Two cars leave the same college gate for the same railway station. The first car takes the long highway, a smooth road with no traffic. The second car takes the shorter route through the busy city. Both cars reach the same station. Each car has an odometer, the meter that counts the kilometres driven. The odometer of the first car shows the larger reading. Yet a map shows the college and the station at the same two points for both journeys.
What distance measures
Suppose you travel from a point A to a point B along a curved road. The length of road that you actually cover is the distance travelled. Distance is a scalar quantity. Distance has a magnitude and no direction. Distance depends on the route taken, not only on the two end points. A longer route between the same two points gives a larger distance.
Watch the reading in Fig 2.1. The reading grows the whole time. The reading grows on every bend. The reading can never become smaller, because the particle keeps covering path. This is a general property. Distance never decreases during motion.
The distance travelled by a body in a time interval is the total length of the path covered by the body in that interval. Distance is a scalar and is never negative.
Two cars start from the same gate and stop at the same station. Car 1 takes the long highway. Car 2 goes through the city, a shorter route. Which statement is correct?
Average speed
Average speed tells how fast, on the whole, a body covers distance. Average speed is defined as the total distance travelled divided by the total time taken.
\[ \text{average speed} = \dfrac{\text{total distance}}{\text{total time}} \]
For example, a bus covers \(120\,\text{km}\) from Pune to Nashik in \(3\,\text{h}\). The average speed of the bus is \(120/3 = 40\,\text{km/h}\). This does not mean that the bus moved at exactly \(40\,\text{km/h}\) at every moment. The bus was slower in the towns and faster on the highway. The average speed describes the whole journey with one number.
The average speed of a body over a time interval is the total distance travelled divided by the total time taken. The total time includes any time during which the body is at rest.
A cyclist rides \(6\,\text{km}\) in \(20\,\text{min}\). The cyclist then rests for \(10\,\text{min}\). Finally the cyclist rides \(9\,\text{km}\) in \(30\,\text{min}\). Find the average speed for the whole trip.
Total distance \(= 6 + 9 = 15\,\text{km}\).
Total time \(= 20 + 10 + 30 = 60\,\text{min} = 1\,\text{h}\).
Therefore the average speed is \(15\,\text{km} \div 1\,\text{h} = 15\,\text{km/h}\).
The rest period is part of the total time.
If the rest period is left out, the total time becomes \(50\,\text{min}\) and the answer becomes \(18\,\text{km/h}\), which is wrong.
The total time includes the time spent at rest. A body covers no distance while at rest, but the clock keeps running. Leaving out the rest time is the most common error in average-speed problems.
Two special cases
Many average-speed problems fall into one of two cases. In the first case, the body spends equal times at two speeds. In the second case, the body covers equal distances at two speeds. When you solve a problem, first check which case it is. The case-wise results below then give the answer quickly, and they warn you of a common trap.
Case 1: equal times at two speeds
Suppose a car moves at speed \(v_1\) for a time \(t\), and then at speed \(v_2\) for the same time \(t\). The distance in the first part is \(v_1 t\). The distance in the second part is \(v_2 t\). The total distance is \((v_1 + v_2)\,t\). The total time is \(2t\). Divide the total distance by the total time.
\[ v_{\text{avg}} = \dfrac{(v_1 + v_2)\,t}{2t} = \dfrac{v_1 + v_2}{2} \]
When the times are equal, the average speed is the arithmetic mean of the two speeds. For example, \(1\,\text{h}\) at \(40\,\text{km/h}\) followed by \(1\,\text{h}\) at \(60\,\text{km/h}\) gives an average speed of \(50\,\text{km/h}\).
Case 2: equal distances at two speeds
Now suppose a car covers a distance \(d\) at speed \(v_1\), and then the same distance \(d\) at speed \(v_2\). The times are no longer equal. The car spends more time at the slower speed. The derivation below shows what happens to the average.
\[ t_1 = \dfrac{d}{v_1}, \qquad t_2 = \dfrac{d}{v_2} \]
Why. each part covers the same distance \(d\). The time for each part is the distance divided by the speed of that part.
\[ v_{\text{avg}} = \dfrac{2d}{t_1 + t_2} = \dfrac{2d}{\dfrac{d}{v_1} + \dfrac{d}{v_2}} \]
Why. average speed is the total distance \(2d\) divided by the total time \(t_1 + t_2\), by definition.
\[ v_{\text{avg}} = \dfrac{2}{\dfrac{1}{v_1} + \dfrac{1}{v_2}} \]
Why. the distance \(d\) cancels from the numerator and the denominator. The answer does not depend on \(d\).
\[ v_{\text{avg}} = \dfrac{2 v_1 v_2}{v_1 + v_2} \]
Read it. combine the two fractions in the denominator and simplify. This expression is called the harmonic mean of \(v_1\) and \(v_2\).
Take \(v_1 = 40\,\text{km/h}\) and \(v_2 = 60\,\text{km/h}\). The arithmetic mean is \(50\,\text{km/h}\). The harmonic mean is \(2 \times 40 \times 60 / 100 = 48\,\text{km/h}\). The harmonic mean is smaller. The reason is simple. The car spends more time at \(40\,\text{km/h}\) than at \(60\,\text{km/h}\), because the slower part takes longer. The slower speed therefore counts for more in the average.
Average speed is never the plain average of the speeds unless the times are equal. Whenever you are not sure, go back to the definition. Write the total distance, write the total time, and divide.
A car covers \(60\,\text{km}\) at \(30\,\text{km/h}\) and the next \(60\,\text{km}\) at \(60\,\text{km/h}\). What is the average speed for the whole trip?
Rank these motions by average speed, fastest first.
A delivery scooter covers \(12\,\text{km}\) in \(30\,\text{min}\), waits \(15\,\text{min}\) at a shop, and covers another \(6\,\text{km}\) in \(15\,\text{min}\). Find the average speed for the whole trip in km/h.
The Mumbai to Delhi Rajdhani Express covers about \(1380\,\text{km}\) in about \(16\,\text{h}\), including stops. Estimate the average speed in km/h.
A road sign near Bengaluru says "Airport 40 km, 55 min". The sign is stating an average speed of about \(44\,\text{km/h}\), although the speed limit on the highway is \(80\,\text{km/h}\). The sign includes the time lost at signals, at toll booths and in slow traffic. The average speed, total distance over total time, is the number a traveller actually needs.
- Averaging the speeds instead of dividing total distance by total time.Write the total distance and the total time. Divide. The plain average of speeds is correct only for equal times.
- Leaving a rest period out of the total time.The clock runs while the body is at rest. The distance stays fixed; the time grows.
- Mixing km/h with m/s in one calculation.Convert every speed to the same unit before adding or comparing. \(1\,\text{m/s} = 3.6\,\text{km/h}\).