Velocity and Acceleration

This lesson covers velocity, acceleration and velocity-time graphs as required by AQA GCSE Combined Science Trilogy (8464), section 6.5.2. These concepts build on the speed, distance and time work from the previous lesson and are essential for understanding Newton's laws and momentum.

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Velocity

Velocity is speed in a stated direction. It is a vector quantity.

$v = \frac{\text{displacement}}{t}$v=tdisplacement​

An object can move at constant speed but have a changing velocity — this happens whenever the direction changes (e.g. moving in a circle).

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Acceleration

Acceleration is the rate of change of velocity. It tells you how quickly an object is speeding up, slowing down, or changing direction.

$a = \frac{v - u}{t}$a=tv−u​

Where:

• a = acceleration (m/s²)

• v = final velocity (m/s)

• u = initial velocity (m/s)

• t = time (s)

• A positive acceleration means the object is speeding up (in the direction of motion).

• A negative acceleration (deceleration) means the object is slowing down.

Exam Tip (AQA 8464): The equation $a = (v - u) / t$a=(v−u)/t is on the AQA equation sheet, but you should practise rearranging it to find $v$v, $u$u or $t$t.

Rearranging the Equation

To find	Rearranged equation
Acceleration	$a = (v - u) / t$a=(v−u)/t
Final velocity	$v = u + at$v=u+at
Time	$t = (v - u) / a$t=(v−u)/a
Initial velocity	$u = v - at$u=v−at

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Worked Examples

Example 1 — Calculating Acceleration

A car accelerates from 5 m/s to 25 m/s in 10 s. Calculate the acceleration.

Solution:

$a = \frac{v - u}{t} = \frac{25 - 5}{10} = \frac{20}{10} = 2$a=tv−u​=1025−5​=1020​=2 m/s²

Example 2 — Calculating Deceleration

A bus travelling at 18 m/s brakes and comes to a stop in 6 s. Calculate the deceleration.

Solution:

$a = \frac{v - u}{t} = \frac{0 - 18}{6} = \frac{-18}{6} = -3$a=tv−u​=60−18​=6−18​=−3 m/s²

The deceleration is 3 m/s² (the negative sign indicates slowing down).

Example 3 — Finding Final Velocity

A bicycle starts from rest and accelerates at 1.5 m/s² for 8 s. What is its final velocity?

Solution:

$v = u + at = 0 + (1.5 \times 8) = 12$v=u+at=0+(1.5×8)=12 m/s

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The Equation of Motion: $v^2 - u^2 = 2as$v2−u2=2as

This equation links velocity, acceleration and distance (without time):

$v^2 - u^2 = 2as$v2−u2=2as

Where:

• v = final velocity (m/s)
• u = initial velocity (m/s)
• a = acceleration (m/s²)
• s = distance (m)

Worked Example

A car travelling at 12 m/s accelerates uniformly at 2 m/s² over a distance of 100 m. Find the final velocity.

Solution:

$v^2 = u^2 + 2as = 12^2 + 2(2)(100) = 144 + 400 = 544$v2=u2+2as=122+2(2)(100)=144+400=544

$v = \sqrt{544} = 23.3$v=544​=23.3 m/s (to 1 d.p.)

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Velocity-Time Graphs

A velocity-time graph plots velocity (y-axis) against time (x-axis). These graphs contain two key pieces of information.

1. Gradient = Acceleration

$\text{acceleration} = \text{gradient} = \frac{\Delta v}{\Delta t}$acceleration=gradient=ΔtΔv​

Feature	Meaning
Straight line sloping upwards	Constant positive acceleration (speeding up)
Straight line sloping downwards	Constant deceleration (slowing down)
Horizontal line	Constant velocity (no acceleration)
Steeper line	Greater acceleration
Curve	Changing acceleration

2. Area Under the Line = Distance Travelled

$\text{distance} = \text{area under the velocity-time graph}$distance=area under the velocity-time graph

For simple shapes:

• Rectangle: area = base × height
• Triangle: area = ½ × base × height
• Trapezium: area = ½ × (a + b) × h

graph TD
    subgraph "Velocity-Time Graph Features"
        A["Positive gradient → Accelerating"]
        B["Zero gradient (horizontal) → Constant velocity"]
        C["Negative gradient → Decelerating"]
        D["Area under graph → Distance travelled"]
    end

    style A fill:#27ae60,color:#fff
    style B fill:#2980b9,color:#fff
    style C fill:#e74c3c,color:#fff
    style D fill:#8e44ad,color:#fff

Worked Example — Velocity-Time Graph

A velocity-time graph shows an object that accelerates uniformly from 0 to 20 m/s in 5 s, then travels at 20 m/s for 10 s.

(a) Calculate the acceleration in the first 5 s.

$a = \frac{20 - 0}{5} = 4$a=520−0​=4 m/s²

(b) Calculate the total distance travelled.

Triangle (first 5 s): $\frac{1}{2} \times 5 \times 20 = 50$21​×5×20=50 m

Rectangle (next 10 s): $10 \times 20 = 200$10×20=200 m

Total distance = 50 + 200 = 250 m

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Comparing Distance-Time and Velocity-Time Graphs

Feature	Distance-Time Graph	Velocity-Time Graph
Gradient gives	Speed	Acceleration
Area under graph gives	Not used	Distance travelled
Horizontal line means	Stationary	Constant velocity
Upward slope means	Moving (constant speed)	Accelerating

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Uniform and Non-Uniform Acceleration

• Uniform (constant) acceleration: velocity changes by equal amounts in equal time intervals. On a v-t graph, this is a straight line.
• Non-uniform acceleration: velocity changes by different amounts. On a v-t graph, this is a curve.

Near the Earth's surface (ignoring air resistance), all objects experience uniform acceleration due to gravity: $g \approx 9.8$g≈9.8 m/s².

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Common Mistakes

• Confusing the gradient of a distance-time graph (speed) with the gradient of a velocity-time graph (acceleration).
• Forgetting to find the area under a v-t graph for distance — students often just read the final velocity.
• Mixing up velocity and speed — velocity is a vector (has direction), speed is a scalar.
• Using $v^2 = u^2 + 2as$v2=u2+2as but forgetting to take the square root at the end.
• Ignoring the sign of acceleration — negative means deceleration.

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Summary