Understanding Motion through Experience (Advanced)

Motion depends on the observer: frames of reference, relative speed, scalars and vectors, adding displacements, and the distance covered in one particular second.

Key ideas

Rest and motion are always relative to a reference point; an inertial frame is at rest or moves at constant velocity.
Two bodies moving the same way see each other at the difference of their speeds; moving towards each other, at the sum.
Scalars (distance, speed, mass, time) have magnitude only; vectors (displacement, velocity, force) also have direction.
Vectors add head to tail: the resultant joins the first tail to the last head; walks at right angles combine by Pythagoras.
Distance is never less than the size of the displacement, which is zero for a round trip.
From rest, the distances in successive seconds are in the ratio 1 : 3 : 5 : 7 …

Remember

Equations of motion	v = u + at; s = ut + ½at²; v² = u² + 2as
Distance in the nth second	s_n = u + a(2n − 1)/2
Right-angled displacements	resultant = √(a² + b²)
Stopping distance	s = u² ÷ 2a, so doubling the speed makes it four times as long

Watch out

Using ut + ½at² (the total in n seconds) when the question asks for the nth second alone.
Adding the speeds of two trains moving the same way.
Adding perpendicular displacements as plain numbers instead of by Pythagoras.

Worked example

A toy car is pushed along a floor and its distance from the start O is marked every second: 2 m at 1 s, 6 m at 2 s, 12 m at 3 s, 20 m at 4 s. It moves with uniform acceleration. Find (a) the distance covered in the 3rd second, (b) the acceleration, (c) the initial velocity and (d) the velocity at 4 s.

Working: (a) 12 − 6; (b) each second's distance grows by a = 2; (c) s₁ = u + a/2 = 2; (d) v = u + 4a

Answer: (a) 6 m; (b) 2 m/s²; (c) 1 m/s; (d) 9 m/s