on the height. For a launch that lands at the same height it started, the time of flight is 2v0y/g and the range is v02sin(2θ)/g. These shortcuts assume equal launch and landing heights. For unequal heights, solve the vertical displacement equation for time, then use that time to find the horizontal range.
Relative velocity in one dimension: vA/B = vA/ground − vB/ground. Two cars heading east at 5 m/s and 3 m/s: the faster one moves at +2 m/s as seen from the slower one, and the slower one moves at −2 m/s as seen from the faster one. Reversing the subscripts reverses the sign.
05 / EXPERIMENT WITH IT
Projectile simulator
Move the sliders and watch what changes. The axes rescale independently, so compare the numeric results rather than the apparent steepness of the curve. Predict how the time, range, and maximum height will change, then adjust one slider to check your prediction.
The horizontal and vertical axes rescale independently; use the labels to read distances. Trajectory updates as you change the controls. The numeric results are listed below the sliders.
Trajectory
Launch velocity and apex
Landing point
Guides to maximum height
At the same angle on level ground, doubling speed quadruples range.
On level ground, 45° gives the greatest range.
Above zero, the level-ground shortcuts stop working.
Watch the four results while you drag this one.
3.28Time in air (s)
62.8Range (m)
13.2Maximum height (m)
25.0Impact speed (m/s)
The mass slider does nothing, and that is the lesson. Drag it from 0.1 kg to 10 kg and every number above holds still. In this uniform-gravity model with no air resistance, mass does not affect acceleration or the trajectory. A bowling ball and a marble launched identically land at the same time in the same place. In force calculations, mass relates net force to acceleration through Fnet = ma.
Three things worth checking with it. First, set the height to zero and compare 30° with 60°: the ranges match, but the times and heights do not. Second, raise the launch height and watch the optimum angle for range drop below 45°. Third, set the angle to zero for a horizontal launch and confirm that changing the speed moves the landing point without changing the time in the air.
06 / PUT IT TO WORK
Kinematics Worked Examples: Motion Graphs and Projectiles
Each one starts with a sign convention and ends with a physical check, which is the habit I try to build in every student before the first test.
Example 1 · Motion graphs
A cart accelerates along a track
A cart's velocity increases uniformly from +2.0 m/s to +8.0 m/s over 3.0 s. Find its acceleration and its displacement during the interval. Take rightward as positive.
The graph in section 3 is this motion. Press play below to watch it happen, then work through the steps.
The numeric values for time, velocity and position are listed below the animation.
Cart
Velocity vector, drawn to scale
Marker dropped every 0.50 s
Scale below the track is in metres from the start
0.00Time (s)
2.00Velocity (m/s)
0.00Position (m)
Watch the gaps between the markers. They are dropped at equal time intervals, so the growing spacing is the acceleration made visible. Equal gaps would mean constant velocity. This runs at about a third of real speed.
List what you know. v0 = +2.0 m/s, v = +8.0 m/s, t = 3.0 s.
Acceleration is the slope. a = (v − v0) / t = (8.0 − 2.0) / 3.0 = +2.0 m/s2
Displacement is the area. Rectangle: (2.0 m/s)(3.0 s) = 6.0 m Triangle: ½(3.0 s)(8.0 − 2.0 m/s) = 9.0 m Δx = 6.0 + 9.0 = +15 m
Check it against an equation. Δx = ½(v0 + v)t = ½(2.0 + 8.0)(3.0) = 15 m. The cart never reverses, so the distance is also 15 m.
Answer: acceleration +2.0 m/s2, displacement +15 m, both rightward.
Example 2 · Projectile motion
A ball rolls off a platform
A ball leaves a platform 1.25 m high, moving horizontally at 4.00 m/s. How long is it in the air, and how far does it travel horizontally? Neglect air resistance, treat the ball as a particle, and use g = 9.8 m/s2.
Key
Launch point Origin, (0, 0)
Short arrow v0x = 4.00 m/s
Downward arrow ay = −g = −9.8 m/s2
Vertical bar Drop height, 1.25 m
Landing point Δy = −1.25 m, Δx = 2.0 m
Put the origin at the launch point and take up as positive. The ground is then at Δy = −1.25 m, which is where the negative sign in step 2 comes from.
