Velocity-Time Graphs and Terminal Velocity

• Step 1: Extract values from the axes to find the change in velocity (y-axis) and time taken (x-axis). $\Delta v = 12\text{ m/s} - 0\text{ m/s} = 12\text{ m/s}$ $t = 4\text{ s} - 0\text{ s} = 4\text{ s}$

• Step 2: Write the formula for the gradient (acceleration). $a = \frac{\Delta v}{t}$ (where $\Delta v$ is the change in velocity)

• Step 3: Substitute the values. $a = \frac{12}{4}$

• Step 4: Calculate the final answer with units. $a = 3\text{ m/s}^2$

For curved graphs (Higher Tier), you must find the instantaneous acceleration by drawing a tangent (a straight line that just touches the curve at a specific point). You then calculate the gradient of this tangent by forming a large right-angled triangle.

Distance from Area Under the Graph (Higher Tier)

You can figure out exactly how far a journey was just by looking at the geometric shape of a velocity-time graph. The area under the graph (the space between the velocity line and the time axis) represents the total distance travelled or displacement.

• For linear graphs with straight lines, split the area into standard shapes like rectangles and triangles.
• For non-linear, curved graphs, estimate the area by counting the squares under the curve.
• To find the distance from squares, multiply the x-axis value of one square by the y-axis value: $\text{Value of 1 square} = (\text{width in s}) \times (\text{height in m/s})$.

Worked Example: Calculating Distance from a Graph

A velocity-time graph shows a diagonal line rising from $(0\text{ s}, 0\text{ m/s})$ to a peak at $(5\text{ s}, 10\text{ m/s})$, then remaining perfectly horizontal until $t = 15\text{ s}$. Calculate the total distance travelled.

• Step 1: Split the area under the line into standard geometric shapes by reading the axes. There is a triangle from $t = 0\text{ s}$ to $t = 5\text{ s}$, and a rectangle from $t = 5\text{ s}$ to $t = 15\text{ s}$.

• Step 2: Read the base and height for the triangle from the axes to calculate its area. $\text{Base} = 5\text{ s} - 0\text{ s} = 5\text{ s}$ $\text{Height} = 10\text{ m/s} - 0\text{ m/s} = 10\text{ m/s}$ $\text{Triangle area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 10 = 25\text{ m}$

• Step 3: Read the base and height for the rectangle from the axes to calculate its area. $\text{Base} = 15\text{ s} - 5\text{ s} = 10\text{ s}$ $\text{Height} = 10\text{ m/s} - 0\text{ m/s} = 10\text{ m/s}$ $\text{Rectangle area} = \text{base} \times \text{height} = 10 \times 10 = 100\text{ m}$

• Step 4: Add the areas together for the total distance. $\text{Total distance} = 25 + 100 = 125\text{ m}$

Terminal Velocity

Why does a skydiver eventually stop speeding up, no matter how long they fall? This happens due to the changing balance of forces acting on an object falling through a fluid (like air or water).

• Initial fall: At release, the only force is weight, acting downwards. This creates a large resultant force, causing an initial downward acceleration of approx $9.8\text{ m/s}^2$.
• Increasing speed: As the object falls faster, air resistance (drag) increases because it collides with more air particles per second.
• Resultant force reduction: The weight does NOT change. However, because upward drag is increasing, the overall resultant force decreases, meaning acceleration decreases (the graph curve becomes less steep).
• Terminal velocity: Eventually, air resistance increases until it exactly equals the weight. The forces are in a state of equilibrium. The resultant force becomes zero ($0\text{ N}$), so the object stops accelerating and reaches a constant maximum speed, called terminal velocity. This appears as a horizontal line on the graph.

Opening a parachute drastically increases surface area, creating a sudden spike in air resistance. The drag temporarily becomes greater than weight, creating an upward resultant force that causes rapid deceleration until a new, lower terminal velocity is reached.

If you drop different stacks of cupcake cases, a heavier stack reaches a higher terminal velocity. The weight is greater, so the cases must fall faster to generate enough air resistance to balance the forces again.