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Recovering Velocity from Acceleration

Integrate acceleration component by component, then use the given velocity at one time to determine and check the complete velocity function.

Intermediate 9 min read

What You'll Learn

  • Use the vector-valued fundamental theorem of calculus to recover velocity from acceleration.
  • Integrate each acceleration component separately.
  • Apply an initial velocity at a specified time to fix the constant vector.
  • Work with polynomial and trigonometric acceleration functions.
  • Check a recovered velocity by differentiating it and testing the initial value.

From Acceleration Back to Velocity

Acceleration describes how velocity changes over time. If the acceleration function is known, integrating it recovers the change in velocity. To determine the velocity itself, we also need its value at one time: without that information, the integral leaves an unknown constant vector.

This is the reverse of the relationship \( \mathbf{a}(t)=\mathbf{v}'(t) \). The calculation is performed component by component, but the result is a single vector-valued velocity function. The initial velocity determines the constant in that function.

Throughout the examples, time \(t\) is measured numerically in seconds, the acceleration components are in metres per second squared, and velocity components are in metres per second. Trigonometric arguments are in radians, with coefficients understood to carry the units needed for the stated acceleration units.

Definition: If a particle has acceleration \(\mathbf{a}(t)\) on an interval, a velocity function \(\mathbf{v}(t)\) is a function satisfying \(\mathbf{v}'(t)=\mathbf{a}(t)\). Recovering velocity means finding such a function that also matches the given velocity at a specified time.

The Recovery Formula

Let \(t_0\) be a time in the interval, and suppose the velocity at that time is \(\mathbf{v}(t_0)=\mathbf{v}_0\). If \(\mathbf{a}\) is continuous between \(t_0\) and \(t\), the vector-valued fundamental theorem of calculus gives the velocity directly:

$$ \mathbf{v}(t)=\mathbf{v}_0+\int_{t_0}^{t}\mathbf{a}(s)\,ds. $$

Here \(s\) is a dummy variable of integration; the resulting function depends on \(t\). For three-dimensional motion, write \(\mathbf{a}(s)=\langle a_x(s),a_y(s),a_z(s)\rangle\). The integral is evaluated in each component:

$$ \mathbf{v}(t) = \left\langle v_{0x}+\int_{t_0}^{t}a_x(s)\,ds,\; v_{0y}+\int_{t_0}^{t}a_y(s)\,ds,\; v_{0z}+\int_{t_0}^{t}a_z(s)\,ds \right\rangle. $$

This definite-integral form automatically uses the initial velocity and avoids having to solve for separate constants afterward. Alternatively, one can find an antiderivative of each acceleration component and then use the initial velocity to determine the constants.

Why the Formula Works

The justification follows by applying the one-variable fundamental theorem of calculus to each component. Write the velocity as \(\mathbf{v}(t)=\langle v_x(t),v_y(t),v_z(t)\rangle\). Since \(\mathbf{v}'(t)=\mathbf{a}(t)\), each coordinate satisfies \(v_x'(t)=a_x(t)\), \(v_y'(t)=a_y(t)\), and \(v_z'(t)=a_z(t)\). Therefore, for example, the first component obeys

$$ v_x(t)-v_x(t_0)=\int_{t_0}^{t}v_x'(s)\,ds =\int_{t_0}^{t}a_x(s)\,ds. $$

The same reasoning applies to the other two components. Combining the three scalar equalities gives the vector equation \( \mathbf{v}(t)-\mathbf{v}(t_0)=\int_{t_0}^{t}\mathbf{a}(s)\,ds \). Substituting \(\mathbf{v}(t_0)=\mathbf{v}_0\) gives the recovery formula.

When it applies: The formula requires a specified initial velocity and an acceleration that is integrable on the interval of interest. Continuity of the acceleration is a standard sufficient condition for applying the fundamental theorem in this form.

Worked Examples

Worked Example: Polynomial Acceleration with an Initial Velocity at Zero

Suppose \( \mathbf{a}(t)=\langle 6t,4,-2\rangle \) m/s² and \( \mathbf{v}(0)=\langle 1,-3,5\rangle \) m/s. Find \(\mathbf{v}(t)\).

Integrate each component from the initial time \(0\) to \(t\), then add the corresponding component of the initial velocity:

$$ \begin{aligned} v_x(t)&=1+\int_0^t 6s\,ds =1+\left[3s^2\right]_0^t =1+3t^2,\\ v_y(t)&=-3+\int_0^t 4\,ds =-3+\left[4s\right]_0^t =-3+4t,\\ v_z(t)&=5+\int_0^t (-2)\,ds =5+\left[-2s\right]_0^t =5-2t. \end{aligned} $$

Thus, in metres per second,

$$ \mathbf{v}(t)=\langle 1+3t^2,\,-3+4t,\,5-2t\rangle. $$

Check by differentiating: \( \mathbf{v}'(t)=\langle 6t,4,-2\rangle=\mathbf{a}(t) \). At the initial time, \( \mathbf{v}(0)=\langle 1,-3,5\rangle \), as required.

Worked Example: Trigonometric Acceleration

Suppose \( \mathbf{a}(t)=\langle 3t^2,2\cos t,-4\sin(2t)\rangle \) m/s² and \( \mathbf{v}(0)=\langle 0,1,-2\rangle \) m/s. Find the velocity function.

