Position and Movement (II): Uniformly Accelerated Motion
FoundationStraight-line motion with constant acceleration is described by four equations: v = u + at, s = ½(u+v)t, s = ut + ½at² and v² = u² + 2as. List the knowns, pick the equation that fits — braking distance and free fall are the classic applications.
- Uniformly accelerated motion = straight-line motion with constant acceleration; the v-t graph is a sloping straight line
- Four equations: v = u + at, s = ½(u+v)t, s = ut + ½at², v² = u² + 2as
- Three-step method: list knowns, pick the equation with no spare unknown, substitute with signs — fix the positive direction first
文體小知識：科學知識說明文——介紹均勻加速運動的四條方程與選用策略，應用於煞車距離與自由落下等計算。
§Text Passage
1Straight-line motion with constant acceleration is called uniformly accelerated motion. Five quantities describe it: initial velocity, final velocity, acceleration, displacement and time. This lesson introduces four equations of motion linking them: knowing any three, you can pick a suitable equation to find the other two — the basic skill of mechanics calculation and a guaranteed exam item.
2Choosing an equation has a knack: write down the knowns and the unknown, then pick the equation that links exactly those. If time is not involved, choose the equation without time; if time is known and displacement is wanted, choose the displacement form containing time. Sign conventions matter too: take the initial velocity's direction as positive, so a braking deceleration enters as a negative value — otherwise the answer goes completely wrong.
3Braking is a classic application. For example, a car travelling at 100 km h⁻¹ moves at about 27.8 m s⁻¹; braking hard with a deceleration of 5 m s⁻², the braking distance works out at about 77.3 m. Braking distance grows with the square of speed: double the speed and the distance quadruples — the physics behind the danger of speeding.
4Vertical fall is another application: ignoring air resistance, a falling object accelerates at the gravitational acceleration g of about 9.8 m s⁻² directed downward. For example, a ball released from rest has a speed of about 9.8 m s⁻¹ after 1 s and 19.6 m s⁻¹ after 2 s; the distance fallen grows with the square of time, about 19.6 m in 2 s. Replace the acceleration with g in all four equations and mind the direction.
5Check answers against graphs: the slope of a v-t graph is the acceleration and its area is the displacement, so graphs verify the equations. In reality air resistance matters: a parachutist's acceleration shrinks after the canopy opens and a steady terminal speed follows; in a vacuum a feather and a coin land together, in air they do not. Master uniformly accelerated motion and the next lesson — how forces produce acceleration — will come naturally.
1Summary 重點摘要
Straight-line motion with constant acceleration is described by four equations: v = u + at, s = ½(u+v)t, s = ut + ½at² and v² = u² + 2as. List the knowns, pick the equation that fits — braking distance and free fall are the classic applications.
2Vocabulary cards Vocabulary
Straight-line motion with constant acceleration; its v-t graph is a sloping straight line.
The velocity at the start of the timed interval, symbol u.
The velocity at the end of the interval considered, symbol v.
The four relations v = u + at, s = ½(u + v)t, s = ut + ½at² and v² = u² + 2as.
Motion under gravity alone starting from rest, with acceleration g.
The acceleration of free fall, about 9.8 m s⁻² directed toward the ground.
The distance travelled from the moment braking starts to a complete stop.
The chosen positive direction for straight-line motion; quantities in the opposite direction take negative values.
Acceleration while slowing down; directed opposite to the motion and entered as a negative value.
The steady speed reached when air resistance balances weight during a fall.
3Step by step Steps
- Step 1: What uniformly accelerated motion isStraight-line motion with constant acceleration is uniformly accelerated motion: the velocity changes uniformly with time, so the v-t graph is a sloping straight line. In physics 'accelerating' covers slowing down too: braking is uniformly accelerated motion with acceleration opposite to the velocity. For example, a train starting from rest gains 1 m s⁻¹ every second and reaches 20 m s⁻¹ after 20 s — uniformly accelerated at 1 m s⁻²; a train dropping from 20 m s⁻¹ by 1 m s⁻¹ each second until it stops is also uniformly accelerated, with a negative a.
- Step 2: Five quantities and the sign ruleUniformly accelerated motion involves five quantities: initial velocity u, final velocity v, acceleration a, displacement s and time t. Step one of any solution is to list the knowns and fix a positive direction — usually the direction of the initial velocity; anything in the opposite direction takes a minus sign. For example, for a ball thrown upward take upward as positive and enter a = −9.8 m s⁻²; for a braking car moving forward, enter a = −5 m s⁻². One wrong sign ruins the whole question, so sketch the situation and mark the directions first — the iron rule of mechanics calculation.
