Motion with constant acceleration

A bicycle gains the same amount of speed each second. Why does it travel farther in the second second than in the first, and why can doubling the acceleration time quadruple the distance? Let us follow how the change in velocity builds up.

Record velocity at equal time intervals: 2, 4, 6, 8 m/s. What stays equal is not the distance per second but the increase in velocity, 2 m/s. If the velocity vector changes by the same amount over any equal time intervals, acceleration is constant. This is a model: a real cyclist may pedal unevenly. It applies to the whole section only when variations in acceleration can reasonably be neglected.

To compare rates of acceleration, divide the change in the velocity component by time: a_x=(v_x-v_{0x})/t. This gives the acceleration when it is constant, or the average over the interval for arbitrary motion. The unit m/s² means a change of that many m/s each second. The same increase in half the time means twice the acceleration. An unchanged velocity gives a zero numerator, so straight-line motion at constant velocity has no acceleration.

Choose an axis pointing right. Then a_x<0 means the change in velocity points left. For a body moving right, this change reduces speed; for one already moving left, it increases speed. For example, going from −2 to −4 m/s is speeding up despite negative acceleration. A minus sign does not always mean braking. To decide whether speed is increasing, compare the directions of velocity and acceleration.

How much velocity change builds up in t seconds of constant acceleration? The same amount is added each second, so the change is a_xt. Add this to the initial component to get v_x=v_{0x}+a_xt; at t=0, the velocity component is v_{0x}. Doubling the time doubles the increase, not necessarily the whole velocity. When a_x=0, the original velocity remains: uniform motion is a special case of the formula.

Suppose a body moves right while acceleration keeps pointing left. Its velocity first decreases to zero. But zero velocity does not switch off acceleration: the next change makes the component negative, and the body moves left. A ball thrown upwards behaves this way under gravity. For a braking car, the model usually changes once it stops; do not automatically continue the earlier acceleration.

How can velocity give displacement? At constant velocity over a short time \Delta t, we get \Delta s_x=v_x\Delta t, the area of a rectangle under the graph. Adding such pieces gives the signed area under v_x(t). Above the axis it is positive; below, negative. Thus s_x=x-x_0, measured in metres, can be zero even after motion: contributions before and after reversal cancel.

With constant acceleration, v_x(t) is a straight line: equal times give equal velocity changes. The area under it is s_x=(v_{0x}+v_x)t/2. Averaging the two endpoint components works specifically because the change is linear. If velocity crosses zero, the formula still gives signed displacement. To find distance, add the absolute areas on both sides of the time axis.

Split the area under the line into a rectangle from the initial velocity and a triangle from its change. This gives s_x=v_{0x}t+a_xt^2/2. The second term contains t twice: acceleration lasts longer, and the accumulated velocity change is larger. Starting from rest, doubling time therefore quadruples displacement. At a_x=0, only v_{0x}t remains; at t=0, there is no displacement yet. Acceleration must be constant throughout the interval.

Sometimes initial and final velocities are known but time is not. Eliminate time from the velocity and displacement relations: v_x^2-v_{0x}^2=2a_xs_x. When braking while moving right, a_x<0 and s_x>0, so the final velocity squared is smaller than the initial one, as expected. Braking to rest at the same acceleration magnitude, doubling initial speed quadruples the distance. Squares do not retain the final velocity's direction.

Imagine a ball rising 5 m and returning to its launch point. Its displacement is zero; its distance is 10 m. When motion reverses, first find when v_x=0, split the motion, and add the magnitudes of the section displacements. For distance, add the absolute areas on both sides of the time axis. This explains 1:3:5:\ldots when accelerating from rest: total distances are proportional to 1^2,2^2,3^2,\ldots, and distances during individual intervals are their differences.

A car brakes uniformly from 20 m/s to rest in 8 s. Find its acceleration and braking distance.

Point the axis along the initial velocity: v₀x=20 м/с, vx=0.

Acceleration ax=(0−20)/8=−2,5 м/с². The minus sign means it points opposite to the axis.

The direction does not change before stopping. Average speed (20+0)/2=10 м/с.

Distance S=10×8=80 м. Keeping the original speed would give 160 m; braking gives a shorter distance.

The acceleration component is −2.5 m/s², its magnitude is 2.5 m/s², and the braking distance is 80 m. Speed decreases by 2.5 m/s each second. The average during braking is half the initial speed, so the distance is half the 160 m that the original speed would give.

Acceleration tells us how velocity changes. Its sign gives the direction of the change; deciding whether speed decreases also requires the velocity sign.

Constant-acceleration formulas describe the accumulation of velocity change and area under the velocity graph; the t² term comes from both processes.

After a reversal, opposite displacements subtract while distances add. Recheck the braking model once the body stops.

Next we apply acceleration to falling and changes in the direction of velocity.