When a velocity function is known, it becomes the main connection between motion quantities. Calculus lets you convert that single function into full descriptions of an object's acceleration and position.

Velocity as the central motion function

In one-dimensional mechanics, velocity often provides the most useful starting point because it links the two other key kinematic quantities. The acceleration function tells how rapidly velocity changes, while the position function tells where the object is at each instant. If $v(t)$ is already known from a model or previous derivation, the remaining task is mathematical: use differentiation to obtain acceleration and use integration to obtain position.

To complete that task, you also need information about the state of the motion at a particular time. Without that extra information, integration produces a family of possible position functions rather than a single physical answer.

Initial conditions usually appear as statements such as $x(0)=0$, $x(2)=5$, or a known value of velocity at a reference time. In this subsubtopic, they are especially important when finding position, because integration introduces an arbitrary constant.

Finding acceleration from velocity

Acceleration is obtained from the time derivative of velocity.

This step answers the question, "How quickly is the velocity function changing at each moment?" If the slope of the velocity-time graph is positive, acceleration is positive.

A velocity–time graph with distinct rising, flat, and falling segments that visually encode acceleration as slope. The steeper the line, the larger the magnitude of $a(t)=\frac{dv}{dt}$ at that time, while a horizontal segment corresponds to $a=0$. Source

If the slope is negative, acceleration is negative. If the velocity function is constant, its derivative is zero, so the acceleration is zero.

When taking this derivative, keep the sign of $v(t)$ separate from the sign of $a(t)$. A particle can have positive velocity and negative acceleration, or negative velocity and positive acceleration. Velocity describes direction of motion; acceleration describes how the velocity is changing. This distinction becomes important when analyzing whether the object is speeding up, slowing down, or changing direction.

Finding position from velocity

Position is obtained from the time integral of velocity. Because velocity is the rate of change of position, integrating velocity reverses that process. Mathematically, this produces position as a function of time, but physically it also shows that the area under a velocity-time graph over an interval gives the change in position during that interval.

This definite-integral form is especially useful because it builds the initial condition directly into the formula. If you instead use an indefinite integral, you would write $x(t)=\int v(t),dt+C$, where $C$ must then be found from a known position value. Both approaches are equivalent when used correctly.

Using initial conditions correctly

A position function is not complete until the constant or reference value is fixed. In AP Physics C, this usually means substituting a known time and position after integrating.

• Choose the reference time carefully, often $t=0$ unless the problem states otherwise.

• Substitute the known position condition exactly as given, such as $x(0)=x_0$ or $x(t_1)=x_1$.

• Solve for the integration constant or use the definite-integral form to avoid a separate constant.

• Check that the final position function has units of meters and satisfies the stated condition.

A common mistake is to confuse displacement with position. The integral of velocity over an interval gives change in position, written as $x(t)-x(t_0)$, not automatically the absolute position unless a reference position is included.

Interpreting results physically

Once you have $a(t)$ and $x(t)$, the three functions $x(t)$, $v(t)$, and $a(t)$ should form a consistent set. The derivative of position should recover velocity, and the derivative of velocity should recover acceleration. This gives a powerful self-check, especially on free-response questions where algebraic errors can easily enter during integration or differentiation.

You should also connect the functions to motion behavior:

• If $v(t)>0$, the object moves in the positive coordinate direction.

• If $v(t)<0$, the object moves in the negative coordinate direction.

• If $a(t)$ has the same sign as $v(t)$, the speed increases.

• If $a(t)$ has the opposite sign from $v(t)$, the speed decreases.

Zeros of $v(t)$ are especially important. They can indicate moments when the object is instantaneously at rest. However, a zero velocity does not always mean the motion reverses; the sign of velocity on each side of that time determines whether a true change of direction occurs. Similarly, a zero of $a(t)$ does not mean velocity is zero; it only means the velocity function has a horizontal tangent there.

In many problems, the final algebraic form of $x(t)$ may look more complicated than the form of $v(t)$ because integration introduces additional constants or different function types. That is normal. The essential requirement is that differentiating your position function gives back the original velocity function exactly.