Differentiate position with respect to the stated time variable.
Syllabus 4.2
Notes and key methods
For straight-line motion, \(\displaystyle (v= \frac{dx}{dt})\) and \(\displaystyle (a= \frac{dv}{dt})\). The derivative gives a signed velocity, not necessarily a speed. Differentiate the complete displacement expression before substituting a time.
Changing the position origin adds a constant to x but does not change v or a. Keep the initial position when a displacement from that position is required. The examples use the power rule from Pure Mathematics 1, within the stated time interval.
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 · Velocity from displacement
For \(0\le t\le 3\), a particle has position \(x=2 t^{3} + 2 t + 4\) metres, with t in seconds.
Find the velocity at \(\displaystyle t=3 s\), in m/s.
Find the acceleration at \(\displaystyle t=3 s\), in \(\displaystyle \frac{m}{s^{2}}\).
Find the displacement from \(\displaystyle t=0\) to \(\displaystyle t=3 s\), in m.
Find the initial velocity in m/s.
Show the reasoning
(a)
Differentiate each time-dependent position term before evaluation.
The derivative \(6 t^{2} + 2\) gives \(56\) m/s at the stated time.
Answer: \(56\).
(b)
Differentiate the position twice with respect to time.
Differentiating the velocity gives \(12 t\) and hence \(36\) \(\displaystyle \frac{m}{s^{2}}\).
Answer: \(36\).
(c)
Calculate the difference of positions, including the origin constant consistently.
Subtract \(\displaystyle x(0)=4\) from the final position to obtain \(60\) m.
Answer: \(60\).
(d)
Evaluate the derivative, not the original position, at the initial time.
The derivative at \(\displaystyle t=0\) is \(\displaystyle \frac{2m}{s}\).
Answer: \(2\).
Example 2 · Velocity from displacement
For \(0\le t\le 4\), a particle has position \(x=3 t^{3} + 3 t + 5\) metres, with t in seconds.
Find the velocity at \(\displaystyle t=4 s\), in m/s.
Find the acceleration at \(\displaystyle t=4 s\), in \(\displaystyle \frac{m}{s^{2}}\).
Find the displacement from \(\displaystyle t=0\) to \(\displaystyle t=4 s\), in m.
Find the initial velocity in m/s.
Show the reasoning
(a)
Differentiate each time-dependent position term before evaluation.
The derivative \(9 t^{2} + 3\) gives \(147\) m/s at the stated time.
Answer: \(147\).
(b)
Differentiate the position twice with respect to time.
Differentiating the velocity gives \(18 t\) and hence \(72\) \(\displaystyle \frac{m}{s^{2}}\).
Answer: \(72\).
(c)
Calculate the difference of positions, including the origin constant consistently.
Subtract \(\displaystyle x(0)=5\) from the final position to obtain \(204\) m.
Answer: \(204\).
(d)
Evaluate the derivative, not the original position, at the initial time.
The derivative at \(\displaystyle t=0\) is \(\displaystyle \frac{3m}{s}\).
Answer: \(3\).
Example 3 · Velocity from displacement
For \(0\le t\le 5\), a particle has position \(x=4 t^{3} + 4 t + 6\) metres, with t in seconds.
Find the velocity at \(\displaystyle t=5 s\), in m/s.
Find the acceleration at \(\displaystyle t=5 s\), in \(\displaystyle \frac{m}{s^{2}}\).
Find the displacement from \(\displaystyle t=0\) to \(\displaystyle t=5 s\), in m.
Find the initial velocity in m/s.
Show the reasoning
(a)
Differentiate each time-dependent position term before evaluation.
The derivative \(12 t^{2} + 4\) gives \(304\) m/s at the stated time.
Answer: \(304\).
(b)
Differentiate the position twice with respect to time.
Differentiating the velocity gives \(24 t\) and hence \(120\) \(\displaystyle \frac{m}{s^{2}}\).
Answer: \(120\).
(c)
Calculate the difference of positions, including the origin constant consistently.
Subtract \(\displaystyle x(0)=6\) from the final position to obtain \(520\) m.
Answer: \(520\).
(d)
Evaluate the derivative, not the original position, at the initial time.
The derivative at \(\displaystyle t=0\) is \(\displaystyle \frac{4m}{s}\).
Answer: \(4\).
Common mistakes
Do not differentiate after first replacing t by a number.
A constant initial position has zero derivative but remains in position evaluations.