- Step 3: Equations one and twoThe first equation, v = u + at, links velocity and time and contains no displacement — ideal for finding a velocity at some moment or the time to reach one. The second, s = ut + ½at², gives the displacement directly from initial velocity and acceleration. For example, a bicycle travelling at 4 m s⁻¹ accelerates at 0.5 m s⁻² for 8 s: final velocity v = 4 + 0.5 × 8 = 8 m s⁻¹, and the displacement s = 4 × 8 + ½ × 0.5 × 8² = 32 + 16 = 48 m. Substitute carefully: square the time first, and carry any minus sign of the acceleration through the whole term.
- Step 4: Equations three and fourThe third equation, s = ½(u + v)t, is the 'average-velocity form': the average velocity of uniformly accelerated motion is (u + v)/2, multiplied by time gives the displacement — fastest when both velocities and time are known. The fourth, v² = u² + 2as, contains no time: use it whenever the question neither gives nor asks for time. For example, an object speeds up from 6 m s⁻¹ to 14 m s⁻¹ in 5 s: s = ½ × (6 + 14) × 5 = 50 m; given instead the acceleration 1.6 m s⁻², use (14² − 6²) = 2 × 1.6 × s and get s = 160 ÷ 3.2 = 50 m — the same answer by both routes.
- Step 5: Three steps to pick an equationChoose an equation in three steps: list which three of u, v, a, s, t are known and what is wanted; pick the equation that contains exactly those quantities and no extra unknown; substitute the numbers with their signs and check the units and the reasonableness of the answer. For example, a train at 10 m s⁻¹ accelerates at 0.4 m s⁻² — find its speed after 15 s: knowns u, a, t; wanted v, so use v = u + at = 10 + 0.4 × 15 = 16 m s⁻¹ directly, without detouring through displacement. That is the efficiency of the 'just right' equation.
- Step 6: Braking distance and road safetyBraking is the classic uniform deceleration: final velocity zero, and v² = u² + 2as gives the braking distance s = u²/(2a). Distance grows with the square of speed — double the speed and the distance quadruples. For example, a car at 100 km h⁻¹ (about 27.8 m s⁻¹) braking at 5 m s⁻² needs s = 27.8² ÷ 10 ≈ 77.3 m; halve the speed to 50 km h⁻¹ and the distance falls to about 19.3 m, less than a quarter. This square law is the physics behind speed limits, safe following distances and the ban on speeding.
- Step 7: Free fall — swap a for gIgnoring air resistance, every object in free fall accelerates at the same gravitational acceleration g = 9.8 m s⁻² directed downward, whatever its mass; all four equations apply with a replaced by g. For example, a stone dropped from a cliff top: after 2 s its speed is v = u + at = 0 + 9.8 × 2 = 19.6 m s⁻¹ and it has fallen s = ½ × 9.8 × 2² = 19.6 m. For a ball thrown upward, take upward as positive and a = −9.8 m s⁻²; at the highest point v = 0, from which the rise time and height follow — a common variation.
4Diagrams Diagrams
5Worked example Example
Example: a car travels at 100 km h⁻¹ and the driver brakes hard, decelerating uniformly at 5 m s⁻² to a stop. Find (a) the initial velocity in m s⁻¹, (b) the braking time, and (c) the braking distance.
- Convert: 100 km h⁻¹ = 100 × 1000 ÷ 3600 ≈ 27.8 m s⁻¹ forward; take forward as positive, so a = −5 m s⁻²
- Braking time: v = u + at → 0 = 27.8 − 5t → t ≈ 5.56 s
- Braking distance: v² = u² + 2as → 0 = 27.8² + 2(−5)s → s = 27.8² ÷ 10 ≈ 77.3 m
- Check: s = ½(u + v)t = ½ × 27.8 × 5.56 ≈ 77.3 m — consistent with the above (rounding aside)
- Answer: (a) About 27.8 m s⁻¹; (b) about 5.6 s; (c) braking distance about 77.3 m. If the question also asks for the reaction distance (travelled at full speed before braking), find it with s = ut and add it to the braking distance for the total stopping distance.
6Common mistakes Mistakes
7Quick check Quick check
Q1Write down the four equations of uniformly accelerated motion.
Q2An object accelerates from 4 m s⁻¹ at 2 m s⁻² for 3 s. Find its final velocity.
Q3Why does doubling the speed quadruple the braking distance?
Q4Ignoring air resistance, how far does an object fall from rest in 2 s?
Q5At the highest point of a vertical throw, what are the velocity and the acceleration?
8Recap Recap
- Uniformly accelerated motion = straight-line motion with constant acceleration; the v-t graph is a sloping straight line
- Four equations: v = u + at, s = ½(u+v)t, s = ut + ½at², v² = u² + 2as
- Three-step method: list knowns, pick the equation with no spare unknown, substitute with signs — fix the positive direction first
- Braking distance s = u²/(2a) grows with the square of speed; convert km h⁻¹ to m s⁻¹ first
- Free fall: replace a with g = 9.8 m s⁻²; in throws g is negative and v = 0 at the top