Kinematics of motion in a straight line
Scalars, vectors and graphs
- Distance and speed are scalars; displacement, velocity and acceleration are vectors.
- On a velocity–time graph: area = displacement, gradient = acceleration.
- Differentiate to go displacement → velocity → acceleration; integrate to go back.
Practice
On a velocity–time graph, the area under the line gives the:
Area = displacement; the gradient gives the acceleration.
The suvat equations
For constant acceleration:
$$v = u + at, \quad s = ut + \tfrac12 at^2,$$
$$v^2 = u^2 + 2as, \quad s = \tfrac12(u+v)t.$$
- Example: from rest ($u=0$), $a=2.5$, $t=4$: $v = 10\,\text{m s}^{-1}$, $s = 20\,\text{m}$.
Practice
A car starts from rest and accelerates at 2.5 m/s² for 4 s. Using v = u + at, what is its speed (m/s)?
v = 0 + 2.5 × 4 = 10 m/s.
Practice
For the same car (u = 0, a = 2.5, t = 4), what distance does it travel? (s = ut + ½at²)
s = 0 + ½ × 2.5 × 4² = ½ × 2.5 × 16 = 20 m.
You've got it
Key idea
- v–t graph: area = displacement, gradient = acceleration
- suvat (constant a): $v = u+at$, $s = ut + \tfrac12 at^2$, $v^2 = u^2 + 2as$
- differentiate s→v→a; integrate to reverse