Mastering Speed, Velocity, and Acceleration for Your Physics Exam

What the examiner is testing

Examiners assess your ability to correctly apply the definitions of speed, velocity, and acceleration to calculate unknown quantities from given data. Marks are awarded for selecting the correct formula, showing clear algebraic manipulation, and presenting your final answer with appropriate units.

The method

1. Identify the known quantities and the unknown quantity. Read the question carefully to extract all numerical values and determine what you need to calculate.
2. Determine if the question involves speed, velocity, or acceleration. This will guide your choice of formula. Remember, velocity includes direction, while speed does not.
3. Select the appropriate formula.
   • For speed: \( \text{speed} = \frac{\text{distance}}{\text{time}} \)
   • For velocity: \( \text{velocity} = \frac{\text{displacement}}{\text{time}} \)
   • For acceleration: \( \text{acceleration} = \frac{\text{change in velocity}}{\text{time taken}} \) or \( a = \frac{v - u}{t} \)
4. Ensure all units are consistent. Convert any units (e.g., km/h to m/s, minutes to seconds) if necessary before calculation.
5. Substitute the known values into the formula.
6. Rearrange the formula and calculate the unknown quantity. Show each step of your algebraic manipulation.
7. State your final answer with the correct units.

Worked example

A car travels a distance of 1500 m in 75 seconds. Calculate its average speed.

$$ \text{Step 1: Identify knowns and unknown.} \\ \text{Known: distance} = 1500 \text{ m} \\ \text{Known: time} = 75 \text{ s} \\ \text{Unknown: average speed} \\ \\ \text{Step 2: Determine type of quantity.} \\ \text{The question asks for average speed.} \\ \\ \text{Step 3: Select formula.} \\ \text{Formula for speed is } \text{speed} = \frac{\text{distance}}{\text{time}} \\ \\ \text{Step 4: Check units.} \\ \text{Units are consistent (metres and seconds).} \\ \\ \text{Step 5: Substitute values.} \\ \text{speed} = \frac{1500 \text{ m}}{75 \text{ s}} \\ \\ \text{Step 6: Calculate.} \\ \text{speed} = 20 \text{ m/s} \\ \\ \text{Step 7: State final answer with units.} \\ \text{Average speed} = 20 \text{ m/s} $$

Sanity check: A car moving at 20 m/s (which is 72 km/h) for 75 seconds would cover 1500 m. This seems reasonable for a car.

Worked example: a harder one

A train accelerates uniformly from rest to a velocity of 30 m/s in 2 minutes. Calculate the acceleration of the train.

$$ \text{Step 1: Identify knowns and unknown.} \\ \text{Known: initial velocity } (u) = 0 \text{ m/s (from rest)} \\ \text{Known: final velocity } (v) = 30 \text{ m/s} \\ \text{Known: time } (t) = 2 \text{ minutes} \\ \text{Unknown: acceleration } (a) \\ \\ \text{Step 2: Determine type of quantity.} \\ \text{The question asks for acceleration.} \\ \\ \text{Step 3: Select formula.} \\ \text{Formula for acceleration is } a = \frac{v - u}{t} \\ \\ \text{Step 4: Check units.} \\ \text{The time is in minutes, but velocities are in m/s. We need to convert minutes to seconds.} \\ t = 2 \text{ minutes} \times 60 \text{ s/minute} = 120 \text{ s} \\ \\ \text{Step 5: Substitute values.} \\ a = \frac{30 \text{ m/s} - 0 \text{ m/s}}{120 \text{ s}} \\ \\ \text{Step 6: Calculate.} \\ a = \frac{30 \text{ m/s}}{120 \text{ s}} \\ a = 0.25 \text{ m/s}^2 \\ \\ \text{Step 7: State final answer with units.} \\ \text{Acceleration} = 0.25 \text{ m/s}^2 $$

Why the obvious first move fails: If you directly substitute 2 minutes into the formula without converting to seconds, you would get \( a = \frac{30}{2} = 15 \text{ m/s/minute} \). While technically a unit of acceleration, \( \text{m/s}^2 \) is the standard unit expected in the exam, and mixing units like this is a common error that leads to incorrect final answers or loss of marks.

Practice

1. A runner completes a 400 m race in 50 seconds. What is their average speed?
2. A car travels at a constant velocity of 25 m/s for 30 seconds. What displacement does it cover?
3. A cyclist increases their velocity from 5 m/s to 15 m/s in 4 seconds. Calculate their acceleration.
4. A ball is dropped from a height. Its velocity increases from 0 m/s to 9.8 m/s in 1.0 second. It then hits the ground and comes to rest in 0.1 seconds. Calculate the acceleration of the ball as it falls, and its deceleration as it hits the ground.

Answers:

1. \( \text{speed} = \frac{400 \text{ m}}{50 \text{ s}} = 8 \text{ m/s} \)
2. \( \text{displacement} = \text{velocity} \times \text{time} = 25 \text{ m/s} \times 30 \text{ s} = 750 \text{ m} \)
3. \( a = \frac{v - u}{t} = \frac{15 \text{ m/s} - 5 \text{ m/s}}{4 \text{ s}} = \frac{10 \text{ m/s}}{4 \text{ s}} = 2.5 \text{ m/s}^2 \)
4. Acceleration while falling:
   $$ u = 0 \text{ m/s} \\ v = 9.8 \text{ m/s} \\ t = 1.0 \text{ s} \\ a = \frac{v - u}{t} = \frac{9.8 \text{ m/s} - 0 \text{ m/s}}{1.0 \text{ s}} = 9.8 \text{ m/s}^2 $$
   Deceleration while hitting the ground:
   $$ u = 9.8 \text{ m/s} \\ v = 0 \text{ m/s} \\ t = 0.1 \text{ s} \\ a = \frac{v - u}{t} = \frac{0 \text{ m/s} - 9.8 \text{ m/s}}{0.1 \text{ s}} = \frac{-9.8 \text{ m/s}}{0.1 \text{ s}} = -98 \text{ m/s}^2 $$
   The negative sign indicates deceleration. So, the deceleration is \( 98 \text{ m/s}^2 \).

The three mistakes that lose marks

1. Confusing speed and velocity: Using 'distance' when 'displacement' is required for velocity calculations, or vice versa. This often leads to incorrect answers when direction changes are involved, e.g., calculating the average speed of a car that drives 100 m east and then 100 m west in 20 seconds as \( \frac{200 \text{ m}}{20 \text{ s}} = 10 \text{ m/s} \) (correct for speed), but stating its average velocity is also \( 10 \text{ m/s} \) (incorrect, as displacement is 0 m, so velocity is 0 m/s).
2. Inconsistent units: Failing to convert all quantities to standard units (metres, seconds, m/s) before calculation. For example, calculating acceleration with time in minutes and velocity in m/s, leading to an answer like \( 0.5 \text{ m/s/minute} \) instead of the required \( 0.0083 \text{ m/s}^2 \).
3. Incorrect formula rearrangement: Making algebraic errors when rearranging the formula to solve for an unknown. For instance, if \( \text{speed} = \frac{\text{distance}}{\text{time}} \), trying to find time by calculating \( \text{time} = \text{speed} \times \text{distance} \) instead of \( \text{time} = \frac{\text{distance}}{\text{speed}} \).

30-second recap

Speed is the rate at which distance is covered, a scalar quantity. Velocity is the rate of change of displacement, a vector quantity including direction. Acceleration is the rate of change of velocity, also a vector. Ensure consistent units and correct formula application.