Rectilinear motion

The idea

The simplest motion: a particle moving along a straight line. Everything is one-dimensional, so position, velocity and acceleration are just numbers with signs. It’s the template for everything that follows, from curves to rotating bodies.

The definitions

Quantity	Definition	Note
Position $s$	Distance from a fixed origin, with a sign
Displacement $\Delta s$	Change in position	Not the same as distance travelled
Velocity $v$	$v=\frac{ds}{dt}$	Its sign shows direction
Acceleration $a$	$a=\frac{dv}{dt}$	Negative $a$ isn’t always “slowing down”: it depends on the sign of $v$

Distance vs displacement: if something goes forward 5 m and back 1 m, the displacement is 4 m but the distance travelled is 6 m. Find where $v=0$ (where it turns round) and add up each leg.

Constant acceleration

Only when $a$ is constant:

$v=v_0+ats=s_0+v_{0}t+\frac{1}{2}a{t}^2{v}^2=v_0^2+2a(s-s_0)$

If $a$ changes with time, integrate instead (see Calculus for kinematics).

Worked examples

1. A delivery robot speeds up from 0.5 m/s at a constant 0.8 m/s² for 3 s. Final speed and distance?

Answer

$v=0.5+0.8\times 3=2.9$ m/s $s=0.5\times 3+\frac{1}{2}\times 0.8\times 3^2=1.5+3.6=5.1$ m Check: $v^2=0.25+2\times 0.8\times 5.1=8.41$ → $v=2.9$ ✓

2. A slider’s position is $s=2t^3-9t^2+12t$ m. When does it stop, and how far does it travel in the first 3 s?

Answer

$v=6t^2-18t+12=6(t-1)(t-2)$, so it stops at t = 1 s and t = 2 s. $a=12t-18$, which is zero at 1.5 s. $s(0)=0$, $s(1)=5$, $s(2)=4$, $s(3)=9$ m. Distance $=5+1+5=$ 11 m, but the displacement is only 9 m.

Common mistakes

• Using the constant-acceleration equations when $a$ depends on $t$.
• Giving displacement when the question asks for distance.
• Dropping signs: pick a positive direction and stick to it.

Check yourself

1. A cart moving at 3 m/s brakes at a constant 1.5 m/s². Stopping distance?

$0=9-2(1.5)s$ → $s=3$ m.