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All lessons Mechanics26 min

Equations of Motion

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← Acceleration and Motion GraphsFree Fall →
01
Hook
02
Explore
03
Formalize
04
Practice
05
Challenge
Interactive simulation
01

Hook

A jet must reach 80 m/s to lift off and the runway is only 1500 m long. You know the engine's acceleration and the takeoff speed — but nobody handed you the time. Can you still prove the plane gets airborne in time?

02

Explore

Complete previous stage
03

Formalize

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04

Practice

Complete previous stage
05

Challenge

Complete previous stage
Spoilers

Equations of Motion — summary and key formula

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The question

A jet must reach 80 m/s to lift off and the runway is only 1500 m long. You know the engine's acceleration and the takeoff speed — but nobody handed you the time. Can you still prove the plane gets airborne in time?

When acceleration is constant, four equations link u, v, a, s and t — and each one skips a different quantity. Pick the equation missing the variable you don't have, and any constant-acceleration problem cracks open.

The key idea

When acceleration is constant, five quantities describe the motion: initial velocity u, final velocity v, acceleration a, displacement s, and time t. Four 'SUVAT' equations connect them, and each equation leaves out exactly one quantity. To solve any problem, list what you know, spot the one variable you neither have nor want, and choose the equation that omits it.

The trick is matching equation to unknowns. No time given? Use v2=u2v^2 = u^2v2=u2 + 2as. Don't know the final velocity yet? Use s = ut + ½at². The equation s=12(u+v)ts = \tfrac{1}{2}(u+v)ts=21​(u+v)t is just average velocity ½(u+v) multiplied by time — handy when a isn't given. Watch signs: pick a positive direction and keep u, v, a and s consistent with it; a deceleration is simply a negative a. For vertical motion under gravity, a=−9.8 m/s2a = -9.8\,\text{m/s}^2a=−9.8m/s2 (the topic of the next lesson, Free Fall). **Limiting case:** set a=0a = 0a=0 and all three equations collapse to s=vts = vts=vt; set u=0u = 0u=0 and they simplify to v=atv = atv=at, s=12at2s = \tfrac{1}{2}at^2s=21​at2 — sanity-check every answer against these. **Connect it:** all three live inside the v–t graph: v=u+atv = u + atv=u+at is the straight line itself, sss is the area under it, and v2=u2+2asv^2 = u^2 + 2asv2=u2+2as is what remains when you eliminate ttt between them. One picture, three equations.

The formula

v=u+ats=ut+12at2v2=u2+2ass=12(u+v)tv = u + at \quad s = ut + \tfrac{1}{2}at^2 \quad v^2 = u^2 + 2as \quad s = \tfrac{1}{2}(u+v)tv=u+ats=ut+21​at2v2=u2+2ass=21​(u+v)t
  • ·u = initial velocity (m/s)
  • ·v = final velocity (m/s)
  • ·a = acceleration (m/s²)
  • ·s = displacement (m)
  • ·t = time (s). v = u + at omits s; s = ut + ½at² omits v; v² = u² + 2as omits t; s = ½(u+v)t omits a. All assume a is constant.

Common mistake

Grabbing a SUVAT equation at random — first list which of u, v, a, s, t you know and want, then pick the equation that omits the variable you don't have.

What to remember

  • ·The four SUVAT equations link u, v, a, s, t; each one leaves a different variable out.
  • ·They apply only to CONSTANT acceleration.
  • ·Pick the equation containing your three knowns plus the unknown.