From Acceleration to a Motion Model
A particle’s acceleration describes how its velocity changes, while its velocity describes how its position changes. If acceleration is given as a function of time, integrating once gives velocity up to a constant. Integrating again gives position up to another constant.
Initial values determine those constants. With them, the resulting functions describe the particle’s motion: its position, velocity, and direction at any time in the interval where the model applies.
The Initial-Value Formulas
Suppose \(a(t)\) is continuous on an interval containing an initial time \(t_0\), and the initial velocity is \(v(t_0)=v_0\). The Fundamental Theorem of Calculus gives velocity by accumulating acceleration from \(t_0\):
If \(v(t)\) is continuous and the initial position is \(x(t_0)=x_0\), then position is obtained by accumulating velocity:
The variable \(s\) in each integral is a dummy variable of integration. It distinguishes the variable being accumulated from the endpoint \(t\). These formulas work whether \(t\) is greater than or less than \(t_0\), provided the functions are defined on the interval between them.
Why the Formulas Work
The formulas follow directly from the Fundamental Theorem of Calculus and the definitions of velocity and acceleration. They also show why the initial values are essential: an indefinite integration introduces constants, and each initial value selects the constant that makes the function match the particle’s stated motion.
By the Fundamental Theorem of Calculus, the derivative of \(\int_{t_0}^{t} a(s)\,ds\) with respect to \(t\) is \(a(t)\). Thus \(v_0+\int_{t_0}^{t}a(s)\,ds\) has derivative \(a(t)\), and its value at \(t_0\) is \(v_0\).
The derivative of \(\int_{t_0}^{t}v(s)\,ds\) is \(v(t)\). Thus \(x_0+\int_{t_0}^{t}v(s)\,ds\) has derivative \(v(t)\), and its value at \(t_0\) is \(x_0\).
Any two antiderivatives of the same function differ by a constant. Requiring \(v(t_0)=v_0\) fixes the velocity constant; requiring \(x(t_0)=x_0\) fixes the position constant.
This also verifies the formulas: differentiating the velocity formula recovers acceleration, and differentiating the position formula recovers velocity. Substitution at \(t=t_0\) checks both initial conditions.
Worked Examples
Worked Example: Acceleration Given with Initial Data at Time Zero
A particle moves on a line with acceleration \(a(t)=6t-4\) meters per second squared, where \(t\) is in seconds. At \(t=0\), its velocity is \(3\) meters per second and its position is \(2\) meters. Find the velocity and position, then describe the motion on \(0\leq t\leq2\).
First integrate acceleration. Since \(v'(t)=6t-4\),
Use \(v(0)=3\): \(3=3(0)^2-4(0)+C\), so \(C=3\). Therefore, \(v(t)=3t^2-4t+3\) meters per second.
Now integrate velocity. Since \(x'(t)=v(t)\),
Use \(x(0)=2\): \(2=0-0+0+D\), so \(D=2\). Thus \(x(t)=t^3-2t^2+3t+2\) meters.
To identify direction, complete the square:
So the particle moves in the positive direction throughout the interval. Its initial position is \(x(0)=2\) meters, and its final position is \(x(2)=8-8+6+2=8\) meters. Its displacement on this interval is \(8-2=6\) meters.
Check: Differentiating \(x(t)\) gives \(x'(t)=3t^2-4t+3=v(t)\); differentiating again gives \(v'(t)=6t-4=a(t)\). Substitution gives \(v(0)=3\) and \(x(0)=2\), as required.
Worked Example: Initial Values Specified at a Later Time
A particle has constant acceleration \(-4\) meters per second squared. At \(t=1\) second, its velocity is \(6\) meters per second and its position is \(5\) meters. Find its motion on \(1\leq t\leq4\).
Integrate acceleration: \(v(t)=-4t+C\). Using \(v(1)=6\), we get \(6=-4+C\), so \(C=10\). Hence \(v(t)=10-4t\) meters per second.
Integrate this velocity: \(x(t)=10t-2t^2+D\). Using \(x(1)=5\), we get \(5=10-2+D\), so \(D=-3\). Equivalently, writing the result relative to the initial time,
The velocity is zero when \(10-4t=0\), so \(t=\frac{5}{2}\) seconds. For \(1\leq t<\frac{5}{2}\), velocity is positive; for \(\frac{5}{2}<t\leq4\), it is negative. The particle moves in the positive direction, is momentarily at rest at \(t=\frac{5}{2}\), and then moves in the negative direction.
