Uniform linear motion or uniform rectilinaer motion is the simplest motion that has the constant velocity. The displacement-time graph of uniform rectilinaer motion is a linear function and the velocity-time graph is a horizontal straight line. Since there is no change in velocity, the acceleration is zero.

**Encounter**

Two cars are S kilometers apart, with velocities of magnitude v1 and v2, in opposite directions.

$t=\frac{s}{\mathrm{v\u2081\; +\; v\u2082}}$

**Pursuit**

The faster car chases the slower car. The two cars are S kilometers apart, with velocities of magnitude is v1 and v2, in the same direction, and v1>v2.

$t=\frac{s}{\mathrm{v\u2081\; -\; v\u2082}}$

**The train passes through a bridge or a tunnel.**

The s₁ meter train passes over the s₂ meter bridge at speed v. It starts when the locomotive enters the bridge and ends when the rear of the train leaves the bridge. The train displacement is s₁ + s₂.

$t=\frac{\mathrm{s\u2081\; +\; s\u2082}}{v}$

Even the simplest uniform linear motion, the most difficult problems can be created from it.

**When does the hour hand form an angle of θ with the minute hand?**

This problem can be treated as a catch-up problem. How much time does it take for the minute hand to go ahead of the hour hand by θ degrees from 00:00? Or, how much time does it take for the minute hand to fall behind the hour hand by θ degrees from 00:00? The number of laps led by the minute hand should be taken into account in this problem. The distance or magnitude of displacement is the angle between minute hand and hour hand. The velocity is replaced by the angular velocity.

The angular velocity of minute hand

$\mathrm{\omega \u2082}=\frac{\mathrm{2\pi}}{\mathrm{1\; hour}}=\mathrm{2\pi h\u207b\xb9}$.

The angular velocity of hour hand

$\mathrm{\omega \u2081}=\frac{\mathrm{2\pi}}{\mathrm{12\; hour}}=\frac{\pi}{6}\mathrm{h\u207b\xb9}$.

When their angle is θ, the minute hand has caught up with the hour hand for the kth time.

S = 2kπ + θ (k=0, 1, 2, •••••,23) or S = 2kπ - θ(k=1, 2, •••••,24).

$t=\frac{\mathrm{2k\pi \; \pm \; \theta}}{\mathrm{\omega \u2082\; -\; \omega \u2081}}$

if$\theta =\; 30\xb0\; =\frac{\pi}{6}$,$t=\frac{\mathrm{1\; +\; 12k}}{\mathrm{11}}$, (k=0, 1, 2, •••••,23)，or$t=\frac{\mathrm{12k\; -\; 1}}{\mathrm{11}}$, (k=1, 2, •••••,24).

This formula also incidentally solves the problem of how many times the hour hand and the minute hand meet in a day. When θ=0, S = 2kπ,$0\le t=\frac{\mathrm{12k}}{\mathrm{11}}\mathrm{\uff1c\; 24}$.

$\mathrm{0\le \; k\uff1c22}$

So the hour hand and the minute meet 22 times in a day.