Split the directions. Horizontal: v0x = +4.00 m/s, ax = 0 Vertical: v0y = 0, ay = −9.8 m/s2, Δy = −1.25 m
Vertical motion gives the time. Δy = v0yt + ½ayt2 −1.25 = 0 − ½(9.8)t2 t = √(2 × 1.25 / 9.8) = 0.505 s, taking the positive root
The same time goes into the horizontal equation. Δx = v0xt = (4.00)(0.5051) = 2.02 m
Check the physics. The fall time came from the height alone. Doubling the launch speed would double the horizontal distance and leave the time in the air untouched. You can confirm that with the simulator above by pressing "Worked example 2" and then dragging the speed slider.
Answer: 0.51 s in the air and 2.0 m horizontally. Both are given to two significant figures, because g = 9.8 m/s2 carries only two.
07 / LEARN FROM OTHER PEOPLE'S ERRORS
Where students lose points
These six errors can lead to incorrect answers. Use the check beneath each one to review your setup.
1. Setting acceleration to zero at the top of a toss
The velocity is zero there, so the acceleration looks like it should be too. It is not. Gravity does not switch off at the top of the arc, and if it did the object would hang there.
Check: acceleration describes how velocity changes. At the top, velocity is momentarily zero, but it continues to change because gravitational acceleration acts downward.
2. Using the level-ground time of flight when the heights differ
T = 2v0y/g is only valid when the projectile lands at the height it launched from. Students apply it to a ball thrown off a cliff or a roof every year, and the answer comes out short.
Check: before using the shortcut, ask whether the start and end heights are equal. If they are not, go back to Δy = v0yt − ½gt2 and solve the quadratic.
3. Taking the area under a position-time graph
Area under a velocity-time graph gives displacement. Area under a position-time graph does not give distance or displacement. This happens when a student memorizes "slope and area" as a pair without attaching them to specific graphs.
Check: multiply the units. Under a v-t graph, (m/s)(s) = m, a displacement. Under an x-t graph, (m)(s) = m·s, which is not a quantity in this course.
4. Changing sign conventions partway through
A student takes down as positive to make g positive, then writes the initial upward velocity as positive too. Every number after that is wrong even though every equation was right.
Check: write your positive direction on the sketch before you write a single number, then confirm the sign of each quantity against that arrow.
5. Keeping only the positive root
Solving v2 = C for positive C gives v = ±√C. A quadratic can have two distinct real roots, one repeated real root, or no real roots. The negative time is usually discarded correctly, but a negative velocity often gets discarded too, and then a falling object is reported as moving upward.
Check: after taking a root, say out loud which direction the object is actually moving and pick the sign that matches. For a problem asking about motion after launch at t = 0, discard negative-time roots. Negative velocity is valid when motion is in the negative direction.
6. Answering distance when the question asked for displacement
Free-response prompts choose the word deliberately. If an object travels in both directions, distance exceeds the magnitude of displacement. Even without reversal, displacement can be negative while distance is positive.
Check: underline the quantity word in the prompt before you start. If the object reverses, compute both and label them.
08 / PRACTICE
24 AP Physics 1 Kinematics Practice Problems With Answers
Work these on paper with a sketch for each one. Answers are in the collapsible key under each part, and the full worked solutions are in the free answer key PDF. Assume straight-line motion for Part B. For Part C, use uniform gravity with g = 9.8 m/s2 and neglect air resistance. Keep unrounded values between steps; report final numerical answers to two significant figures. Intermediate values are shown in the key where useful.
Part A: Reading motion and graphs
Definitions, motion diagrams, and getting information out of a graph.
A runner jogs 120 m east in 30 s, turns around, and jogs 40 m west in 10 s. Taking east as positive, find the displacement, the distance, the average velocity, and the average speed for the whole trip. The trip is sketched below.
Key
Left dot Start, at the origin
Upper arrow First leg, 30 s
Lower arrow Second leg, 10 s
Green dot Finish
Use the graph below. Find the velocity and the displacement over the interval shown.
Key
Blue line Position, +6.0 m to −2.0 m
Orange point Passes the origin
Dashed guides Read the endpoint values
Slope The quantity you want first
Use the graph below. Find the total displacement over the full 7.0 s and the acceleration during the second phase.