At the turning time, \(x\left(\frac{5}{2}\right)=5+6\left(\frac{3}{2}\right)-2\left(\frac{3}{2}\right)^2=5+9-\frac{9}{2}=\frac{19}{2}\) meters. At the endpoint, \(x(4)=5+6(3)-2(3)^2=5\) meters. The final position equals the initial position, so the displacement over the full interval is \(5-5=0\) meters.
Check: Differentiating \(x(t)\) gives \(x'(t)=6-4(t-1)=10-4t=v(t)\). Then \(v'(t)=-4=a(t)\). At \(t=1\), \(v(1)=6\) and \(x(1)=5\).
Worked Example: Finding a Turning Time from a Nonlinear Velocity
A particle has acceleration \(a(t)=2t\) meters per second squared. At \(t=0\), its velocity is \(-2\) meters per second and its position is \(4\) meters. Find its position and velocity, then describe its motion on \(0\leq t\leq2\).
Integrating acceleration gives \(v(t)=t^2+C\). Since \(v(0)=-2\), we have \(C=-2\), so \(v(t)=t^2-2\) meters per second.
Integrating velocity gives \(x(t)=\frac{t^3}{3}-2t+D\). Since \(x(0)=4\), we have \(D=4\). Thus
Velocity is zero when \(t^2-2=0\), which within the interval occurs at \(t=\sqrt{2}\) seconds. For \(0\leq t<\sqrt{2}\), \(t^2-2<0\), so the particle moves in the negative direction. For \(\sqrt{2}<t\leq2\), \(t^2-2>0\), so it moves in the positive direction.
The position at the turning time is \(x(\sqrt{2})=\frac{2\sqrt{2}}{3}-2\sqrt{2}+4=4-\frac{4\sqrt{2}}{3}\) meters. At \(t=2\), \(x(2)=\frac{8}{3}-4+4=\frac{8}{3}\) meters. The particle therefore ends at a position lower than its initial position, even though it changes direction during the interval.
Check: Differentiating \(x(t)\) gives \(x'(t)=t^2-2=v(t)\); differentiating \(v(t)\) gives \(v'(t)=2t=a(t)\). The initial values are \(v(0)=-2\) and \(x(0)=4\).
Using the Functions to Describe Motion
Once \(v(t)\) and \(x(t)\) have been found, motion over a specified interval can be described systematically. Evaluate the functions at relevant times to find positions and velocities. To determine direction, inspect the sign of velocity: a particle changes direction only if velocity changes sign. A time at which \(v(t)=0\) is a candidate for a change of direction, but it is not enough by itself; check the signs on either side.
Displacement from time \(a\) to time \(b\) is the change in position, \(x(b)-x(a)\). It is also the integral of velocity over that interval. Displacement is signed, so it can be zero even when the particle moves during the interval, as in the second example. Total distance is a different quantity: it measures the full length of the path. To find it when velocity changes sign, the interval must be split at the relevant times. The distinction between displacement and total distance is important when interpreting motion.
A Common Pitfall: Losing the Initial Values
One common error is to integrate correctly but omit the constants or fail to use the initial conditions. For example, if \(a(t)=4t\) and \(v(0)=3\), an incorrect calculation is \(v(t)=2t^2\). That function differentiates to \(4t\), but it gives \(v(0)=0\), not \(3\).
The correct general antiderivative is \(v(t)=2t^2+C\). Substituting \(v(0)=3\) gives \(3=2(0)^2+C\), so \(C=3\), and therefore \(v(t)=2t^2+3\). Differentiation confirms \(v'(t)=4t\), and substitution confirms \(v(0)=3\). The derivative check alone is not sufficient: the initial condition must also hold.
Check Your Understanding
For each question, use the initial values as well as differentiation to check your result.
- If \(a(t)=6\), \(v(0)=-1\), and \(x(0)=3\), find \(v(t)\) and \(x(t)\).
- If \(a(t)=2t\) and \(v(1)=5\), find \(v(t)\). At what time in \(1\leq t\leq3\) is the particle momentarily at rest, if any?
- What additional check is needed after finding an antiderivative of acceleration?
- A particle has \(v(t)=t-2\) on \(0\leq t\leq4\). At what time does it change direction, and what is the sign of velocity on each side?
- How can a particle have zero displacement over an interval while still moving during it?
