Uniform Linear Motion

Uniform linear motion is motion at constant speed, like a car with cruise control on. Acceleration is zero, so equal times mean equal distances, and distance is speed times time. On a v-t graph the path is a horizontal line, and the area under it is the displacement. Here you slide speed and time to see how far the object goes and how that area changes.

What is Constant-Velocity Motion?

🚗 Intuition for Uniform Motion

①Imagine a car with cruise control on a highway

②Velocity does not change, so acceleration = 0

③It always covers the same distance over the same interval

Visualizing Object Motion

Velocity v5 m/s

Time t4 s

🔑 Key Observations

①Higher velocity → more distance in the same time

②Longer time → distance grows proportionally

③Distance = velocity × time — the most basic relation

v-t Graph and Displacement

Uniform Motion Formula

s = vt

Distance (s) = velocity (v) × time (t)

📊 Reading the v-t Graph

①Horizontal line on v-t graph = uniform motion (velocity unchanged)

②Area under the graph = distance (displacement)

③Rectangle area = base (t) × height (v) = vt = s

Worked Examples

Example 1

A car travels at a constant 20 m/s for 5 seconds. How far does it go?

For uniform motion, apply s = vt.

s = vt

Substitute v = 20 m/s and t = 5 s.

s = 20 × 5 = 100 m

▸ 100 m

Distance in uniform motion is speed × time, valid only when the speed is constant.

Example 2

A person walks the same path out at 4 m/s and back at 6 m/s. What is the average speed for the round trip?

Average speed = total distance ÷ total time. Let the one-way distance be d.

t = d4 + d6 = 5d12

Divide the total distance 2d by the total time.

v = 2d5d/12 = 245 = 4.8 m/s

▸ 4.8 m/s

Average speed is total distance ÷ total time, not the arithmetic mean (5 m/s). More time is spent on the slow leg, so it is below 5.

Summary

Uniform Linear Motion

s = vt, v = const, a = 0

Constant-velocity motion — distance is proportional to time

Average Speed

v = st

Total distance / elapsed time

CSAT-style

A 200 m long train moving at a constant 20 m/s passes completely through a 300 m tunnel. How long does this take?

①10 s

②15 s

③20 s

④25 s

⑤30 s

▸ ④ 25 s

To pass completely, the train must travel (tunnel length + train length).

s = 300 + 200 = 500 m

For uniform motion, t = s / v.

t = sv = 50020 = 25 s

🎯 Exam Points

①Uniform motion: a = 0, v = constant

②Distance s = vt (velocity × time)

③v-t graph: horizontal line, area = displacement

④x-t graph: straight line (slope = velocity)

⑤Net force = 0 → uniform linear motion (Newton 1)

Uniformly Accelerated Motion

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