Form 4 · Chapter 7

Solving Problems with Graphs of Motion

Read speed from a distance-time gradient, acceleration from a speed-time gradient, and distance from the area under a speed-time graph.

Two graphs, three ideas

In a distance-time graph, the gradient equals the speed. A steeper line means faster motion; a horizontal line means the object is at rest. In a speed-time graph, the gradient equals the acceleration, and the area under the graph equals the distance travelled.

Key formula

Speed $=\dfrac{\text{distance}}{\text{time}}$ (gradient of distance-time). Acceleration $=\dfrac{v-u}{t}$ (gradient of speed-time). Distance $=$ area under speed-time graph.

Working with the area

The area under a speed-time graph is split into simple shapes. A rectangle gives $\text{base}\times\text{height}$; a triangle gives $\tfrac{1}{2}\times\text{base}\times\text{height}$; a trapezium gives $\tfrac{1}{2}(a+b)\times h$. Add the parts to get total distance. Average speed for a whole journey is $\dfrac{\text{total distance}}{\text{total time}}$.

Worked example

A car accelerates from rest to $20\ \text{m s}^{-1}$ in 4 s, holds $20\ \text{m s}^{-1}$ for 6 s, then stops in 2 s.
Acceleration (first stage) $=\dfrac{20-0}{4}=5\ \text{m s}^{-2}$. Distance $=$ triangle $+$ rectangle $+$ triangle $=\tfrac{1}{2}(4)(20)+(6)(20)+\tfrac{1}{2}(2)(20)=40+120+20=180\ \text{m}$. Average speed $=\dfrac{180}{12}=15\ \text{m s}^{-1}$.

Remember

  • Distance-time gradient $=$ speed; flat line $=$ at rest.
  • Speed-time gradient $=$ acceleration; negative gradient $=$ deceleration.
  • Area under speed-time $=$ distance.

Stuck on this topic? A verified JomKelas tutor can walk you through it.

Find a verified tutor