Motion: Uniform, Non-Uniform, Speed and Average Speed

D · Uniform motion

Equal distances in equal time intervals is the definition of uniform motion, so the speed stays constant.

A · Non-uniform motion

Unequal distances in equal time intervals means non-uniform motion — the speed is changing.

A · A car on a clear straight highway

A car cruising a clear straight highway at a steady speed covers equal distances in equal time intervals.

D · A car moving in city traffic

City traffic forces constant changes in speed, so the car covers unequal distances in equal time intervals.

A · $0.8\ \text{m/s}$

Speed $= 8/10 = 0.8\ \text{m/s}$, and since the gap is constant the motion is uniform.

A · True

**Uniform motion** requires equal distances in equal time intervals, so the speed stays constant; **non-uniform motion** has unequal distances in equal time intervals.

D · Equal distances in equal time intervals

Uniform motion is defined as covering equal distances in equal intervals of time.

C · Non-uniform motion

A car stuck in traffic covers unequal distances in equal time intervals, indicating non-uniform motion.

D · Non-uniform motion

Unequal distance gaps in equal time intervals confirm the object is in non-uniform motion.

A · Total distance divided by total time

Average speed is calculated as total distance divided by total time.

B · Variable speed

Non-uniform motion is characterized by variable speed, as distances covered are unequal in equal time intervals.

B · The vehicle moved non-uniformly

An average speed of 60 km/h does not guarantee uniform motion; the vehicle may have moved non-uniformly.

D · Equal distances covered in equal time intervals

Uniform motion requires that equal distances are covered in equal time intervals.

D · Convert units before calculating

The first step is to convert units to ensure consistency before performing calculations.

D · Speed is the ratio of distance to time

The formula indicates that speed is determined by the ratio of distance traveled to the time taken.

C · Assuming it indicates uniform motion

A common mistake is to assume that average speed indicates uniform motion, which is not necessarily true.

C · $60\ \text{km/h}$

Average speed $= (60+70+50)/3 = 180/3 = 60\ \text{km/h}$.

A · Non-uniform motion

The distances covered each hour are unequal, so the motion is non-uniform even though the average speed is 60 km/h.

B · An average can come from changing speeds

Average speed only gives the overall pace. Uniform motion requires equal distances in equal time intervals, which an average does not prove.

A · $20\ \text{m/s}$

$72 \times \frac{5}{18} = 20\ \text{m/s}$.

B · The train, by $2\ \text{m/s}$

The train is $72 \times \frac{5}{18} = 20\ \text{m/s}$, which is $2\ \text{m/s}$ faster than the horse's $18\ \text{m/s}$.

A · $16.67\ \text{m/s}$

Speed $= 180/3 = 60\ \text{km/h}$, and $60 \times \frac{5}{18} = 16.67\ \text{m/s}$.

C · $4$ hours

Distance $= 360000\ \text{m}$, so time $= 360000/2.5 = 144000\ \text{s} = 4$ hours.

D · $150\ \text{s}$

First leg $= 500/10 = 50\ \text{s}$; second leg $= 500/5 = 100\ \text{s}$; total $= 150\ \text{s}$.

C · $20\ \text{m/s}$

Remaining time $= 200 - 150 = 50\ \text{s}$; remaining distance $= 1000\ \text{m}$; speed $= 1000/50 = 20\ \text{m/s}$.

A · $10\ \text{m/s}$

Average speed $= 2000/200 = 10\ \text{m/s}$, even though no single leg was travelled at exactly 10 m/s except the first.

D · 20 m/s

To finish the journey in 200 s, the vehicle has 50 s left after the first leg. The remaining distance is 1000 m, so the required speed is $\frac{1000\text{ m}}{50\text{ s}} = 20\text{ m/s}$.

A · Uniform motion

The distance increases by 8 m for every 5 s interval, indicating constant speed. Thus, this represents uniform motion.

B · 60 km/h; No

The average speed is $\frac{60 + 70 + 50}{3} = 60\text{ km/h}$. However, since the distances are not equal for each hour, it does not indicate uniform motion.