Key
Blue line Velocity
Shaded region Area = displacement
First phase Constant, 0 to 3.0 s
Second phase Straight to zero at 7.0 s
Use the graph below. Find the acceleration, the displacement, and the distance traveled over the 4.0 s.
Key
Blue line Velocity, +6.0 to −6.0 m/s
Orange point Direction reverses here
Upper region Positive area
Lower region Negative area
An object has negative velocity and positive acceleration. Is it speeding up or slowing down? Explain in one sentence.
An object starts from rest and has a constant acceleration of +2.0 m/s2 for 5.0 s. Find its final velocity and its displacement.
An object moves along a straight line in the positive direction. A motion diagram shows dots every 0.10 s with successive gaps of 2.0, 4.0, 6.0, and 8.0 cm. Are these data consistent with constant acceleration? Under that model, find the acceleration. Can the dots alone rule out acceleration changes between measurements?
An object starts from rest under constant acceleration. Measurements give Δx = 0.75 m at 1.0 s, 3.0 m at 2.0 s, and 6.75 m at 3.0 s. Plot Δx against t2 conceptually, find the slope, and find the acceleration.
Show answers for Part A
Displacement = 120 − 40 = +80 m; distance = 120 + 40 = 160 m. Over 40 s, average velocity = 80/40 = +2.0 m/s and average speed = 160/40 = 4.0 m/s.
Area = (4)(3) + ½(4)(4) = +20 m. During the second phase, a = (0 − 4)/(7 − 3) = −1.0 m/s2.
a = (−6 − 6)/4 = −3.0 m/s2. The two triangular areas are +6 m and −6 m: displacement = 0 m, distance = 12 m.
Slowing down at that instant: positive acceleration makes the negative velocity less negative, reducing its magnitude.
v = at = (2.0)(5.0) = +10 m/s; Δx = ½at2 = ½(2.0)(5.0)2 = +25 m.
The interval-average velocities are 0.20, 0.40, 0.60, and 0.80 m/s. Under constant acceleration they equal velocities at interval midpoints, spaced 0.10 s apart. Thus a = 0.20/0.10 = +2.0 m/s2. The data are consistent with this model but do not rule out variation between dots.
Plot points (t2, Δx) = (1, 0.75), (4, 3.0), and (9, 6.75). The line passes through the origin with slope 0.75 m/s2. Since slope = a/2, a = 1.5 m/s2.
Part B: Constant acceleration in one dimension
Choosing the right equation, and handling motion that comes in stages.
A car starts from rest and accelerates at 3.0 m/s2 for 5.0 s. Find its final speed and the distance covered.
A car traveling at 25 m/s brakes uniformly to a stop in 40 m. Find the acceleration.
For the car in problem 10, how long does the stop take?
An object is launched along a track at +8.0 m/s with a constant acceleration of −2.0 m/s2. When does it return to its starting point, and what is its velocity then?
A cyclist accelerates from rest at 2.0 m/s2 for 6.0 s, then holds that speed for 10 s. Find the total distance.
A train moving at 30 m/s decelerates at 1.2 m/s2. How far does it travel in the first 5.0 s, and how far does it travel during the fifth second alone?
Object A starts from rest at the origin with a = 2.0 m/s2. Object B passes the origin at the same instant moving at a constant 6.0 m/s. When and where does A catch B?
A rocket sled accelerates uniformly from 50 m/s to 150 m/s over 500 m. Find the acceleration and the elapsed time.
Show answers for Part B
v = (3.0)(5.0) = 15 m/s. Distance = ½(3.0)(5.0)2 = 37.5 m, or 38 m to two significant figures.
a = (v2 − v02)/(2Δx) = (0 − 252)/(2 × 40) = −7.8125 m/s2, or −7.8 m/s2.
Use the unrounded acceleration: t = (0 − 25)/(−7.8125) = 3.2 s.
Set Δx = 8t − t2 = 0. The roots are 0 and 8 s; return occurs at 8.0 s. Then v = 8 − 2(8) = −8.0 m/s.
After acceleration, v = (2)(6) = 12 m/s and Δx1 = ½(2)(6)2 = 36 m. The next stage covers (12)(10) = 120 m. Total = 156 m, or 1.6 × 102 m.
