Kinematics of motion in a straight line

Scalars and vectors in kinematics Paper 4

Kinematics describes motion without asking what causes it. In Paper 4 the motion is always in a straight line, in one dimension, so a direction is just a sign: choose one direction as positive and stick to it.

• Distance and speed are scalars: they have size only and are never negative.
• Displacement, velocity and acceleration are vectors: they have a direction, shown by the sign.

A ball thrown up $5\text{m}$ and caught again has travelled a distance of $10\text{m}$ but has displacement $0$. A particle moving at $-3{\text{m s}}^{-1}$ has speed $3{\text{m s}}^{-1}$ and is moving in the negative direction.

Acceleration is the rate of change of velocity. A negative acceleration does not always mean slowing down: a particle moving in the negative direction with negative acceleration is speeding up. The word deceleration is used for a decrease in speed: "a deceleration of $2{\text{m s}}^{-2}$" means the speed falls by $2{\text{m s}}^{-1}$ every second, so in your equations $a=-2$ if the motion is in the positive direction.

Displacement-time and velocity-time graphs Paper 4

Graphs carry three facts that examiners test again and again.

• The gradient of a displacement-time graph is the velocity. A straight section means constant velocity; a horizontal section means the particle is at rest; a downward slope means it is moving in the negative direction.
• The gradient of a velocity-time graph is the acceleration. A horizontal section means constant velocity, zero acceleration.
• The area under a velocity-time graph is the displacement.

Why is the area the displacement? Over a short time $\delta t$ the velocity is nearly constant, so the displacement is about $v\times \delta t$, the area of a thin strip. Adding the strips gives the area. Where the graph is below the time axis the velocity is negative, so that area counts as negative displacement. The distance travelled is the total area, counting every part as positive.

Most velocity-time graphs in Paper 4 are made of straight lines, so the areas are triangles, rectangles and trapezia. When you sketch one, label the key velocities and times; a sketch need not be to scale, but it should show the shape and the values you know.

The constant acceleration formulae Paper 4

When the acceleration is constant, five quantities describe a stage of motion: $s$ (displacement), $u$ (initial velocity), $v$ (final velocity), $a$ (acceleration) and $t$ (time). These formulae are printed in the list of formulae (MF19):

$$v=u+ats=\frac{1}{2}(u+v)ts=ut+\frac{1}{2}a{t}^2{v}^2=u^2+2as$$

They all come from the straight-line velocity-time graph. The gradient is $a$, so $v=u+at$. The area is a trapezium, so $s=\frac{1}{2}(u+v)t$. Substituting one into the other gives the other two. You may also use $s=vt-\frac{1}{2}a{t}^2$, which is not printed.

Method: list the five letters, fill in the three you know, mark the one you want, and choose the formula that contains those four. Each formula leaves out one of the five letters.

Vertical motion under gravity alone is constant acceleration of $g=10{\text{m s}}^{-2}$ downwards. Take upwards as positive and $a=-10$ throughout, even while the particle is rising; this way one equation covers the whole flight and the signs sort themselves out. At the highest point $v=0$.

These formulae are only for constant acceleration. If the acceleration changes, use them separately for each stage, or use calculus.

Two particles: setting up more than one equation Paper 4

Some questions follow two particles, or one particle over two stages. The approach is the same each time:

• measure every displacement from the same point and every time from the same instant;
• write an expression for each particle's displacement (or velocity) in terms of $t$;
• turn the event into an equation: "meet" or "catch up" means equal displacements; "same speed" means equal velocities; "$20\text{m}$ apart" means the displacements differ by $20$.

If one particle starts later, its time is $t-t_0$, not $t$. Draw a quick diagram of the two positions; it shows which displacement is bigger.

Variable acceleration: using calculus Paper 4

When the acceleration is not constant, the formulae above do not apply. Instead use the definitions:

$$v=\frac{\mathrm{d}s}{\mathrm{d}t}a=\frac{\mathrm{d}v}{\mathrm{d}t}$$

These are the gradient facts from the graphs, written exactly. Going the other way, integrate: $s=\int v\mathrm{d}t$ and $v=\int a\mathrm{d}t$. Each integration brings a constant, which you find from a known value, such as "starts from rest" ($v=0$ when $t=0$) or "starts at $O$" ($s=0$ when $t=0$). Alternatively, the displacement between two times is the definite integral ${\int }_{t_1}^{t_2}v\mathrm{d}t$, the area under the velocity-time graph.

The calculus needed is restricted to the Pure Mathematics 1 techniques: powers of $t$, including fractional and negative powers.

• Greatest or least velocity: set $a=\frac{\mathrm{d}v}{\mathrm{d}t}=0$.
• Instantaneous rest: set $v=0$.
• Distance travelled: find when $v=0$ inside the interval. If the particle changes direction, integrate each piece separately and add the sizes; a single integral over the whole interval gives the displacement, in which the backward motion cancels some of the forward motion.