POP

E2.9 Graphs in Practical Situations: Kinematics & Rates of Change

Image · 914 × 270

Distance-time and speed-time graphs turn a moving object's journey into a picture, and reading the right feature off each type of graph is the whole skill.

What does the gradient of a distance-time graph tell me?

A straight section's gradient gives constant speed; the gradient of a tangent to a curved section gives the instantaneous speed at that exact moment. A horizontal section (zero gradient) means the object has stopped — distance is not changing even though time is passing.

How do I estimate the gradient of a tangent at a point on a curve?

Draw a straight line that just touches the curve at the given point without crossing through it nearby, then pick two points that are easy to read on that tangent line and calculate its gradient using $\dfrac{\text{change in }y}{\text{change in }x}$ as normal. The accuracy of this method depends entirely on how carefully the tangent is drawn, so use a ruler and make the line as long as the grid allows to reduce reading error.

What does the gradient of a speed-time graph tell me?

It gives acceleration. A positive gradient means the object is speeding up; a negative gradient means it's slowing down (decelerating).

Exam Tip

A falling section of a speed-time graph has a negative gradient. State the deceleration as a positive number, but describe it explicitly as a deceleration — don't just call it "acceleration."

How do I find the distance travelled from a speed-time graph?

Distance travelled equals the area under the graph. Depending on the shape, this might be a triangle ( $\tfrac12\times\text{base}\times\text{height}$ ), a rectangle (base $\times$ height), or a trapezium ( $\tfrac12(a+b)h$ ).

How do I handle a journey with several stages?

Split the graph into its separate sections (acceleration, constant speed, deceleration), find the area of each section using the appropriate formula, and add the areas together for the total distance.

A cyclist accelerates uniformly from rest to $8\text{ m/s}$ in $10\text{ s}$ , holds that speed for $20\text{ s}$ , then decelerates to rest in $6\text{ s}$ . The graph is a trapezium shape: rising from $(0,0)$ to $(10,8)$ , flat to $(30,8)$ , then falling to $(36,0)$ .

\text{Distance} = \frac{1}{2}(20+36)\times8 = \frac{1}{2}(56)(8) = 224\text{ m}

Here the two parallel sides of the trapezium are the duration at constant speed ( $20\text{ s}$ ) and the total journey time ( $36\text{ s}$ ).

Exam Tip

It's just as safe (and often clearer) to split a composite speed-time graph into a triangle, a rectangle, and a second triangle, and add the three separate areas, rather than trying to spot the trapezium shape directly. Both methods give the same total.

Worked Example

A car decelerates uniformly from $24\text{ m/s}$ to rest in $8$ seconds.

(a) Find the deceleration.

(b) Find the distance travelled during this braking period.

(a) Deceleration is the size of the gradient:

\text{Acceleration} = \frac{\Delta v}{\Delta t} = \frac{0-24}{8} = -3\text{ m/s}^2 \implies \text{deceleration } = 3\text{ m/s}^2

(b) The speed-time graph here is a single triangle with base $8$ and height $24$ :

\text{Distance} = \frac{1}{2}\times8\times24 = 96\text{ m}

Exam Tip

Always double-check which pair of values on the graph form the "parallel sides" of a trapezium before applying the area formula — using the wrong two time intervals is the single most common error in composite-journey questions. Sketching the shape roughly first, and labelling each vertex with its coordinates, catches this before it costs marks.