Acceleration, speed and consistent units — 9709 P4
Calculate a signed velocity change before deciding whether speed decreases.
Syllabus 4.2
Notes and key methods
Average acceleration is (final velocity−initial velocity)/elapsed time, with all velocities measured along the same positive axis. A particle can have negative acceleration while speeding up if it already moves in the negative direction.
When velocity and acceleration have opposite signs, speed is decreasing. With matching signs, speed is increasing. Convert km/h to m/s by dividing by 3.6 before using seconds. A deceleration magnitude is positive; its signed acceleration depends on the chosen axis.
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, speed and consistent units
Take right as positive. A particle’s signed velocity changes uniformly from \(\displaystyle \frac{11m}{s}\) to \(\displaystyle \frac{3m}{s}\) in 4 s. Separately, a speed reading is 54 km/h.
Find the signed acceleration in \(\displaystyle \frac{m}{s^{2}}\).
Find the positive deceleration magnitude in \(\displaystyle \frac{m}{s^{2}}\).
Convert the separate speed reading to m/s.
If instead the velocity changes from \(\displaystyle - \frac{11m}{s}\) to \(\displaystyle - \frac{3m}{s}\) over the same time, find the signed acceleration in \(\displaystyle \frac{m}{s^{2}}\).
Show the reasoning
(a)
Subtract the original velocity from the final velocity and divide by elapsed time.
The signed change -8 divided by 4 gives \(-2\) \(\displaystyle \frac{m}{s^{2}}\).
Answer: \(-2\).
(b)
Take the magnitude of this negative acceleration only after interpreting the motion.
Both velocities are positive and speed decreases, so the deceleration magnitude is \(2\).
Answer: \(2\).
(c)
Convert kilometres to metres and hours to seconds.
Divide 54 by 3.6 to obtain \(15\) m/s.
Answer: \(15\).
(d)
Apply the same subtraction to the two negative signed velocities.
The change is \(\displaystyle \frac{8m}{s}\), so acceleration is \(2\) \(\displaystyle \frac{m}{s^{2}}\). Speed still decreases.
Answer: \(2\).
Example 2 · Acceleration, speed and consistent units
Take right as positive. A particle’s signed velocity changes uniformly from \(\displaystyle \frac{12m}{s}\) to \(\displaystyle \frac{4m}{s}\) in 5 s. Separately, a speed reading is 72 km/h.
Find the signed acceleration in \(\displaystyle \frac{m}{s^{2}}\).
Find the positive deceleration magnitude in \(\displaystyle \frac{m}{s^{2}}\).
Convert the separate speed reading to m/s.
If instead the velocity changes from \(\displaystyle - \frac{12m}{s}\) to \(\displaystyle - \frac{4m}{s}\) over the same time, find the signed acceleration in \(\displaystyle \frac{m}{s^{2}}\).
Show the reasoning
(a)
Subtract the original velocity from the final velocity and divide by elapsed time.
The signed change -8 divided by 5 gives \(\displaystyle - \frac{8}{5}\) \(\displaystyle \frac{m}{s^{2}}\).
Answer: \(\displaystyle - \frac{8}{5}\).
(b)
Take the magnitude of this negative acceleration only after interpreting the motion.
Both velocities are positive and speed decreases, so the deceleration magnitude is \(\displaystyle \frac{8}{5}\).
Answer: \(\displaystyle \frac{8}{5}\).
(c)
Convert kilometres to metres and hours to seconds.
Divide 72 by 3.6 to obtain \(20\) m/s.
Answer: \(20\).
(d)
Apply the same subtraction to the two negative signed velocities.
The change is \(\displaystyle \frac{8m}{s}\), so acceleration is \(\displaystyle \frac{8}{5}\) \(\displaystyle \frac{m}{s^{2}}\). Speed still decreases.
Answer: \(\displaystyle \frac{8}{5}\).
Example 3 · Acceleration, speed and consistent units
Take right as positive. A particle’s signed velocity changes uniformly from \(\displaystyle \frac{13m}{s}\) to \(\displaystyle \frac{5m}{s}\) in 2 s. Separately, a speed reading is 90 km/h.
Find the signed acceleration in \(\displaystyle \frac{m}{s^{2}}\).
Find the positive deceleration magnitude in \(\displaystyle \frac{m}{s^{2}}\).
Convert the separate speed reading to m/s.
If instead the velocity changes from \(\displaystyle - \frac{13m}{s}\) to \(\displaystyle - \frac{5m}{s}\) over the same time, find the signed acceleration in \(\displaystyle \frac{m}{s^{2}}\).
Show the reasoning
(a)
Subtract the original velocity from the final velocity and divide by elapsed time.
The signed change -8 divided by 2 gives \(-4\) \(\displaystyle \frac{m}{s^{2}}\).
Answer: \(-4\).
(b)
Take the magnitude of this negative acceleration only after interpreting the motion.
Both velocities are positive and speed decreases, so the deceleration magnitude is \(4\).
Answer: \(4\).
(c)
Convert kilometres to metres and hours to seconds.
Divide 90 by 3.6 to obtain \(25\) m/s.
Answer: \(25\).
(d)
Apply the same subtraction to the two negative signed velocities.
The change is \(\displaystyle \frac{8m}{s}\), so acceleration is \(4\) \(\displaystyle \frac{m}{s^{2}}\). Speed still decreases.
Answer: \(4\).
Common mistakes
A negative acceleration does not always imply decreasing speed.
Convert both velocity readings to the same units before subtraction.