Position & Velocity vs Time Graph Simulator

The simulator models one object moving along a line with constant acceleration for the first 10 seconds. The top graph is position $x$ against time (from −20 m to +20 m), and the bottom graph is velocity $v$ against time (from −15 m/s to +15 m/s). Both share the same time axis, marked every 2 s.

Below the two equations, a readout reports where the velocity is zero, for example “Velocity = 0 at t = 6 s (turning point on x–t)”. If the velocity never reaches zero between 0 and 10 s it says there is no turning point in that window, and with a = 0 it notes that the position graph is a straight line.

If a curve runs past the top or bottom of a graph, it is drawn flat along the edge. That flat stretch is just the plot running out of room, not the object stopping, so pick smaller values if you need to see the whole motion.

For constant acceleration, the two graphs are described by the equations shown in the simulator: $$x = x_0 + v_0 t + \tfrac{1}{2} a t^2 \qquad v = v_0 + a t$$ On the AP Physics 1 equation sheet these appear with an $x$ subscript, $v_x = v_{x0} + a_x t$ and $x = x_0 + v_{x0} t + \tfrac{1}{2} a_x t^2$, along with $v_x^2 = v_{x0}^2 + 2a_x(x - x_0)$.

The graphs are linked by two rules. The slope of the position–time graph at any instant is the velocity, and the slope of the velocity–time graph is the acceleration. Going the other way, the area between the velocity–time graph and the time axis equals the displacement, with area below the axis counting as negative.

When $a$ is constant and not zero, the position graph is a parabola. It curves upward (concave up) when $a > 0$ and downward when $a < 0$. The vertex of that parabola is the turning point, where $v = 0$, at $$t = -\frac{v_0}{a}$$

An object speeds up when $v$ and $a$ have the same sign and slows down when they have opposite signs. The sign of the acceleration alone does not tell you whether the object is speeding up.

An object starts at $x_0 = 2$ m with $v_0 = -4$ m/s and a constant acceleration of $a = +0.8$ m/s². Describe its motion over the first 10 s.

Turning point. $t = -v_0/a = -(-4)/0.8 = 5$ s. At that moment $x = 2 + (-4)(5) + \tfrac{1}{2}(0.8)(5)^2 = 2 - 20 + 10 = -8$ m.

At t = 10 s. $x = 2 + (-4)(10) + \tfrac{1}{2}(0.8)(10)^2 = 2 - 40 + 40 = 2$ m, and $v = -4 + (0.8)(10) = +4$ m/s.

Check with area. From 0 to 5 s the velocity graph is a triangle below the axis with area $\tfrac{1}{2}(5)(-4) = -10$ m. From 5 to 10 s it is a triangle above the axis with area $\tfrac{1}{2}(5)(4) = +10$ m. The total displacement is 0, which matches the object ending where it started, but the distance traveled is 20 m.

Speeding up or slowing down? For the first 5 s, $v$ is negative and $a$ is positive, so the object slows down even though the acceleration is positive. After 5 s both are positive, so it speeds up.

To see this in the simulator, set x₀ = 2, v₀ = −4 and a = 0.8. The readout should say velocity = 0 at t = 5 s, the position parabola should open upward with its lowest point at −8 m, and the velocity line should cross zero at 5 s and finish at +4 m/s.

Converting between representations of motion is central to Unit 1 (Topic 1.3, Representing Motion). Multiple-choice questions often show one graph and ask you to pick the matching one, and the Translation Between Representations free-response question can ask you to sketch $v$–$t$ from $x$–$t$ or to justify a sketch with slope and area reasoning. The same ideas return in later units whenever you read a graph of a quantity against time.