B. Kinematics · Kinematics: Motion in One and Two DimensionsDeriving the Equations of Uniformly Accelerated Motion
B2.3 B2.2 B3.1About 70 minutes
Learning goals- I can derive v₂ = v₁ + aΔt from the slope of a velocity-time graph.
- I can derive Δd = ½(v₁ + v₂)Δt, Δd = v₁Δt + ½aΔt² and Δd = v₂Δt − ½aΔt² from areas under a velocity-time graph.
- I can derive v₂² = v₁² + 2aΔd by combining two equations.
- I can state which variable each equation leaves out and when the equations apply.
Words to know
uniformly accelerated motion
Motion with constant acceleration; its v–t graph is a straight line.
v₁ (initial velocity)
The velocity at the start of the time interval.
v₂ (final velocity)
The velocity at the end of the time interval.
derive
To build a new equation step by step from facts or equations you already accept.
trapezoid
A four-sided figure with one pair of parallel sides. Area = ½(sum of parallel sides) × distance between them.
Five equations from one graph
You could memorize five equations for motion with constant acceleration. Or you could understand where they come from, so that you can rebuild any of them from a single sketch. Physicists prefer the second way. Everything in this lesson comes from one straight-line velocity-time graph and two facts you already know: slope = acceleration and area = displacement.
The set-up
An object has constant acceleration a. At the start of a time interval its velocity is v₁; after a time interval Δt its velocity is v₂. Its displacement in that interval is Δd. Because a is constant, the v–t graph is a straight line from (0, v₁) to (Δt, v₂).
That gives five variables: Δd, v₁, v₂, a and Δt.
Equation 1: from the slope
The slope of the v–t graph is the acceleration:
a = (v₂ − v₁) ÷ Δt
Multiply both sides by Δt and add v₁:
(1) v₂ = v₁ + aΔt
In words: the final velocity is the starting velocity plus the change (aΔt).
Equations 2 and 3: from the area
The region under the line is a trapezoid, with parallel sides v₁ and v₂ and width Δt:
(2) Δd = ½(v₁ + v₂)Δt
This also tells you that, for constant acceleration only, the average velocity is the average of the starting and final velocities: vₐᵥ = (v₁ + v₂) ÷ 2.
Now split the same region into a rectangle and a triangle:
- rectangle: height v₁, width Δt, area v₁Δt
- triangle: height (v₂ − v₁), width Δt, area ½(v₂ − v₁)Δt
From Equation 1, v₂ − v₁ = aΔt. Substitute this into the triangle: ½(aΔt)(Δt) = ½aΔt². Adding the two pieces:
(3) Δd = v₁Δt + ½aΔt²
Equation 5: rectangle minus triangle
You can also think of the area as a big rectangle of height v₂ and width Δt, minus the triangle that sits above the line. That triangle has height v₂ − v₁ = aΔt and width Δt, so its area is ½aΔt²:
(5) Δd = v₂Δt − ½aΔt²
(We call this Equation 5 because Equation 4, below, needs a little algebra instead of a picture.)
Worked exampleExample 1: Deriving Equation 4 (no Δt)
Combine Equations 1 and 2 to get an equation that does not contain Δt.
- From Equation 1: v₂ − v₁ = aΔt, so Δt = (v₂ − v₁) ÷ a.
- Substitute into Equation 2: Δd = ½(v₁ + v₂) × (v₂ − v₁) ÷ a.
- Multiply both sides by 2a: 2aΔd = (v₂ + v₁)(v₂ − v₁).
- The right side is a difference of squares (from Grade 10): (v₂ + v₁)(v₂ − v₁) = v₂² − v₁².
- So 2aΔd = v₂² − v₁², which rearranges to (4) v₂² = v₁² + 2aΔd.
| Equation | Leaves out |
|---|
| (1) v₂ = v₁ + aΔt | Δd |
| (2) Δd = ½(v₁ + v₂)Δt | a |
| (3) Δd = v₁Δt + ½aΔt² | v₂ |
| (4) v₂² = v₁² + 2aΔd | Δt |
| (5) Δd = v₂Δt − ½aΔt² | v₁ |
Worked exampleExample 2: Checking that the equations agree
An object starts at v₁ = 4.0 m/s and accelerates at a = 2.0 m/s² for Δt = 5.0 s. Find v₂ and then find Δd with Equations 2, 3 and 5. Check with Equation 4.
