Motion in a Straight Line · Lesson 5 of 10
Deriving the Equations of Uniformly Accelerated Motion
“Derive and apply the constant-acceleration equations using graphs, algebra and calculus.”
• Derive the velocity, displacement and velocity-squared equations. • Include a nonzero initial position correctly. • Understand the calculus derivations and their assumptions. • Choose an equation and split distance calculations at reversals.
State the assumptions before choosing a formula
Uniform acceleration means that acceleration remains constant in magnitude and direction during the interval. Choose one axis, set elapsed time t = 0 at the interval’s start, and let initial position be x₀ and initial velocity be u. At time t, position is x and velocity is v. Write s = x − x₀ for signed displacement. The symbol s here does not mean total distance if a reversal occurs.
Deriving displacement from a graph
On a velocity-time graph, constant acceleration gives a straight line from u to v. First consider positive velocities with v larger than u. The area under the line consists of a rectangle of area ut and a triangle of area ½(v − u)t. Their sum is displacement. Substituting v − u = at gives the second equation. The same algebra remains valid with signed areas when velocity is negative or crosses zero.
Eliminating time
Multiply s = (u + v)t/2 by 2a, and substitute at = v − u. This gives 2as = (u + v)(v − u). The product is a difference of squares, v² − u², so the third equation follows. This multiplication derivation also avoids dividing by a when acceleration is zero.
The same derivations using calculus
Integration accumulates a rate over an interval. To integrate a constant with respect to time, multiply it by the interval. To integrate t, use t²/2. The limits enforce initial conditions, so no arbitrary starting value is lost. These three derivations connect the definitions of velocity and acceleration to the familiar equations.
For the third relation, use the chain rule on a smooth segment: a = dv/dt = (dv/dx)(dx/dt) = v dv/dx. Rearrange as v dv = a dx and integrate. At a turning point it is safer to work segment by segment because velocity as a function of position may have two branches. The final constant-acceleration identity nevertheless holds across the complete trajectory.
Choosing the shortest valid route
| Wanted relationship | Useful equation | Quantity absent |
|---|---|---|
| Velocity after an elapsed time | v = u + at | Displacement |
| Displacement after an elapsed time | s = ut + ½at² | Final velocity |
| Velocity and displacement | v² = u² + 2as | Time |
| Displacement from endpoint velocities and time | s = (u + v)t/2 | Acceleration |
Problem
At t = 0, a cart is at x₀ = −3 m, moving at u = +2 m/s. Its constant acceleration is +4 m/s². Find its position and velocity after 3 s.
- 1.Velocity is v = 2 + 4 × 3 = +14 m/s.
- 2.Displacement is s = 2 × 3 + ½ × 4 × 3² = 24 m.
- 3.Position is x = −3 + 24 = 21 m. Writing x = 24 m would incorrectly ignore the initial position.
- 4.Check with v² = u² + 2as: 196 = 4 + 2 × 4 × 24. All terms have units m²/s².
When displacement is not distance
Problem
A particle starts with u = 10 m/s and a = −2 m/s². Find displacement and distance after 8 s.
- 1.It turns when 0 = 10 − 2t, giving t = 5 s.
- 2.Whole-interval displacement is 10 × 8 − ½ × 2 × 8² = 16 m.
- 3.During the first 5 s it advances 25 m. During the remaining 3 s it starts from rest and travels 9 m backwards.
- 4.Distance is 25 + 9 = 34 m. Constant acceleration allows one equation for displacement across a reversal, but distance requires adding segment lengths.
What changes when acceleration varies
The definitions v = dx/dt and a = dv/dt remain valid for differentiable motion even when acceleration changes. The three constant-acceleration equations do not. For instance, if a = kt with k constant, integrating gives v = u + ½kt² and then s = ut + kt³/6. Replacing a in ut + ½at² by its final value kt would give the wrong coefficient. If acceleration is piecewise constant, apply the usual equations separately to each phase and carry final position and velocity into the next phase.
Quiz
Which assumption is needed to use s = (u + v)t/2 for an arbitrary interval?
Key Takeaways
• Derive v = u + at from acceleration, then derive displacement from graph area. • Use s = x − x₀, with consistent signs. • The calculus definitions are general; the standard three equations require constant acceleration. • Check for a reversal whenever total distance is requested.