Uniformly Accelerated Motion (Distance-Time Law)

Insert acceleration, time and initial velocity into s = ½at² + v₀t. Example: a car starts from rest (v₀ = 0) with a = 2 m/s²; after t = 5 s, s = ½ × 2 × 25 = 25 m. If the body already moves at v₀, the term v₀t is added: with v₀ = 10 m/s it would be 25 + 50 = 75 m. Units matter: time in seconds, acceleration in m/s², velocity in m/s; divide km/h by 3.6 first. The quadratic term means the body covers more distance in the second second than in the first, so the distance-time diagram is a parabola.

Only when the body starts from rest, meaning v₀ = 0. That is the case when pulling away at traffic lights, in free fall from your hand, or at a rocket launch. As soon as an initial velocity exists, a braking car or a train speeding up, you must use the full form s = ½at² + v₀t, otherwise the entire uniform-motion part is missing. When braking, additionally insert a as negative: a car with v₀ = 20 m/s and a = −5 m/s² stops after t = 4 s and covers s = ½·(−5)·16 + 20·4 = −40 + 80 = 40 m of braking distance.

Then the time-free equation v² = v₀² + 2·a·s helps, linking speed and distance directly. It arises by inserting t = (v−v₀)/a from the velocity law into the distance law. Braking example: a car travels at v₀ = 27.8 m/s (100 km/h) and brakes at a = −7 m/s² to a standstill (v = 0). Then 0 = 27.8² + 2·(−7)·s, so s = 772.8/14 ≈ 55 m. This equation also explains why braking distance grows quadratically with speed: twice the speed, four times the braking distance. In exams it is the standard route whenever speeds at the end of a stretch are asked for.

They describe the same motion on two levels: v(t) = a·t + v₀ is the derivative of s(t) = ½at² + v₀t. Intuitively, the distance is the area under the v-t diagram. For constant acceleration this area is a trapezoid made of a rectangle (v₀·t) and a triangle (½·a·t·t = ½at²), exactly the two terms of the distance law. Conversely, the slope of the s-t parabola at any moment is the instantaneous velocity, and the slope of the v-t line is the acceleration. If you master this diagram logic, you can solve many problems graphically without rearranging a single formula, often the fastest route in exams.

Starting from rest, the time appears squared: s = ½at². Substituting 2t gives ½a·(2t)² = ½a·4t², four times as much. Physically this is because the body not only travels longer but is also faster at the end: it accumulates far more distance in the second half of the time than in the first. Concretely, a uniformly accelerating body covers distances in the ratio 1 : 3 : 5 : 7 in successive equal time intervals (odd numbers); Galileo already measured this pattern on his inclined plane. For exams this is a quick sanity check: ratio problems can often be solved via the square law without any number crunching.