Motion Calculator: Uniform, Accelerated, Free Fall

The two equations everything follows from

All of first-year kinematics lives in two equations: v = v₀ + a·t, which says how the speed changes, and s = v₀·t + ½a·t², which says how much ground is covered. Uniform motion is the case where the acceleration is zero and the second reduces to s = v·t; free fall is the case where the initial speed is zero and the acceleration is g. These are not different formulas to memorise but the same pair with one term vanishing.

The third relation, v² = v₀² + 2a·s, comes from eliminating time between the other two. It is needed exactly when the problem never mentions time: “a car at 30 m/s brakes at 5 m/s², in how many metres does it stop”. Solving it via time works, but takes two steps instead of one.

In projectile motion the two directions are handled separately, and that is the whole point of the topic: horizontally the motion is uniform, because no force acts that way, while vertically it is uniformly accelerated downwards. Time is the only shared quantity, and it is what links the range to the maximum height.

Common mistakes

Using km/h instead of m/s: 90 km/h is 25 m/s, and mixing the two units throws the answer off by a factor of 3.6. Divide by 3.6 to go from km/h to m/s.
Getting the sign of the acceleration wrong when braking: if the speed falls, the acceleration opposes the motion and is negative. With a positive sign the car speeds up and there is no stopping distance.
Confusing distance travelled with displacement: a body thrown up and caught again has zero displacement although it covered twice the maximum height. The calculator reports displacement, which is the quantity in the equations.

Frequently asked questions

What is the formula for uniformly accelerated motion?

There are two: v = v₀ + a·t for the speed and s = v₀·t + ½a·t² for the distance. If time is unknown, use v² = v₀² + 2a·s, obtained by eliminating t between the first two.

How long does a body take to fall from a given height?

t = √(2h/g). From 45 metres, with g = 9.81 m/s², it takes about 3.03 seconds and lands at nearly 30 m/s. Mass does not appear: with no air, all bodies fall in the same time.

Why is the range greatest at 45 degrees?

Because the range is v₀²·sin(2α)/g, and the sine peaks when its argument is 90°, that is when the launch angle is 45°. It also follows that complementary angles, such as 30° and 60°, give the same range with different flight times.

Do these calculations account for air resistance?

No. They are the formulas of ideal kinematics, the ones used in school problems. For a stone or a ball over short distances the approximation is good; for a skydiver or a feather it is not.

How this calculation works

Uniform motion: s = v·t. Uniformly accelerated motion: v = v₀ + a·t and s = v₀·t + ½a·t². Time-free relation (Torricelli): v² = v₀² + 2a·Δs, so Δs = (v² − v₀²)/(2a). Free fall from rest: t = √(2h/g), v = g·t, with g = 9.81 m/s². Projectile motion with speed v₀ and angle α: components v₀ₓ = v₀·cos α and v₀ᵧ = v₀·sin α; time of flight T = 2v₀ᵧ/g; range R = v₀ₓ·T = v₀²·sin(2α)/g; maximum height H = v₀ᵧ²/(2g). Every formula assumes constant acceleration and no air resistance.

Dynamics and Newton's laws

Second law, weight, friction, inclined plane, centripetal force and impulse, with the formula inverting itself.

Collisions and momentum

Head-on collisions, elastic, inelastic and in between: final velocities, energy lost and impulse exchanged.

Rotational dynamics

Moment of inertia, the parallel axis theorem, torque, angular momentum and rolling, with the formula that rearranges itself.

Universal gravitation

Newton's law, surface gravity, orbital and escape speed, Kepler's third law and gravitational potential energy.