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Physics 1/Kinematics

Describing motion

How position, velocity and acceleration relate — and the constant-acceleration equations that follow. Anchored to OpenStax University Physics Vol. 1, Ch. 3 (§3.3–3.6) and Ch. 4 (§4.3).

Before you start — give these a try

Attempting first primes your brain for the lesson — even if you miss. Nothing is graded or saved; it's just a warm-up.

For that same car (from rest, 3.0 m/s23.0\ \text{m/s}^2, 5.0 s5.0\ \text{s}), how far does it travel (m)?

At the highest point of a projectile's path, which statement is correct?

What you’ll be able to do

  • Distinguish distance from displacement, and speed from velocity.
  • Use the three constant-acceleration equations to solve 1-D motion.
  • Treat projectile motion as independent horizontal and vertical motion.
1

Position, velocity, acceleration

Displacement Δx=xf−xi\Delta x = x_f - x_i is a vector (it has sign/direction); distance is the path length and is never negative. Average velocity is vˉ=Δx/Δt\bar v = \Delta x/\Delta t and average acceleration is aˉ=Δv/Δt\bar a = \Delta v/\Delta t. Instantaneous values are the derivatives v=dx/dtv = dx/dt and a=dv/dta = dv/dt. (OpenStax §3.1–3.3.)

2

Constant-acceleration equations

When aa is constant, three equations connect x, v, tx,\,v,\,t: v=v0+atv = v_0 + at,   x=x0+v0t+12at2\;x = x_0 + v_0 t + \tfrac12 a t^2, and   v2=v02+2a Δx\;v^2 = v_0^2 + 2a\,\Delta x. Pick the one missing the variable you don't have. Free fall is just a=−ga = -g with g=9.8 m/s2g = 9.8\ \text{m/s}^2. (OpenStax §3.4, §3.6.)

3

Projectile motion

Horizontal and vertical motion are independent: horizontally ax=0a_x = 0 so vx=v0cos⁡θv_x = v_0\cos\theta stays constant; vertically ay=−ga_y = -g. The only thing they share is the clock tt. On level ground the range is R=v02sin⁡(2θ)/gR = v_0^2\sin(2\theta)/g. (OpenStax §4.3.)

Kinematic equations (constant a)

v=v0+at,x=x0+v0t+12at2,v2=v02+2a Δx.v = v_0 + at,\quad x = x_0 + v_0 t + \tfrac12 a t^2,\quad v^2 = v_0^2 + 2a\,\Delta x.

Intuition. Velocity grows linearly with time; position grows with a t2t^2 term; and the third equation is the first two with time eliminated — handy when you don't know tt.
Range of a projectile

On level ground, R=v02sin⁡(2θ)gR = \dfrac{v_0^2 \sin(2\theta)}{g}.

Intuition. Range is largest at θ=45°\theta = 45° (where sin⁡2θ=1\sin 2\theta = 1) and is the same for complementary angles like 30°30° and 60°60°.

Worked examples

Example 1

A car starts from rest and accelerates at 3.0 m/s23.0\ \text{m/s}^2 for 5.0 s5.0\ \text{s}. Find its final speed and the distance covered.

  1. 1

    Final speed: v=v0+at=0+(3.0)(5.0)=15 m/sv = v_0 + at = 0 + (3.0)(5.0) = 15\ \text{m/s}.

  2. 2

    Distance: x=v0t+12at2=0+12(3.0)(5.0)2=37.5 mx = v_0 t + \tfrac12 a t^2 = 0 + \tfrac12(3.0)(5.0)^2 = 37.5\ \text{m}.

Answer. v=15 m/sv = 15\ \text{m/s},   x=37.5 m\;x = 37.5\ \text{m}.