- (1) v₂ = 4.0 + (2.0)(5.0) = 14 m/s.
- (2) Δd = ½(4.0 + 14)(5.0) = ½(18)(5.0) = 45 m.
- (3) Δd = (4.0)(5.0) + ½(2.0)(5.0)² = 20 + 25 = 45 m.
- (5) Δd = (14)(5.0) − ½(2.0)(5.0)² = 70 − 25 = 45 m.
- (4) v₂² = 4.0² + 2(2.0)(45) = 16 + 180 = 196, so v₂ = √196 = 14 m/s ✓. All five equations agree.
v₂ = 14 m/s and Δd = 45 m by every method.
Worked exampleExample 3: A first application
A car travelling at 12 m/s [E] accelerates at 1.5 m/s² [E] for 6.0 s. How far does it travel in that time?
- East positive. Given: v₁ = 12 m/s, a = 1.5 m/s², Δt = 6.0 s. Required: Δd. Not given and not needed: v₂.
- Use the equation that leaves out v₂, Equation 3: Δd = v₁Δt + ½aΔt².
- Δd = (12)(6.0) + ½(1.5)(6.0)² = 72 + 27 = 99 m.
- Check another way: v₂ = 12 + 1.5 × 6.0 = 21 m/s; Δd = ½(12 + 21)(6.0) = 99 m ✓.
When do these equations work?
- Only when the acceleration is constant for the whole interval. If the motion has stages (speed up, cruise, brake), use the equations one stage at a time.
- If a = 0, every equation reduces to Δd = vΔt (uniform motion).
- All quantities except Δt are vectors in one dimension: choose a positive direction and give each one the correct sign.
- vₐᵥ = (v₁ + v₂) ÷ 2 is true only for constant acceleration. In general, vₐᵥ = Δd ÷ Δt.
Just for funGalileo had no stopwatch. To time balls rolling down a ramp, he let water flow out of a container while the ball rolled, then weighed the water. More water meant more time. With this "water clock" he found that the distance rolled from rest grows with the square of the time, the same idea as Δd = ½aΔt².
Hands-on activityCount the squares
You need: graph paper, a ruler and a pencil.
- Draw a v–t graph with 1 square = 1 s across and 1 square = 1 m/s up. Draw a straight line from (0, 2 m/s) to (6 s, 8 m/s).
- Count the squares under the line (count half squares where the line cuts squares diagonally). Each square is 1 m/s × 1 s = 1 m of displacement.
- Find the area three ways: as a trapezoid, as a rectangle plus a triangle, and as a big rectangle minus a triangle.
- Find the acceleration from the slope and use each of Equations 2, 3 and 5 to check your count.
What to look for: a = (8 − 2) ÷ 6 = 1 m/s². Trapezoid: ½(2 + 8)(6) = 30 m. Rectangle 2 × 6 = 12 m plus triangle ½(6)(6) = 18 m gives 30 m. Big rectangle 8 × 6 = 48 m minus triangle 18 m gives 30 m. Your square count should be 30 (or very close).
Your turn
1Which equation leaves out the acceleration?
Δd = ½(v₁ + v₂)Δt (Equation 2).
2Which equation would you use if you know v₁, a and Δd and want v₂?
v₂² = v₁² + 2aΔd (Equation 4); it leaves out Δt.
3A cart starts at 3.0 m/s and accelerates at 0.50 m/s² for 8.0 s. Find v₂.
v₂ = 3.0 + (0.50)(8.0) = 7.0 m/s.
4For the same cart, find Δd using Equation 3, then check with Equation 2.
Δd = (3.0)(8.0) + ½(0.50)(8.0)² = 24 + 16 = 40 m. Check: ½(3.0 + 7.0)(8.0) = 40 m ✓.
5Show that if a = 0, Equation 3 becomes Δd = vΔt.
With a = 0, Δd = v₁Δt + ½(0)Δt² = v₁Δt, and v₁ = v₂ = v because the velocity does not change.
6In your own words, why does the area of the triangle in the derivation equal ½aΔt²?
Its height is the change in velocity, v₂ − v₁ = aΔt, and its base is Δt, so its area is ½ × base × height = ½(Δt)(aΔt) = ½aΔt².
