2.8–2.10Acceleration
The average acceleration of an object can be calculated using the equation:
acceleration = change in velocity / time taken a = ∆v / t
acceleration, a, in metres per second squared, m/s²; change in velocity, ∆v, in metres per second, m/s; time, t, in seconds, s.
You should be able to recall and apply this equation. An object that slows down is decelerating, which gives a negative acceleration.
You should be able to estimate the magnitude of everyday accelerations — a car pulling away is a few m/s², and a passenger lift perhaps 1 m/s².
The acceleration of an object can be calculated from the gradient of a velocity–time graph.
You should be able to draw velocity–time graphs from measurements and interpret lines and slopes to determine acceleration.
The following equation applies to uniform acceleration:
final velocity² − initial velocity² = 2 × acceleration × distance v² − u² = 2 a s
final velocity, v, in m/s; initial velocity, u, in m/s; acceleration, a, in m/s²; distance, s, in metres, m.
This equation is given on the Physics equation sheet. Use it when a question gives you two velocities and a distance but no time.
Near the Earth's surface any object falling freely under gravity has an acceleration of about 9.8 m/s².
Worked example
A car accelerates uniformly from rest at 3.0 m/s² over a distance of 54 m. Calculate its final velocity.
- No time is given, so use v² − u² = 2 a s.
- The car starts from rest, so u = 0.
- Substitute: v² − 0 = 2 × 3.0 × 54 = 324.
- Take the square root: v = √324.
- Give the unit: metres per second.
Answer: 18 m/s
Common mistake
In v² − u² = 2 a s, both velocities are squared and the final answer needs a square root. Forgetting the square root gives 324 m/s, which should look obviously wrong.
- Acceleration
- the change in velocity divided by the time taken, measured in metres per second squared
- Deceleration
- a negative acceleration, which occurs when an object slows down
2.10Distance from a velocity–time graphHigher tier
Higher tier: the distance travelled by an object, or its displacement, can be calculated from the area under a velocity–time graph.
You should be able to interpret enclosed areas in velocity–time graphs to determine distance travelled or displacement, and to measure the area under a velocity–time graph by counting squares where appropriate.
Counting squares is the method to use when the line is curved and the area cannot be split into neat triangles and rectangles. Work out what one square represents from the axes, count the whole squares, then count part-squares as a half each.
Worked example
A car accelerates uniformly from 8.0 m/s to 20.0 m/s in 6.0 s. Calculate the acceleration and the distance travelled.
- Change in velocity: ∆v = 20.0 − 8.0 = 12.0 m/s.
- a = ∆v ÷ t = 12.0 ÷ 6.0 = 2.0 m/s².
- For the distance, find the area under the velocity–time graph — a rectangle plus a triangle.
- Rectangle: 8.0 × 6.0 = 48 m. Triangle: ½ × 6.0 × 12.0 = 36 m.
- Total area = 48 + 36.
Answer: An acceleration of 2.0 m/s², and a distance of 84 m.
2.11Measuring speed in the laboratory
Speed is distance divided by time, so measuring it means measuring those two things well. Three methods are named, and questions ask you to choose between them and to say why.
Two things improve any of them. Measure over the longest distance you sensibly can, because a fixed timing error matters less when the total time is larger. And repeat the measurement and take a mean, because that reduces the effect of random error.
A light gate measures the average speed across the width of the card that breaks the beam. Using a narrower card gets closer to the speed at an instant; using two gates a known distance apart gives the average speed between them, and the difference between the two readings gives the acceleration.
Common mistake
Reaction time is a systematic problem with a stopwatch, not a reason to call the result wrong. Say what it does to the measurement — it adds an uncertainty of around 0.2 s — and how light gates remove it.
2.13Terminal velocity
An object falling through a fluid initially accelerates due to the force of gravity. Eventually the resultant force will be zero and the object will move at its terminal velocity.
Physics only students should be able to draw and interpret velocity–time graphs for objects that reach terminal velocity, and interpret the changing motion in terms of the forces acting.
The sequence of forces, which is what the marks are for:
1. At the moment of release the only significant force is weight, so the resultant force is large and the acceleration is greatest. 2. As the object speeds up, air resistance increases, because it depends on speed. 3. The resultant force is therefore smaller, so the acceleration decreases — the object is still speeding up, but less quickly. 4. Eventually air resistance equals weight. The resultant force is zero, so by Newton's First Law the object continues at a constant velocity — its terminal velocity.
On a velocity–time graph this appears as a line that starts steep, gradually becomes shallower, and finally levels off horizontally.
Worked example
A skydiver opens a parachute after reaching terminal velocity. Describe what happens to the forces and the motion.
- Opening the parachute greatly increases the surface area, so air resistance increases sharply.
- Air resistance is now greater than weight, so there is a resultant force upwards.
- A resultant force upwards while moving downwards means the skydiver decelerates.
- As the speed falls, air resistance falls too.
- When air resistance again equals weight the resultant force is zero, and the skydiver falls at a new, much lower terminal velocity.
Answer: Air resistance jumps above weight, giving an upward resultant force and a deceleration; as speed drops, air resistance falls until it again balances weight, and a new lower terminal velocity is reached.
Common mistake
At terminal velocity the object is still moving, and moving fast. Zero resultant force means zero acceleration, not zero speed.
- Terminal velocity
- the steady velocity reached by an object falling through a fluid when the resultant force on it becomes zero
Before you practise
- acceleration = change in velocity ÷ time taken, in m/s².
- (final velocity)² − (initial velocity)² = 2 × acceleration × distance.
- On a velocity-time graph the gradient is the acceleration and the area underneath is the distance travelled.
- A negative gradient means deceleration; count squares or split the area into triangles and rectangles.
- Acceleration in free fall is about 10 m/s², whatever the mass.
- Light gates remove reaction time; a stopwatch adds an uncertainty of roughly 0.2 s at each end.
- Terminal velocity is reached when the drag force has grown equal to the weight, so the resultant force is zero.