Kinematics Calculator: Solve for Any Variable

Enter any three of the five motion variables (initial velocity, final velocity, acceleration, time, displacement) and the calculator above solves the constant-acceleration equations for the rest. Pick equations by the variable each one leaves out, and keep one sign convention throughout.

final velocity (v)

11 m/s

Full solution: v0 = 5 m/s, v = 11 m/s, a = 2 m/s^2, t = 3 s, dx = 24 m.

Steps

  1. 1.Given: v0 = 5 m/s, a = 2 m/s^2, t = 3 s, dx = 24 m. Solve for v.
  2. 2.v = v0 + a t = 5 m/s + 2 m/s^2 x 3 s = 11 m/s

AP Physics: Unit 1 (topics 1.2 Displacement, Velocity, and Acceleration). Constant-acceleration motion is Topic 1.2 in Unit 1 of AP Physics 1, and Unit 1 carries 10 to 15 percent of the multiple-choice section. The same constant-acceleration model also opens Unit 1 of AP Physics C: Mechanics.

How to use the calculator above

Type any three of the five variables into the calculator above and it solves the constant-acceleration equations for whatever is missing. The five inputs match the symbols on the AP Physics 1 equation sheet: initial velocity v0v_0 (written vx0v_{x0} on the sheet), final velocity vv, acceleration aa, time tt, and displacement Δx\Delta x (the sheet writes it as xx0x - x_0).

Every kinematics equation connects four of these five quantities, so three known values are enough to pin down the remaining two. The calculator picks the equation that skips your unknown, substitutes your numbers, and reports each result with units. Use it to check homework, not to replace the setup: on the AP free-response section you still need to write the equation, show the substitution, and do the algebra yourself.

The three equations on the AP sheet

The AP Physics 1 equation sheet prints exactly three kinematics equations:

vx=vx0+axtv_x = v_{x0} + a_x t
x=x0+vx0t+12axt2x = x_0 + v_{x0} t + \frac{1}{2} a_x t^2
vx2=vx02+2ax(xx0)v_x^2 = v_{x0}^2 + 2 a_x (x - x_0)

All three assume constant acceleration. Each one is missing exactly one variable: the first has no displacement, the second has no final velocity, and the third has no time.

A fourth equation you may have learned, Δx=v0+v2t\Delta x = \frac{v_0 + v}{2} t, is not printed on the sheet. It follows from the definition of average velocity: displacement equals average velocity times time, and with constant acceleration the average velocity is the average of v0v_0 and vv. You are free to use it, just know where it comes from. For full derivations, see the kinematic equations guide.

Pick the equation by its missing variable

The fastest reliable method: list what you know, note what you want, and find the one variable that is neither. Then use the equation that does not contain that variable.

Variable you neither know nor needEquation to use
Displacement Δx\Delta xv=v0+atv = v_0 + at
Final velocity vvΔx=v0t+12at2\Delta x = v_0 t + \frac{1}{2} a t^2
Time ttv2=v02+2aΔxv^2 = v_0^2 + 2a\Delta x
Acceleration aaΔx=v0+v2t\Delta x = \frac{v_0 + v}{2} t (not on the sheet)

Say a problem gives v0v_0, aa, and Δx\Delta x and asks for vv. Time is neither given nor requested, so the no-time equation v2=v02+2aΔxv^2 = v_0^2 + 2a\Delta x finishes the job in one step. The calculator above runs this same selection logic, which makes it a quick way to confirm you picked the equation a grader would expect.

Sign conventions: the biggest source of wrong answers

Every variable in these equations except time is a vector component, so signs carry direction. Before substituting anything, pick a positive direction and apply it to all five variables at once.

  • Ball thrown upward with up as positive: v0v_0 is positive and a=9.8 m/s2a = -9.8 \ \text{m/s}^2 for the entire flight, including at the very top.
  • Car braking while moving in the positive direction: v0v_0 is positive and aa is negative. Deceleration just means the acceleration points opposite the velocity.
  • A negative Δx\Delta x in your result means the object ended up on the negative side of its starting point. That is direction information, not an algebra mistake.

