UARM Calculator

The three UARM equations and when to use each

UARM is described by three equations derived from one another. The first, v = v₀ + a·t, relates velocity to time and is used when you want to know how fast the object is moving at a given instant. The second, d = v₀·t + ½·a·t², gives the distance travelled as a function of time and is the one that appears in most free-fall problems. The third, v² = v₀² + 2·a·d, is especially useful because it does not include time: it relates velocity and distance directly, which is very handy in braking problems where time is neither known nor asked for. Choosing the right equation is the step that saves the most time when solving an exercise.

Worked example

A car travels at 20 m/s and brakes with a constant acceleration of −4 m/s². How long does it take to stop and how far does it travel? For the time we use v = v₀ + a·t with v = 0: 0 = 20 + (−4)·t, giving t = 5 s. For the distance the third equation is better, as it does not depend on time: v² = v₀² + 2·a·d → 0 = 400 + 2·(−4)·d → d = 400/8 = 50 m. The car covers 50 metres before stopping, which explains why braking distance grows so quickly with speed: since it depends on v², doubling the speed quadruples the distance.

How to read UARM graphs

In UARM the graph of velocity against time is a straight line whose slope is the acceleration: rising if the object speeds up and falling if it brakes. The area under that line equals the distance travelled, a very useful trick for solving problems graphically. The position-time graph, by contrast, is a parabola: curvature upwards indicates positive acceleration and downwards, deceleration. If the parabola reaches a maximum and then descends, it means the object stopped and reversed direction, as happens in a vertical throw when the ball reaches its highest point.

Braking and free fall, solved

Braking distance: a car at 100 km/h (27.8 m/s) brakes with a deceleration of 7 m/s², a reasonable value on dry asphalt. The distance to a stop is v² / (2a) = 27.8² / 14 ≈ 55 metres, to which you must add the reaction distance: if the driver takes one second to hit the brake, another 27.8 metres pass before braking begins. That is about 83 metres in total. The key point is that distance grows with the square of speed: at 120 km/h the braking distance alone goes from 55 to about 79 metres. Free fall: an object dropped from 45 metres, ignoring air resistance, takes t = √(2h/g) = √(90/9.8) ≈ 3.03 seconds and reaches the ground at 29.7 m/s, that is over 100 km/h. This is why a fall from a third floor is so dangerous.

Signs: the mistake that costs the most marks

Almost every error in accelerated-motion problems comes from signs, not formulas. The rule is simple: choose a positive direction before starting and keep it to the end. If you take the direction of motion as positive, braking has negative acceleration, and if an object goes up and then falls, gravitational acceleration is negative throughout, including on the way up. That is the point that confuses most: at the highest point the velocity is zero, but the acceleration is still 9.8 m/s² downwards, because gravity never stops acting. Another detail: if a time comes out negative, the problem is not wrong — that mathematical solution corresponds to an instant before the motion started and must be discarded.