With Constant Acceleration, v and x Follow Simple Formulas
Pin the acceleration to a single value. Set a with the slider, and the particle gets faster by the same amount every second; the speed piles up by a each second. Motion whose acceleration never changes is called uniformly accelerated motion. A free fall, a child going down a slide, a car braking at a steady rate all belong here. With the single condition that acceleration is constant, you no longer have to redo the differentiation and integration of the last lesson every time, because velocity and position now follow fixed formulas: v = v0 + at, x = x0 + v0t + ½at², and, with time removed, v² = v0² + 2a·Δx. These formulas are almost the whole of constant-acceleration problems. This lesson walks through why each one is true, one picture at a time.
Move the acceleration a with the slider. You see the particle's position at equal time intervals, with the velocity arrow at each instant. Watch the velocity arrows. When a is positive, each arrow grows by the same amount from one step to the next, by a every second, steadily. This is what constant acceleration really means: not that the velocity is constant, but that the amount the velocity grows is constant. That is why the position marks spread out more and more toward the end, since it keeps getting faster. Turn a up and the arrows grow more steeply and the particle travels much farther. Set a to zero and the arrows are all the same, which is exactly the constant-velocity motion from the previous lesson.
When acceleration is constant, velocity is a straight line against time. Drag time t and watch how the velocity v changes. The speed at the start, the height at t=0, is the initial velocity v0. On top of that, a is added for every second that passes. The slope of the line is exactly a. Written as a formula, v = v0 + at. Because it is a straight line, there is nothing to memorize: start at v0 and pile on a each second, that is all. If a is positive the line rises to the right (speeding up); if negative it falls to the right (slowing down). In the last lesson velocity was the slope; here that slope stays the same the whole time, which is why velocity itself comes out as a straight line.
Position, as we saw in the last lesson, is the area under the velocity graph. With constant acceleration the velocity is a straight line, so the area below is a trapezoid. Drag time t and watch how the shaded area builds up. Cutting the trapezoid in two makes its meaning clear. The rectangle along the bottom is the distance you would have covered at the initial speed alone, v0t. The triangle stacked on top is the extra distance won by accelerating, ½at². Add them and x = x0 + v0t + ½at². The starting position x0, plus the distance you would cover at constant speed, plus the bonus distance the acceleration adds. If a is zero the triangle vanishes and only the rectangle remains, giving x = x0 + v0t, plain constant velocity.
Sometimes you do not care about the time. How far must I go to reach this speed, or what speed will I have after going this far? For these, the formula with time removed is handy. Eliminating t between v = v0 + at and the x formula gives v² = v0² + 2a·Δx. As a graph, v² is a straight line in direct proportion to the distance Δx. Drag Δx. The height at the start line is v0², and the slope of the line is 2a. Double the distance and the gain in v² doubles too. The key is that it is the square of the speed, not the speed itself, that is proportional to distance. That is why a car's braking distance grows fourfold when the speed doubles. It is the formula that links speed and distance directly, without ever using time.
The most famous case of constant acceleration is gravity. Ignoring air resistance, every falling object is pulled downward with the same acceleration g, about 9.8 m/s², regardless of its weight. Switch between the three cases with the buttons: released from rest, thrown down, thrown up. Taking up as positive, the acceleration is always the same, -g. The only thing that changes is the initial velocity v0. That is why the three velocity-time lines differ only in their starting height and all share the same slope, parallel lines. For the upward throw, the moment the velocity reaches zero is the highest point, and even there the acceleration is still -g. In the end, free fall is just constant-acceleration motion with a = g, so the formulas above all carry over unchanged.