KinexiaLEARN PHYSICS

SUVAT Displacement — Elevator Rising (Worked Example)

Easy
Problem

An elevator starts from rest at the lobby and accelerates upward at 1.2 m/s² for 3.0 s. How far does it rise during this acceleration phase?

Given

  • u = 0 m/s

  • a = 1.2 m/s²

  • t = 3.0 s

Find

The distance risen s.

Principle

Even vertically, constant acceleration obeys s = ut + ½at². Direction only matters for the sign convention.

Formula

Solution

  1. Write the equation: s=ut+12at2s = ut + \frac{1}{2}at^2

  2. Substitute: s=0+12(1.2)(3.0)2s = 0 + \frac{1}{2}(1.2)(3.0)^2

  3. Calculate: s=0.5×1.2×9=5.4 ms = 0.5 \times 1.2 \times 9 = 5.4\ \mathrm{m}

5.4 m

Unit check

m/s² × s² = m — a distance, as required.

What this means physically

5.4 m is roughly one and a half floors — the elevator covers this while still speeding up, before settling into its constant cruising speed.

Common mistake

Treating the 3.0 s as a distance or mixing up a and t. Label every quantity with its unit before substituting.