Use x(t) = 30t − 0.6t2. x(5) = 135 m, or 1.4 × 102 m. The fifth second is t = 4 to 5 s: x(5) − x(4) = 135 − 110.4 = 24.6 m, or 25 m.
xA = t2 and xB = 6t. Set them equal: t(t − 6) = 0. Excluding the initial meeting, A catches B at 6.0 s and x = 36 m.
a = (1502 − 502)/(2 × 500) = 20 m/s2. t = (150 − 50)/20 = 5.0 s.
Part C: Free fall and projectiles
Two directions, one clock. Sketch the axes before you write anything.
A ball is dropped from rest from a height of 20 m. Find the time to fall and the speed on impact.
A ball is thrown straight up at 15 m/s. Find the time to reach the top, the maximum height above the launch point, and the total time to return to launch height.
A ball is thrown straight up at 12 m/s from a hand 1.5 m above the ground. Find its speed when it hits the ground and the total time in the air.
A stone is thrown horizontally at 20 m/s from a 45 m cliff. Find the time in the air, the horizontal distance, and the impact speed.
A projectile is launched at 25 m/s at 40° above the horizontal from level ground. Find the time of flight, the range, and the maximum height.
Two projectiles are launched at 25 m/s from level ground, one at 30° and one at 60°. Compare their ranges and their maximum heights, with numbers.
A plane flying horizontally at 60 m/s at an altitude of 500 m releases a package. Find the time to fall and the horizontal distance. Where is the plane when the package lands, assuming it holds its velocity?
At the same instant and from a 3.0 m height, one ball is dropped from rest and another is launched horizontally at 5.0 m/s. Which lands first, and how far from the base does the second one land?
Show answers for Part C
t = √(2h/g) = √(40/9.8) = 2.0203 s → 2.0 s. Impact speed = √(2gh) = 19.799 m/s → 20 m/s.
ttop = 15/9.8 = 1.5306 s → 1.5 s. Height gain = 152/(2 × 9.8) = 11.480 m → 11 m. Return time = 2(15)/9.8 = 3.0612 s → 3.1 s.
Impact speed = √(122 + 2 × 9.8 × 1.5) = 13.168 m/s → 13 m/s. Solve 0 = 1.5 + 12t − 4.9t2: the positive root is 2.5682 s → 2.6 s. Impact velocity is downward; speed has no direction.
t = √(2 × 45/9.8) = 3.0305 s → 3.0 s. Range = 20t = 60.609 m → 61 m. Impact speed = √(202 + 2 × 9.8 × 45) = 35.805 m/s → 36 m/s.
v0x = 25 cos 40° = 19.151 m/s; v0y = 25 sin 40° = 16.070 m/s. T = 2v0y/g = 3.2795 s → 3.3 s. R = v0xT = 62.807 m → 63 m. H = v0y2/(2g) = 13.175 m → 13 m.
R = 252 sin(2θ)/9.8. Since sin 60° = sin 120°, both ranges are 55.231 m → 55 m. H = (25 sin θ)2/(2 × 9.8): 8.0 m at 30° and 24 m at 60°. The 60° launch reaches three times the height.
t = √(2 × 500/9.8) = 10.1015 s → 10 s. Use that unrounded time: R = 60t = 606.09 m → 610 m. The plane is directly above the package at landing because their horizontal velocities remain equal.
They land together: each starts with zero vertical velocity and falls the same height. t = √(2 × 3.0/9.8) = 0.78246 s → 0.78 s. The launched ball travels x = 5.0t = 3.9123 m → 3.9 m.
FREE PRINTABLES · NO SIGNUP
Free AP Physics 1 Kinematics Worksheets and Cheat Sheet
The printable set includes a cheat sheet, a 24-question worksheet, and an answer key, sized for US Letter. Teachers and study groups are welcome to print and distribute these in class. A link back to this page is appreciated but not required.
AP Physics 1 Unit 1 · everything on one page · g = 9.8 m/s2
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Definitions
Displacement, Δx
xf − xi. Has direction. Can be negative.
Distance
Total path length. Never negative.
Average velocity
Δx / Δt
Average speed
distance / Δt
Average acceleration
Δv / Δt
Instantaneous v
Slope of the tangent to x versus t
Constant acceleration
v = v0 + atLeaves out Δx
Δx = v0t + ½at2Leaves out v
v2 = v02 + 2aΔxLeaves out t
Δx = ½(v0 + v)tLeaves out a
Valid only while a is constant. If a changes, split the motion into stages and carry the end conditions forward.
