Kinematics Equations Calculator

The Four Equations and the Variable Each One Skips

Every constant-acceleration problem uses five quantities: initial velocity $v_0$, final velocity $v$, acceleration $a$, time $t$ and displacement $\Delta x$. Each kinematic equation contains four of them and skips one.

OpenStax writes the same set with $x - x_0$ in place of $\Delta x$. Some courses add a fifth, $\Delta x = vt - \tfrac{1}{2}at^2$, which skips $v_0$. In SI units, velocities are in m/s, acceleration in m/s², time in seconds and displacement in meters. Mixing km/h with seconds is the fastest way to get an answer that's off by a factor of 3.6, so convert first or set the unit menus above.

Choosing an Equation by What the Problem Never Mentions

A problem gives you three of the five quantities and asks for a fourth. The fifth one goes unmentioned, and that tells you which equation to use. A braking problem gives a starting speed, a final speed of zero and a deceleration, then asks how far the car travels. Time never comes up, so you reach for $v^2 = v_0^2 + 2a\Delta x$, the one equation without $t$.

The wording hides some of the givens. "From rest" means $v_0 = 0$. "Comes to a stop" means $v = 0$. "Dropped" means $v_0 = 0$, and with up as positive, $a = -9.8\ \mathrm{m/s^2}$. At the top of a vertical throw, $v = 0$.

The six modes above follow this logic. Each one starts from a set of three knowns and returns the other two quantities, so a stopping-distance problem also gives you the stopping time.

Worked Example: A Car Merging Onto the Highway

OpenStax gives a car merging into traffic. It starts at 10.0 m/s, accelerates at 2.00 m/s² and covers 200 m. How long does that take?

1. Final velocity isn't given and time is the unknown, so use $\Delta x = v_0 t + \tfrac{1}{2}at^2$.
2. Substitute: $200 = 10.0t + \tfrac{1}{2}(2.00)t^2$, which rearranges to $t^2 + 10t - 200 = 0$.
3. Factor or use the quadratic formula: $t = 10.0$ s or $t = -20.0$ s.
4. Drop the negative root. It would mean the merge happened 20 s before the car started accelerating.

The car merges after 10.0 s, moving at $10.0 + 2.00 \times 10.0 = 30.0$ m/s. Enter $v_0 = 10$, $a = 2$ and $\Delta x = 200$ in the "Find v and t" mode and both numbers come out without a quadratic. The tool finds $v$ from $v^2 = v_0^2 + 2a\Delta x$ first, then gets the time from $t = (v - v_0)/a$.

Signs in Free Fall

Choose a positive direction before you write any number, and keep it for the full problem. OpenStax takes up as positive, so $a = -g = -9.8\ \mathrm{m/s^2}$ even while a ball is still rising. A ball thrown downward at 4.9 m/s from the top of a 98 m building has $v_0 = -4.9$ m/s and $\Delta x = -98$ m. It hits the ground after 4.0 s at $-44.1$ m/s, and the minus sign says it's moving down.

A baseball hit straight up and caught 5.0 s later at the same height left the bat at 24.5 m/s. It climbs for 2.5 s to a peak of 30.6 m, falls for another 2.5 s, and comes down at $-24.5$ m/s. At the peak its velocity is zero. Its acceleration is still 9.8 m/s² downward.

When the Square Root Gives Two Answers

$v^2 = v_0^2 + 2a\Delta x$ hands you $v^2$, and every positive number has two square roots. OpenStax's rocket booster leaves its release point moving up at 200.0 m/s with only gravity acting on it. At 1.0 × 10³ m above release, $v = \pm 142.8$ m/s. Both roots describe the booster: +142.8 m/s on the way up and $-142.8$ m/s on the way back down. When a point is passed twice like this, the calculator reports the first pass and gives the time and velocity of the second in the note under the result.

Where the Equations Stop Applying

The equations hold only while the acceleration stays constant. When it changes, split the motion into phases and let the final velocity of one phase be $v_0$ for the next. OpenStax's car at 30.0 m/s covers 15.0 m during a 0.500 s reaction time, at constant speed, before the brakes engage. Braking at $-7.00$ m/s² on dry concrete adds 64.3 m, for 79.3 m in total. On wet concrete at $-5.00$ m/s², braking takes 90.0 m and the total grows to 105 m.

Free-fall problems also leave out air resistance. That works for a ball or a booster, and less well for a feather on Earth. The Physics Classroom drops its feather on the Moon instead, where $g = 1.67\ \mathrm{m/s^2}$ and a 1.40 m fall takes 1.29 s.

Kinematics on the Chem/Phys Section

Introductory physics makes up about 25 percent of Chem/Phys, according to the AAMC outline summarized in the MCAT sections guide, and there's no calculator. Prep guides such as Shemmassian's use $g \approx 10\ \mathrm{m/s^2}$ for projectile work so the arithmetic stays clean. With that value, a jump to a height of 1.25 m needs a takeoff speed of $\sqrt{2 \times 10 \times 1.25} = 5$ m/s, and the hang time is $2 \times 5/10 = 1.0$ s. The Physics Classroom gets 5.03 m/s and 1.03 s for a 1.29 m leap using 9.8, close enough that the rounding rarely changes which answer choice you pick.

Square roots are the slow step without a calculator. The no-calculator math guide covers estimating them, and the Arithmetic Trainer drills the same skills. The kinematic equations aren't among the 18 equations the AAMC outline writes out, as the formula memorization guide counts them, but motion problems still need them from memory. The Formula Writing Drill has you write each one from its name.