Summary — What This Chapter Covers
- Distance and speed are scalars; displacement, velocity and acceleration are vectors (with a sign in one dimension).
- Gradient of a displacement–time graph = velocity; gradient of a velocity–time graph = acceleration.
- Area under a velocity–time graph = displacement.
- Constant acceleration: the five suvat equations.
- Variable acceleration: v = ds/dt, a = dv/dt; integrate the other way and use the initial conditions.
1. Displacement–Time and Velocity–Time Graphs
Graphs
Gradient and area
On a displacement–time graph the gradient is the velocity, so a horizontal section means the particle is at rest. On a velocity–time graph the gradient is the acceleration and the area between the graph and the time axis is the displacement. Area below the axis counts as negative displacement; for distance add the areas as positive values.
Most graph questions are trapezium questions: the motion is a sequence of straight-line sections, so the area is a combination of triangles and rectangles. If the question gives the total distance or total time, set up an equation in the unknown.
Reading a v–t graph
acceleration = gradient displacement = signed area distance = total area
Worked Example
A train accelerates uniformly from rest to 12 m s−1 in 4 s, travels at 12 m s−1 for 10 s, then decelerates uniformly to rest in 6 s. Find the acceleration, the deceleration and the total distance.
Practice Question 1
A car starts from rest and accelerates at 2 m s−2 to a speed V, travels at V for 20 s, then decelerates at 1 m s−2 to rest. The whole journey takes 50 s. Find V and the total distance.
2. Constant Acceleration Formulae
suvat
Five equations, three knowns
When acceleration is constant, list s, u, v, a, t, mark the three you know and the one you want, and pick the equation that leaves out the fifth. Choose a positive direction first. For a particle moving vertically under gravity only, a = −10 m s−2 if up is positive.
Questions often involve two particles: write an expression for each one's displacement in terms of t (being careful if they start at different times), then equate positions or set their separation.
suvat (constant a only)
v = u + at s = ½(u + v)t s = ut + ½at2 s = vt − ½at2 v2 = u2 + 2as
Worked Example
A ball is thrown vertically upwards at 15 m s−1 from a point 20 m above the ground. Find the time it takes to reach the ground and its greatest height above the ground.
Practice Question 2
Car P starts from rest at a point O and accelerates at 2 m s−2. At the same instant car Q passes O at a constant 8 m s−1 in the same direction. When and where does P catch Q?
3. Variable Acceleration Using Calculus
Calculus
Differentiate down, integrate up
not
distance
Calculus links
v = ds/dt, a = dv/dt s = ∫v dt, v = ∫a dt
Memorise