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Common mistake doing average speed calculation

Two most common errors you should avoid when you see an average speed calculation

Mistake #1

The most common mistakes in average speed problems occur when the question asks you to calculate the average speed of an object moving at two different speeds for different parts of the journey. Students are generally tempted to calculate the arithmetic mean (average) of the two speeds and select a corresponding answer.  This averaging approach is wrong.

Incorrect way

$\text {Average speed} = \text {Arithmetic mean of speeds} = \frac {v_1 + v_2}{2}$

Correct Way
$\text {Average Speed} =\frac {\text{ Total distance }}{\text {Total Time taken}}$

Mistake #2

Another problem that happens Quite often is with the questions where we have different units of measurement in one question. For example, if a Truck takes 60 minutes to travel 60 miles, and then takes 6 hours to travel 60 miles, you calculate the average speed of the car for the entire journey expressing 60 minutes as 1 hour or by converting 6 hours to 360 minutes. Either way, you never calculate time, distance, or speed using different units of measurement for the different parts of the journey. In this problem, if you had wrongly used two different units of time for calculating average speed, you would obtain the wrong answers even with the correct formula.

There are Shortcuts for average speed calculation

There are two situations in which you can save time when dealing with average speed questions:

-When the distances are equal

Average speed= $ \frac {2(ab)}{(a+b)}$

Here a and b are the speed in two parts

-When times are equal
Average Speed= $ \frac {(a+b)}{2}$

Example 1
A man moves from Home to Shop at speed 10 km/hr and then comes back through same path with speed 20 km/hr. What is the average speed of the Journey
Solution
Incorrect way will be
Average Speed = $ \frac {10 + 20}{2} = 15$ km/hr
Correct Way will be
Average speed= $ \frac {2(ab)}{(a+b)} = \frac { 2 \times 10 \times 20}{10 + 20} =13.33$ km/hr

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