The calculator above uses exactly the signs you type, so enter 9.8-9.8 for a rising ball when up is positive. On free-response work, state your positive direction before the first substitution so every sign in your solution has a defined meaning.

When constant acceleration applies

These equations work only while the acceleration stays constant in magnitude and direction. That covers the classic Unit 1 situations: free fall near Earth's surface with a=9.8 m/s2a = 9.8 \ \text{m/s}^2 downward, cars speeding up or braking at a steady rate, and objects pushed by a constant net force. They do not cover a rocket burning off fuel or any interval where the acceleration changes value. For those, split the motion into constant-acceleration segments and chain them: the final velocity of one segment becomes the initial velocity of the next.

Projectiles are the special case you will meet most often: constant 9.8 m/s29.8 \ \text{m/s}^2 downward vertically and zero acceleration horizontally. Split the motion into axes and these same equations handle each axis separately, which is exactly what the projectile motion calculator automates. All of this sits in Topic 1.2 of AP Physics 1 Unit 1, which carries 10 to 15 percent of the multiple-choice section.

Common mistakes the calculator can catch

Solve by hand first, then compare with the calculator's output. When the two disagree, the cause is usually one of these:

  • Unit mismatches: 72 km/h must become 20 m/s before it touches an equation. Convert everything to meters and seconds first.
  • Dropping the 12\frac{1}{2} in Δx=v0t+12at2\Delta x = v_0 t + \frac{1}{2} a t^2, or forgetting to square the time.
  • Entering +9.8+9.8 for gravity after declaring up as positive.
  • Treating the top of a vertical arc as v=0v = 0 in two-dimensional problems. Only the vertical component is zero there; the horizontal component never changes.
  • Ignoring the second root when solving for time. Both roots of the quadratic can be physical, as the third worked example below shows.

If the setup itself feels shaky, the kinematic equations guide walks through choosing knowns and unknowns, and the projectile motion guide covers the two-dimensional version. Kinematics is the foundation the rest of the course builds on, so a fast start here is a real part of how hard AP Physics 1 actually is to manage.

Braking distance from the no-time equation

A car moving at 24.0 m/s24.0 \ \text{m/s} brakes with a constant acceleration of 6.0 m/s2-6.0 \ \text{m/s}^2 (direction of motion taken as positive). How far does it travel before stopping?

  1. List the knowns: v0=24.0 m/sv_0 = 24.0 \ \text{m/s}, v=0v = 0, a=6.0 m/s2a = -6.0 \ \text{m/s}^2. Time is neither given nor asked for, so use the no-time equation v2=v02+2aΔxv^2 = v_0^2 + 2a\Delta x.

  2. Substitute: 0=(24.0 m/s)2+2(6.0 m/s2)Δx0 = (24.0 \ \text{m/s})^2 + 2(-6.0 \ \text{m/s}^2)\Delta x, which becomes 0=576 m2/s2(12.0 m/s2)Δx0 = 576 \ \text{m}^2/\text{s}^2 - (12.0 \ \text{m/s}^2)\Delta x.

  3. Solve: Δx=(576 m2/s2)/(12.0 m/s2)=48.0 m\Delta x = (576 \ \text{m}^2/\text{s}^2) / (12.0 \ \text{m/s}^2) = 48.0 \ \text{m}.

The car travels 48.0 m48.0 \ \text{m} before stopping. Entering v0=24.0v_0 = 24.0, v=0v = 0, and a=6.0a = -6.0 into the calculator above returns the same displacement plus the stopping time, t=4.0 st = 4.0 \ \text{s}.

Time to the top and maximum height

A volleyball leaves a player's hand moving straight up at 14.7 m/s14.7 \ \text{m/s}. Taking up as positive, find the time to reach the highest point and the maximum height above the release point.

  1. At the highest point the vertical velocity is zero, so the knowns are v0=+14.7 m/sv_0 = +14.7 \ \text{m/s}, v=0v = 0, and a=9.8 m/s2a = -9.8 \ \text{m/s}^2.

