Interpret a derivative and a change of velocity as different quantities.
Syllabus 4.2
Notes and key methods
The instantaneous acceleration is the derivative of velocity with respect to time. For a polynomial velocity, apply the power rule term by term. Average acceleration over an interval instead uses its endpoint velocity difference divided by its duration.
An acceleration function can change sign without the particle reversing direction; reversal depends on velocity. The examples retain a positive velocity throughout their intervals, so a zero acceleration is an extremum of speed, not a stopping point.
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 · Acceleration from velocity
The signed velocity is \(v=3 t^{2} - 24 t + 56\) m/s for \(0\le t\le 7\), with t in seconds.
Find the acceleration at \(\displaystyle t=7 s\), in \(\displaystyle \frac{m}{s^{2}}\).
Find the time when acceleration is zero, in s.
Find the minimum speed on the stated interval, in m/s.
Find the initial acceleration in \(\displaystyle \frac{m}{s^{2}}\).
Show the reasoning
(a)
Differentiate the given velocity rule and use the requested time.
The derivative \(6 t - 24\) evaluates to \(\displaystyle \frac{18m}{s^{2}}\).
Check that velocity stays positive and then minimise its square-term form.
Velocity is \(3 \left(t - 4\right)^{2} + 8\), always positive; its square term vanishes at \(\displaystyle t=4\), leaving \(\displaystyle \frac{8m}{s}\).
Answer: \(8\).
(d)
Evaluate the derivative with its signed constant term retained.
Substitution of \(\displaystyle t=0\) into \(6 t - 24\) gives \(\displaystyle - \frac{24m}{s^{2}}\).
Answer: \(-24\).
Example 2 · Acceleration from velocity
The signed velocity is \(v=4 t^{2} - 40 t + 109\) m/s for \(0\le t\le 6\), with t in seconds.
Find the acceleration at \(\displaystyle t=6 s\), in \(\displaystyle \frac{m}{s^{2}}\).
Find the time when acceleration is zero, in s.
Find the minimum speed on the stated interval, in m/s.
Find the initial acceleration in \(\displaystyle \frac{m}{s^{2}}\).
Show the reasoning
(a)
Differentiate the given velocity rule and use the requested time.
The derivative \(8 t - 40\) evaluates to \(\displaystyle \frac{8m}{s^{2}}\).
Check that velocity stays positive and then minimise its square-term form.
Velocity is \(4 \left(t - 5\right)^{2} + 9\), always positive; its square term vanishes at \(\displaystyle t=5\), leaving \(\displaystyle \frac{9m}{s}\).
Answer: \(9\).
(d)
Evaluate the derivative with its signed constant term retained.
Substitution of \(\displaystyle t=0\) into \(8 t - 40\) gives \(\displaystyle - \frac{40m}{s^{2}}\).
Answer: \(-40\).
Example 3 · Acceleration from velocity
The signed velocity is \(v=t^{2} - 4 t + 7\) m/s for \(0\le t\le 4\), with t in seconds.
Find the acceleration at \(\displaystyle t=4 s\), in \(\displaystyle \frac{m}{s^{2}}\).
Find the time when acceleration is zero, in s.
Find the minimum speed on the stated interval, in m/s.
Find the initial acceleration in \(\displaystyle \frac{m}{s^{2}}\).
Show the reasoning
(a)
Differentiate the given velocity rule and use the requested time.
The derivative \(2 t - 4\) evaluates to \(\displaystyle \frac{4m}{s^{2}}\).
Check that velocity stays positive and then minimise its square-term form.
Velocity is \(\left(t - 2\right)^{2} + 3\), always positive; its square term vanishes at \(\displaystyle t=2\), leaving \(\displaystyle \frac{3m}{s}\).
Answer: \(3\).
(d)
Evaluate the derivative with its signed constant term retained.
Substitution of \(\displaystyle t=0\) into \(2 t - 4\) gives \(\displaystyle - \frac{4m}{s^{2}}\).
Answer: \(-4\).
Common mistakes
A zero acceleration does not imply zero velocity.
Instantaneous acceleration and interval-average acceleration need not agree.