More practice: warm-up
1What does the slope of a v–t graph give?
2What shape is the region under a straight v–t line that starts above zero?
3Write the equation that leaves out Δt.
4Write the equation that leaves out v₁.
5From rest, a = 4.0 m/s² for 3.0 s. Find Δd.
Δd = 0 + ½(4.0)(3.0)² = 18 m.
More practice: on your own
1A bike slows from 10 m/s to 4.0 m/s in 3.0 s. Find Δd without finding a first.
Δd = ½(10 + 4.0)(3.0) = 21 m (Equation 2).
2A ball reaches 6.0 m/s after accelerating at 1.5 m/s² for 2.0 s. Find Δd without finding v₁.
Δd = v₂Δt − ½aΔt² = (6.0)(2.0) − ½(1.5)(2.0)² = 12 − 3.0 = 9.0 m.
3Starting from Equations 1 and 3, eliminate v₁ to derive Equation 5.
From (1): v₁ = v₂ − aΔt. Substitute into (3): Δd = (v₂ − aΔt)Δt + ½aΔt² = v₂Δt − aΔt² + ½aΔt² = v₂Δt − ½aΔt².
4A cart starts from rest and travels 8.0 m in 4.0 s with constant acceleration. Find a and v₂.
Δd = ½aΔt²: 8.0 = ½a(16), so a = 1.0 m/s². v₂ = 0 + 1.0 × 4.0 = 4.0 m/s. (Check: ½(0 + 4.0)(4.0) = 8.0 m ✓.)
5Why can't you use vₐᵥ = (v₁ + v₂) ÷ 2 for a car that sped up, cruised and then braked?
That shortcut works only when the acceleration is constant. Here the acceleration changes between stages, so you must use vₐᵥ = Δd ÷ Δt for the whole trip, or apply the equations stage by stage.
Challenge
1Starting from rest, show that the distances travelled in the 1st, 2nd and 3rd seconds of constant acceleration are in the ratio 1 : 3 : 5.
Δd = ½at², so positions at t = 1, 2, 3 s are ½a(1), ½a(4), ½a(9). Distances in each second: ½a(1 − 0), ½a(4 − 1), ½a(9 − 4) = ½a × 1, 3, 5. (Galileo noticed this odd-number pattern.)
2Derive Equation 4 using a different route: square Equation 1 and use Equation 3.
v₂² = (v₁ + aΔt)² = v₁² + 2v₁aΔt + a²Δt² = v₁² + 2a(v₁Δt + ½aΔt²). The bracket is Δd (Equation 3), so v₂² = v₁² + 2aΔd.
3An object's v–t graph goes from v₁ = 6.0 m/s to v₂ = −2.0 m/s in 4.0 s with constant acceleration. Find Δd with Equation 2, and explain why it is smaller than the distance travelled.
Δd = ½(6.0 + (−2.0))(4.0) = 8.0 m. The object turns around (at t = 3.0 s, where v = 0), so the area below the axis (−1.0 m) cancels part of the area above it (+9.0 m). The distance is 9.0 + 1.0 = 10 m.
Quiz
1. The equation v₂ = v₁ + aΔt comes from:
2. Which equation does NOT contain Δt?
3. In the derivation of Δd = v₁Δt + ½aΔt², the term v₁Δt is the area of a:
4. From rest, a car accelerates at 3.0 m/s² for 4.0 s. Its displacement is:
5. The five equations of motion are valid only when:
6. v₁ = 2.0 m/s, v₂ = 10.0 m/s, Δt = 4.0 s, constant a. Δd is:
Big idea: Expectation B2.3 asks students to derive the equations of uniformly accelerated motion from a velocity-time graph, not just memorize them. Every equation is either the slope (Equation 1) or an area (Equations 2, 3, 5), plus one algebraic combination (Equation 4).
How to help: Ask your teen to rebuild one equation from a sketch without notes. If they can explain the rectangle-plus-triangle picture, they own Equation 3.
Common difficulties:
- Forgetting to square Δt, or squaring the ½ or the a.
- Using vₐᵥ = (v₁ + v₂) ÷ 2 when the acceleration is not constant.
- Algebra slips when rearranging (especially Equation 4). Encourage writing each step on its own line.