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Grade 10-11 (age 15-17)

high school physics Uniform Linear Motion

Like a car on the highway with cruise control on, motion whose velocity never changes is called uniform linear motion. The acceleration is 0, so the object always covers the same distance in each equal stretch of time, and the distance is velocity times time, s = vt. On a v-t graph it is a horizontal line at height v, and the area of the rectangle under it is the displacement. Move the velocity and time sliders and the evenly spaced positions marked each second change along with the final position, while in the graph below the rectangle's area follows with the same value.

Motion at Constant Velocity

Before using s = vt, first check that the velocity is constant, that is, that the acceleration is 0. When the net force is 0, Newton's first law gives exactly this motion, and the position grows by the same distance in each equal time. When the v-t graph is horizontal, the rectangle under it has area v × t, equal to the displacement, which is why this chapter sets the formula and the area side by side. If the velocity changes partway, multiplying t by a single v no longer works.

🚗 The idea of uniform linear motion
①It is motion whose velocity does not change, like a car with cruise control on
②Since the velocity does not change, the acceleration is 0
③The object covers the same distance in every equal stretch of time
④The x-t graph is a straight line whose slope is the velocity
⑤When the net force is 0, an object moves in uniform linear motion (Newton's first law)

How the Object Moves

5 m/s
4 s

In the motion figure a green circle sits on a horizontal line marked from 0 to 100 m, at the position s = vt computed from the velocity and time chosen on the sliders. Faint dots mark where it was each second from the start, and the gaps between them all being v is exactly what uniform motion means. The orange arrow shows the direction and size of the velocity, and the bottom line writes out s = v × t. In the v-t figure the rectangle under the horizontal line at the same height is shaded, and its area is read as the displacement. Both figures use the same velocity and time, and changing a slider redraws the dots, circle, arrow and rectangle to match.

🔑 What to notice
①The greater the velocity, the farther it goes in the same time
②Doubling the time doubles the distance
③The gaps between the faint dots are all v
④Distance = velocity × time is the most basic relation

The v-t Graph and Displacement

The sliders below are the same velocity and time as in the previous step. Changing them here also updates the motion figure and the v-t area.

5 m/s
4 s
Uniform linear motion
s = vt
Distance s = velocity v × time t

A common slip is to write the average speed of a round trip as the arithmetic mean of the two speeds. When the speeds out and back differ, the slower leg takes longer, so the average speed is the total distance divided by the total time, which is smaller than the arithmetic mean. In motion whose velocity changes partway, t cannot be multiplied by a single v in s = vt. The safe order is to find the distance and time of each leg separately and divide at the end.

📊 Reading the v-t graph
①A horizontal v-t graph means uniform motion
②The area under the graph is the displacement
③The rectangle's area = base t × height v = vt = s
④Changing the velocity and time sliders changes this graph too

Worked Examples

Before finding a distance, check whether the velocity is constant. If it is, use s = vt; when asked how long a train takes to clear a tunnel completely, take the distance as the tunnel length plus the train length and then use t = s/v. The extra train length is needed because the back of the train must leave the tunnel too. Dividing distance by speed gives the time directly only when the speed is constant.

Example 1
A car travels at a constant 20 m/s for 5 seconds. How far does it go?
1
The velocity is constant, so this is uniform motion and s = vt applies.
s = vt
2
Put in v = 20 m/s and t = 5 s.
s = 20 × 5 = 100 m
▸ 100 m
Finding distance as velocity × time works only when the velocity 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?
1
Average speed is total distance ÷ total time. Let the one-way distance be d.
t = d4 + d6 = 5d12
2
Divide the total distance 2d by the total time 5d/12.
v = 2d5d/12 = 245 = 4.8 m/s
▸ 4.8 m/s
The average speed is not the arithmetic mean of the two speeds, 5 m/s. More time is spent walking at the slower 4 m/s, so it comes out below 5, at 4.8 m/s.

Summary

Learning a = 0, a constant v, s = vt, the horizontal v-t graph and the area under it gives a baseline for the next chapter, uniformly accelerated motion, where you can compare what changes once the velocity does. The fact that the x-t graph is a straight line whose slope is the velocity pairs with how that graph bends once there is an acceleration. Starting from the rectangle under a horizontal v-t graph leads naturally to reading the trapezoid under a sloped one.

Uniform linear motion
s = vt, v = const, a = 0
Motion at constant velocity; the distance is proportional to the time
Average speed
v = st
Total distance ÷ time taken
CSAT-style problem, adapted
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
1
To clear the tunnel completely, the train must travel the tunnel length plus its own length.
s = 300 + 200 = 500 m
2
The speed is constant, so t = s/v.
t = sv = 50020 = 25 s
🎯 Exam Points
①Uniform linear motion has acceleration a = 0 and a constant velocity v
②The distance is s = vt
③The v-t graph is a horizontal line, and the area under it is the displacement
④The x-t graph is a straight line whose slope is the velocity
⑤When the net force is 0, the motion is uniform and linear (Newton's first law)

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