  2. Time from the no-displacement equation: v=v0+atv = v_0 + at gives 0=14.7+(9.8)t0 = 14.7 + (-9.8)t, so t=(14.7 m/s)/(9.8 m/s2)=1.50 st = (14.7 \ \text{m/s}) / (9.8 \ \text{m/s}^2) = 1.50 \ \text{s}.

  3. Height from the no-time equation: v2=v02+2aΔxv^2 = v_0^2 + 2a\Delta x gives 0=(14.7 m/s)2(19.6 m/s2)Δx=216.09 m2/s2(19.6 m/s2)Δx0 = (14.7 \ \text{m/s})^2 - (19.6 \ \text{m/s}^2)\Delta x = 216.09 \ \text{m}^2/\text{s}^2 - (19.6 \ \text{m/s}^2)\Delta x.

  4. Solve: Δx=(216.09 m2/s2)/(19.6 m/s2)=11.0 m\Delta x = (216.09 \ \text{m}^2/\text{s}^2) / (19.6 \ \text{m/s}^2) = 11.0 \ \text{m}.

The ball rises for 1.50 s1.50 \ \text{s} and peaks 11.0 m11.0 \ \text{m} above the release point.

Two times at the same height (a quadratic in t)

A ball is thrown straight up at 19.6 m/s19.6 \ \text{m/s}. At what times is it 14.7 m14.7 \ \text{m} above the launch point? Take up as positive, so a=9.8 m/s2a = -9.8 \ \text{m/s}^2.

  1. Final velocity is neither given nor wanted, so use Δx=v0t+12at2\Delta x = v_0 t + \frac{1}{2} a t^2: substituting gives 14.7=19.6t+12(9.8)t2=19.6t4.9t214.7 = 19.6t + \frac{1}{2}(-9.8)t^2 = 19.6t - 4.9t^2 (meters throughout).

  2. Rearrange into standard form: 4.9t219.6t+14.7=04.9t^2 - 19.6t + 14.7 = 0. Divide every term by 4.9: t24.00t+3.00=0t^2 - 4.00t + 3.00 = 0.

  3. Factor: (t1.00)(t3.00)=0(t - 1.00)(t - 3.00) = 0, so t=1.00 st = 1.00 \ \text{s} or t=3.00 st = 3.00 \ \text{s}.

  4. Interpret both roots: the ball passes 14.7 m14.7 \ \text{m} on the way up at 1.00 s1.00 \ \text{s} and again on the way down at 3.00 s3.00 \ \text{s}. Neither root is extraneous.

t=1.00 st = 1.00 \ \text{s} on the way up and t=3.00 st = 3.00 \ \text{s} on the way down. Whenever a height is reached twice, the quadratic produces two positive roots; match the root to the moment the question asks about.

Frequently asked questions

How do I know which kinematic equation to use?

Find the variable that is neither given nor requested, then use the equation that leaves it out. The first AP equation has no displacement, the second has no final velocity, and the third has no time. This missing-variable check works on nearly every constant-acceleration problem.

Why are there only three kinematics equations on the AP equation sheet?

The AP Physics 1 sheet prints exactly three. The fourth classic equation, displacement equals average velocity times time, is not printed because it follows directly from the definition of average velocity. You may still use it on the exam; it is just not handed to you.

What sign should I use for gravity?

The AP sheet uses a magnitude of 9.8 meters per second squared for g. If you choose up as positive, the acceleration during free fall is negative 9.8; if you choose down as positive, it is positive 9.8. Pick one convention per problem and never switch midway.

Can I use a calculator on the AP Physics 1 exam?

Yes. A four-function, scientific, or graphing calculator is allowed on both the multiple-choice and free-response sections. This page's tool is for practice and checking; on free-response questions you still earn points by writing the equation and showing the substitution.

Does this calculator handle projectile motion?

One axis at a time. Projectile problems split into a horizontal axis with zero acceleration and a vertical axis with constant downward acceleration, and this tool can solve either axis. For the full two-dimensional treatment, use the projectile motion calculator.