Acceleration, Velocity-Time Graphs and Terminal Velocity (Part 2/3) Revision Notes | GCSE AQA Higher Physics

1. Reading Velocity-Time Graphs Like a Detective

You've seen distance-time graphs before, but velocity-time graphs tell a different story. Instead of showing where something is, they show how fast it's going at each moment.

The key insight is this: the gradient (steepness) of a velocity-time graph tells you the acceleration. A steep upward slope means rapid acceleration. A gentle slope means gradual acceleration. A horizontal line means constant velocity - no acceleration at all.

Here's where students often get confused: a curved line on a velocity-time graph shows changing acceleration. The steeper the curve gets, the greater the acceleration becomes at that moment.

Think about a car pulling away from traffic lights. Initially, the acceleration is high as the engine works hard. As the car reaches its cruising speed, the acceleration decreases to zero. This would show as a curve that starts steep then levels off horizontally.

2. Calculating Acceleration from Gradients

Now comes the mathematical bit - but it's simpler than you think. To find acceleration from a velocity-time graph, you calculate the gradient using the same method as any straight line graph.

The formula is: acceleration = change in velocity ÷ change in time. In symbols, that's $a = \frac{v - u}{t}$, where v is final velocity, u is initial velocity, and t is time taken.

**Trap**: Many students forget that deceleration gives a negative answer. If your calculation gives -3 m/s², don't panic - this just means the object is slowing down at 3 m/s².

Worked example: Finding Acceleration from a Velocity-Time Graph

A car accelerates from 10 m/s to 30 m/s in 4 seconds. Calculate its acceleration.

1. Initial velocity u = 10 m/s Final velocity v = 30 m/s Time t = 4 s — Always write down what you know first
2. $a = \frac{v - u}{t}$ — This is the definition of acceleration [M1]
3. $a = \frac{30 - 10}{4}$ — Always include units in your working [M1]
4. $a = \frac{20}{4} = 5$ m/s² — Don't forget the units - m/s² for acceleration [A1]

Answer: 5 m/s²

Notice how we kept track of units throughout. In exams, you'll lose marks if you forget units in your final answer. The unit m/s² makes sense: it's metres per second, per second - how much the velocity changes each second.

3. The Power Equation: v² - u² = 2as

Sometimes you need to find acceleration but don't know the time taken. Or you know the acceleration and want to find the final velocity. That's where this equation becomes your best friend.

The equation $v^2 - u^2 = 2as$ connects velocity, acceleration, and distance without needing time. It's particularly useful for objects falling under gravity or vehicles braking to a stop.

The equation works because it's derived from the basic definitions of acceleration and average velocity. When acceleration is constant, the maths works out beautifully to give us this powerful tool.

Worked example: Using v² - u² = 2as

A car travelling at 20 m/s brakes and comes to a stop in 50 metres. Calculate the deceleration.

1. Initial velocity u = 20 m/s Final velocity v = 0 m/s Distance s = 50 m Acceleration a = ? — The car stops, so final velocity is zero
2. $v^2 - u^2 = 2as$ — We have u, v, and s, but need a [M1]
3. $0^2 - 20^2 = 2 \times a \times 50$ — Be careful with the signs [M1]
4. $-400 = 100a$ — The negative shows this is deceleration
5. $a = \frac{-400}{100} = -4$ m/s² — Negative acceleration = deceleration [A1]

Answer: -4 m/s² (deceleration of 4 m/s²)

The negative answer tells us this is deceleration - the car is slowing down. In exam questions, you might be asked to state this as "deceleration of 4 m/s²" rather than "-4 m/s²".

4. Terminal Velocity - When Forces Balance

Here's a question that puzzles many students: if gravity constantly pulls objects downward, why don't falling objects keep accelerating forever? The answer lies in air resistance.

When an object first starts falling, gravity is the only significant force. The object accelerates downward at 9.8 m/s². But as it speeds up, air resistance increases. Eventually, air resistance equals the gravitational force, and the net force becomes zero.

At terminal velocity, the object continues falling but at constant speed - no more acceleration. A skydiver reaches terminal velocity at about 50 m/s (180 km/h), while a feather might reach just 1 m/s.

The terminal velocity depends on the object's shape, size, and mass. Streamlined objects have higher terminal velocities because they experience less air resistance. This is why skydivers can control their falling speed by changing their body position.

5. Distance from Velocity-Time Graphs (Higher Tier)

Here's where velocity-time graphs become really powerful. Just as the gradient gives you acceleration, the area under the graph gives you distance travelled. This works because distance = average velocity × time.

For a rectangular area (constant velocity), it's straightforward: distance = velocity × time. For triangular areas (constant acceleration), you calculate the area of the triangle: ½ × base × height.

More complex shapes can be broken down into rectangles and triangles, or you can count squares if the graph has a grid. Each square represents a specific distance depending on the scale.

Worked example: Finding Distance from Area Under Graph

A car accelerates uniformly from rest to 20 m/s in 10 seconds. Find the distance travelled.

1. Velocity-time graph shows a straight line from (0,0) to (10,20) — This forms a right-angled triangle
2. Triangle with base = 10 s and height = 20 m/s — Area of triangle = ½ × base × height
3. Area = ½ × 10 × 20 = 100 — This area represents distance travelled
4. Distance = 100 m — Area under v-t graph always gives distance [A1]

Answer: 100 m

6. Exam Success - Avoiding Common Traps

Velocity-time graph questions are exam favourites because they test multiple skills at once. Here's how to maximise your marks and avoid the traps that catch most students.

**The Unit Trap**: Always check your units match what the question asks for. If the graph shows velocity in km/h but asks for acceleration in m/s², you'll need to convert. Remember: 1 km/h = 0.28 m/s approximately.

**The Sign Trap**: Negative gradients mean deceleration, not negative motion. An object moving forward but slowing down still has positive velocity but negative acceleration. Don't confuse direction of motion with direction of acceleration.

**The Area Trap**: When calculating distance from area under the graph, make sure you're finding the area of the right section. If the question asks for distance in the first 5 seconds, don't calculate the area for the entire 10-second journey.

The key to success is practice with real exam questions. Start with simple constant acceleration problems, then move on to more complex scenarios involving changing acceleration and terminal velocity. Remember, the physics concepts are straightforward - it's the mathematical manipulation that needs practice.