Integrate the three components, keeping the initial value for each component:

$$ \begin{aligned} v_x(t)&=0+\int_0^t 3s^2\,ds =\left[s^3\right]_0^t =t^3,\\ v_y(t)&=1+\int_0^t 2\cos s\,ds =1+\left[2\sin s\right]_0^t =1+2\sin t,\\ v_z(t)&=-2+\int_0^t -4\sin(2s)\,ds\\ &=-2+\left[2\cos(2s)\right]_0^t =-2+2\cos(2t)-2 =2\cos(2t)-4. \end{aligned} $$

The \(z\)-component uses the chain rule in reverse: the derivative of \(2\cos(2s)\) is \(-4\sin(2s)\). The recovered velocity is

$$ \mathbf{v}(t)=\langle t^3,\;1+2\sin t,\;2\cos(2t)-4\rangle. $$

Differentiate to check: \( \mathbf{v}'(t)=\langle 3t^2,2\cos t,-4\sin(2t)\rangle \). At \(t=0\), the components are \(0\), \(1+2\sin 0=1\), and \(2\cos 0-4=-2\). Both the acceleration and the initial velocity match the data.

Worked Example: Initial Velocity at a Nonzero Time

Suppose \( \mathbf{a}(t)=\langle 2t,3t^2-1,4\rangle \) m/s² and \( \mathbf{v}(2)=\langle 5,-1,3\rangle \) m/s. Find the velocity for times in the interval containing \(2\).

Use the specified initial time \(2\) as the lower limit. The integration variable \(s\) keeps the limit and the variable \(t\) distinct:

$$ \begin{aligned} v_x(t)&=5+\int_2^t 2s\,ds =5+\left[s^2\right]_2^t =5+t^2-4 =t^2+1,\\ v_y(t)&=-1+\int_2^t (3s^2-1)\,ds\\ &=-1+\left[s^3-s\right]_2^t =-1+(t^3-t)-(8-2) =t^3-t-7,\\ v_z(t)&=3+\int_2^t 4\,ds =3+\left[4s\right]_2^t =3+4t-8 =4t-5. \end{aligned} $$

Therefore,

$$ \mathbf{v}(t)=\langle t^2+1,\;t^3-t-7,\;4t-5\rangle. $$

Differentiate: \( \mathbf{v}'(t)=\langle 2t,3t^2-1,4\rangle \), which is the given acceleration. Check the initial value at \(t=2\): the components are \(2^2+1=5\), \(2^3-2-7=-1\), and \(4(2)-5=3\). Hence \( \mathbf{v}(2)=\langle 5,-1,3\rangle \), as required.

Why an Initial Velocity Is Necessary

Integrating acceleration determines how velocity changes, but acceleration alone does not say what the velocity was at a particular time. For instance, if two velocity functions have the same derivative, their difference has derivative zero and is therefore constant on the interval. In three dimensions, that undetermined constant is a vector: each component can have its own constant value.

The initial velocity supplies all three component values at one time. It is important to use the stated initial time, not automatically assume that the data are given at \(t=0\). The definite-integral formula handles a nonzero initial time directly.

Common Pitfall: Using One Constant for All Components

Consider \( \mathbf{a}(t)=\langle 2t,3\rangle \) with \( \mathbf{v}(0)=\langle 1,4\rangle \). An antiderivative has the form \( \langle t^2+C_x,3t+C_y\rangle \). A common error is to set both constants equal to \(1\), perhaps using only the first component of the initial velocity. That gives \( \langle t^2+1,3t+1\rangle \), whose initial value is \( \langle 1,1\rangle \), not \( \langle 1,4\rangle \).

Instead, use the initial value component by component. From \(v_x(0)=1\), \(C_x=1\); from \(v_y(0)=4\), \(C_y=4\). Thus the correct function is

$$ \mathbf{v}(t)=\langle t^2+1,\,3t+4\rangle. $$

Its derivative is \( \langle 2t,3\rangle \), and its value at zero is \( \langle 1,4\rangle \). The check catches the error immediately.

A Reliable Procedure

For a smooth acceleration function and a known velocity at time \(t_0\), the calculation can be organized as follows:

1
Record the data.
Write the acceleration components and the full initial velocity vector, including the time at which that velocity is specified.
2
Integrate component by component.
Use either the definite-integral formula from \(t_0\) to \(t\), or antiderivatives with separate constants for each component.
3
Apply the initial velocity.
In the antiderivative approach, substitute \(t=t_0\) and match each component to its given initial value.
4
Check both conditions.
Differentiate the proposed velocity to recover the acceleration, then evaluate it at \(t_0\) to recover the initial velocity.

These two checks address the two pieces of information in the problem. A function can have the right derivative but the wrong initial value, or the right initial value but the wrong derivative. A valid recovered velocity must satisfy both.

Check Your Understanding

For each question, keep the velocity components separate and check your result by differentiation and evaluation at the initial time.

  1. If \( \mathbf{a}(t)=\langle 4t,2\rangle \) and \( \mathbf{v}(0)=\langle -1,3\rangle \), what is \(\mathbf{v}(t)\)?
  2. If \( \mathbf{a}(t)=\langle \cos t,-\sin t\rangle \) and \( \mathbf{v}(0)=\langle 2,0\rangle \), find the velocity function.
  3. Explain why an acceleration function by itself does not determine a unique velocity function.
  4. For an initial velocity given at \(t=3\), what lower limit should appear in the definite-integral recovery formula?
  5. What two checks should be made on a proposed velocity function?

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