Motion graphs
Graph
Slope
Signed area
x versus t
Velocity
Nothing
v versus t
Acceleration
Displacement
a versus t
Not assessed
Δv
Signed area gives displacement. Add the magnitudes instead to get distance. From rest at constant a, a plot of Δx against t2 is linear with slope a/2.
Signs
Same signs on v and a: speeding up
Opposite signs on v and a: slowing down
Negative a does not mean slowing down
Pick a positive direction and never change it mid-problem
Free fall
Up positive gives ay = −g = −9.8 m/s2
At the top of a toss, v = 0 but a is still −g
Time up equals time down, for equal heights
Maximum height above launch = v0y2 / 2g
Mass appears nowhere. It does not matter.
Projectiles
Two directions, one clock. The shared elapsed time is what links them.
Horizontal
v0x = v0cosθ, ax = 0 Δx = v0xt
Vertical
v0y = v0sinθ, ay = −g Δy = v0yt − ½gt2
Level ground only: T = 2v0y/g and R = v02sin(2θ)/g. Maximum range at 45°. Complementary angles give equal ranges.
Horizontal launch: v0y = 0, so the fall time depends on the height alone.
Relative velocity, 1D
vA/B = vA/ground − vB/groundReversing the subscripts reverses the sign
Six checks before you hand it in
Did I mark a positive direction on a sketch?
Is a constant over the interval I used?
Does the question want distance or displacement?
Did I pick the right root, and does its sign match the motion?
Are the units right after the arithmetic?
Is the answer physically sensible in size?
The mistake I see most. Setting a = 0 at the top of a toss because v = 0 there. Ask what the object is doing one instant later. It is moving downward, so its velocity changed, so a is not zero.
Print it, or keep this page open while you work. Free to use in class.Download the PDF ↓
09 / CLEAR UP THE CONFUSION
Common kinematics questions
When can I use the kinematic equations?
Use the four equations above when acceleration is constant throughout the interval. If motion has several constant-acceleration stages, solve each stage separately and carry its final position and velocity into the next. If acceleration varies within a stage, these equations do not apply to that whole stage; use the supplied graph or data.
Can velocity be zero while acceleration is not?
Yes. At the highest point of a straight-up toss, velocity is zero but acceleration is still downward. Near Earth's surface, neglecting air resistance, its magnitude is approximately 9.8 m/s2. For an angled projectile, only the vertical velocity is zero at the highest point; its horizontal velocity remains constant.
Is the area under a velocity graph the distance?
Signed area gives displacement. To find distance, add the magnitudes of the areas above and below the time axis. If the object never reverses direction, distance equals the magnitude of displacement. For example, moving at −2 m/s for 3 s gives −6 m displacement and 6 m distance.
Which direction should I choose as positive?
Either direction works if you use it consistently. With upward positive, free-fall acceleration is ay = −g; with downward positive, it is ay = +g. The symbol g denotes the positive magnitude, 9.8 m/s2 in this guide. Changing axes changes component signs, not the physical motion.
Does mass affect projectile motion?
In the model used here, air resistance is neglected and gravitational acceleration is uniform. Objects released from the same position with the same initial velocity then follow the same trajectory regardless of mass. With air resistance, motion can also depend on mass, shape, cross-sectional area, and the surrounding air.
Should I use 9.8 or 10 for g?
Use the value specified in the problem or its instructions. This guide and its answer key use g = 9.8 m/s2. If you use 10 m/s2 instead, your numerical answers will differ slightly; keep the same value throughout your calculation.
Working through Unit 1 with a tutor
I tutor AP Physics 1 in person across Fremont, Newark, and Union City. Sessions can focus on choosing equations, keeping signs consistent, and connecting diagrams with graphs. Bring a recent assignment so we can work through the steps you find difficult.
Call or text to talk about your student's situation.
Scope checked against the College Board AP Physics 1 course page and the Course and Exam Description. This guide covers Unit 1 only and is an independent study resource. AP is a trademark registered by the College Board, which is not affiliated with and does not endorse this resource.