Use the initial position to determine the integration constant.
Syllabus 4.2
Notes and key methods
Integrating velocity gives position by the definite-integral relation below. An indefinite antiderivative requires an integration constant, fixed by a supplied position condition.
A definite integral of velocity gives displacement. It gives distance only when velocity has no sign change, or after splitting at every relevant zero. State the time domain before choosing roots or using a position rule beyond a modelled stage.
\(x(t)=x(0)+\int_0^t v(\tau)\,d\tau\)
Check the conditions
Choose and retain a positive direction; distinguish signed quantities from magnitudes.
Use SI units and \(\displaystyle g= \frac{10m}{s^{2}}\) wherever gravity is relevant.
Apply only the model and time interval stated in the question.
Three worked examples
Example 1 · Position from a velocity integral
For \(0\le t\le 5\), velocity is \(v=4 t^{2} + 5\) m/s and \(\displaystyle x(0)=-4 m\).
Find the complete position rule x(t) on the stated interval.
Find the displacement from \(\displaystyle t=0\) to \(\displaystyle t=5 s\), in m.
Find the average velocity over that interval, in m/s.
Find the total distance over that interval, in m.
Show the reasoning
(a)
Integrating term by term gives \(\displaystyle \frac{4 t^{3}}{3} + 5 t\)+constant. Using \(\displaystyle x(0)=-4\) sets the constant to -4.
Answer: \(\displaystyle x=\frac{4 t^{3}}{3} + 5 t - 4\) m, with t in seconds.
(b)
Integrate over the stated time interval with the correct bounds.
The definite integral of velocity gives \(\displaystyle \frac{575}{3}\) m.
Answer: \(\displaystyle \frac{575}{3}\).
(c)
Divide the definite velocity integral by the elapsed time.
Divide displacement \(\displaystyle \frac{575}{3}\) by duration 5 s to obtain \(\displaystyle \frac{115}{3}\) m/s.
Answer: \(\displaystyle \frac{115}{3}\).
(d)
Check the velocity sign before identifying distance with displacement.
The velocity \(4 t^{2} + 5\) is strictly positive, so distance equals the positive displacement \(\displaystyle \frac{575}{3}\) m.
Answer: \(\displaystyle \frac{575}{3}\).
Example 2 · Position from a velocity integral
For \(0\le t\le 2\), velocity is \(v=t^{2} + 6\) m/s and \(\displaystyle x(0)=-3 m\).
Find the complete position rule x(t) on the stated interval.
Find the displacement from \(\displaystyle t=0\) to \(\displaystyle t=2 s\), in m.
Find the average velocity over that interval, in m/s.
Find the total distance over that interval, in m.
Show the reasoning
(a)
Integrating term by term gives \(\displaystyle \frac{t^{3}}{3} + 6 t\)+constant. Using \(\displaystyle x(0)=-3\) sets the constant to -3.
Answer: \(\displaystyle x=\frac{t^{3}}{3} + 6 t - 3\) m, with t in seconds.
(b)
Integrate over the stated time interval with the correct bounds.
The definite integral of velocity gives \(\displaystyle \frac{44}{3}\) m.
Answer: \(\displaystyle \frac{44}{3}\).
(c)
Divide the definite velocity integral by the elapsed time.
Divide displacement \(\displaystyle \frac{44}{3}\) by duration 2 s to obtain \(\displaystyle \frac{22}{3}\) m/s.
Answer: \(\displaystyle \frac{22}{3}\).
(d)
Check the velocity sign before identifying distance with displacement.
The velocity \(t^{2} + 6\) is strictly positive, so distance equals the positive displacement \(\displaystyle \frac{44}{3}\) m.
Answer: \(\displaystyle \frac{44}{3}\).
Example 3 · Position from a velocity integral
For \(0\le t\le 3\), velocity is \(v=2 t^{2} + 7\) m/s and \(\displaystyle x(0)=-2 m\).
Find the complete position rule x(t) on the stated interval.
Find the displacement from \(\displaystyle t=0\) to \(\displaystyle t=3 s\), in m.
Find the average velocity over that interval, in m/s.
Find the total distance over that interval, in m.
Show the reasoning
(a)
Integrating term by term gives \(\displaystyle \frac{2 t^{3}}{3} + 7 t\)+constant. Using \(\displaystyle x(0)=-2\) sets the constant to -2.
Answer: \(\displaystyle x=\frac{2 t^{3}}{3} + 7 t - 2\) m, with t in seconds.
(b)
Integrate over the stated time interval with the correct bounds.
The definite integral of velocity gives \(39\) m.
Answer: \(39\).
(c)
Divide the definite velocity integral by the elapsed time.
Divide displacement \(39\) by duration 3 s to obtain \(13\) m/s.
Answer: \(13\).
(d)
Check the velocity sign before identifying distance with displacement.
The velocity \(2 t^{2} + 7\) is strictly positive, so distance equals the positive displacement \(39\) m.
Answer: \(39\).
Common mistakes
An integration constant cannot be silently set to zero when initial position is nonzero.
The integral of signed velocity is displacement, not automatically distance.