# Average Speed || How to calculate average speed

## What is Average Speed?

Average Speed is the total distance covered by object divided by total time taken.

Explanation

Suppose you take out a car from garage to travel to your office which is 30 km away.

Your car is capable to touch maximum speed of 180 km/hr

For first 10 km the road was empty, so you drove at 80 km/hr

After that you found heavy traffic on the road and then you have to drive at the speed of 40km/hr for the rest of 20 kms.

In the above example you can see that the car travelled at different speed on the entire journey.

Speed 1 = 80 km/hr for 10 kms
Speed 2 = 40 km/hr for 20 kms

So what is the average speed for the whole distance?

It is given by formula:
\mathtt{Average\ Speed\ =\ \frac{Total\ Distance\ travelled}{Total\ Time\ Taken}}\\\ \\ \mathtt{Average\ Speed\ =\ \frac{D1\ +\ D2\ +\ D3\ .\ .\ .}{T1\ +\ T2\ +\ T3\ .\ .\ .}} \\\ \\

For the above example:
D1 = 10 Km
D2 = 20 Km

Now, we have to calculate individual time for different speed
T1 = Distance/Speed = 10/80 = 0.125 hours
T2 = Distance/Speed = 20/40 = 0.5 hours

Putting all data in average speed formula:

\mathtt{Average\ Speed\ =\ \frac{D1\ +\ D2\ }{T1\ +\ T2\ }}\\\ \\ \mathtt{Average\ Speed\ =\ \frac{10\ +\ 20\ }{0.125\ +0.5} \ }\\\ \\ \mathtt{Average\ Speed\ =\ \frac{30}{0.625} \ =\ 48\ km/hr}

Hence, the average speed of car for 30 km travel is 48 km/hr

## Why do we calculate Average Speed?

We already know the formula for speed.

\mathtt{Speed\ =\ \frac{Distance}{Time}}

So why there is a need to calculate average speed?

Because in practical life no object moves in constant speed.

Even the train which moves in straight line has to change its speed multiple times during the journey.

The train has to stop in multiple station then pick up the speed in the outskirts of the city then again slow down for next station.

In this type of case we don’t have any fixed data for speed, that’s why we have to calculate the average speed for our convinience.

## Average Speed Shortcut Formulas

Shortcut Formula 01

When object travel at two different speed for same time then formula is given as:

\mathtt{Average\ Speed\ =\ \frac{S1\ +\ S2}{2}}

Shortcut Formula 02

When a constant distance is travelled at two different speed, then the average speed formula is given as:

\mathtt{Average\ Speed\ =\ \frac{2\ \times \ S1\ \times \ S2}{S1\ +\ S2}}

Formula Derivation

A cyclist travels from A to B at speed of x m/s and then returns back at speed y m/s.
The distance between A & B is D meter

Speed 1 (S1) = x meter/second
Speed 2 (S2) = y meter/second

Distance A to B = D meter

We know that;
\mathtt{Average\ Speed\ =\ \frac{D1\ +\ D2\ }{T1\ +\ T2\ }} \\\ \\

\mathtt{Time\ taken\ to\ travel\ A\ to\ B}\\\ \\ \mathtt{Time\ 1\ ( T\ 1) \ =\ \frac{Distance}{Speed\ 1} \ =\ \frac{D}{S1}}\\\ \\ \\\ \\ \mathtt{Time\ taken\ to\ travel\ B\ to\ A}\\\ \\ \mathtt{Time\ 2\ ( T2) \ =\ \frac{Distance}{Speed\ 2} =\ \frac{D}{S2}} \\\ \\

Putting the value in average speed formula, we get:

\mathtt{Average\ Speed\ =\ \frac{D\ +\ D}{\frac{D}{S1} +\frac{D}{S2}} \ \ }\\\ \\ \mathtt{Average\ Speed=\ \frac{2D}{D\left(\frac{S1+S2}{S1.S2}\right)} \ =\ \frac{2\ S1.\ S2}{S1+S2}}

Shortcut Formula 03
Average Speed when three equal distance travelled

Similarly, If the person travels three equal distance then the formula becomes

\mathtt{Average\ Speed=\ \frac{3.\ S1.\ S2.\ S3}{S1.S2+S2.S3\ +\ S3.S1}}

## Average Speed Question

(01) A car travelled at the speed of 40 km/hr for one hour and 60 km/hr for another one hour. Fine the average speed of the car.

Given
Speed 1 (S1) = 40 km/hr
Speed 2 (S2) = 60 km/hr

Here both the speed is used for same time (i.e. 1 hour)

In this case;
\mathtt{Average\ Speed\ =\ \frac{S1\ +\ S2}{2}}\\\ \\ \mathtt{Putting\ the\ values;}\\\ \\ \mathtt{Average\ Speed\ =\ \frac{40\ +\ 60}{2} \ =\ 50\ km/hr} \\\ \\

Hence the average speed of car is 50km/hr

(02) The distance between home and school is 55 km. A cyclist travels from home to school at 15 km/hr and return back at the speed of 25 km/hr. Find the average speed of the cyclist

Solution
Distance D = 55 km

Speed (S1) = 15 km/hr
Speed (S2) = 25 km/hr

Here the distance is same for both the travel, so we will use formula:

\mathtt{Putting\ the\ values,\ we\ get}\\\ \\ \mathtt{Average\ Speed\ =\ \frac{2\ .15.\ 25\ }{15+25}}\\\ \\ \mathtt{Average\ Speed\ =\ \frac{750}{40} \ =\ 18.75\ km/hr}

Hence, the average speed of cyclist is 18.75 km/hr

(03) A train travels at the speed of 120 km/hr for 60 km and then change its speed to 150 km/hr for next 100 km distance. Find the average speed of train for whole journey

Speed 1 (S1) = 120 km/hr
Distance 1 (D1) = 60 km

Speed 2 (S2) = 150 km/hr
Distance 2 (D2) = 100 km

Here first we have to calculate individual time taken in each distance

\mathtt{Time\ 1\ =\ \frac{Distance\ 1\ }{Speed\ 1} \ =\ \frac{60}{120} =0.5\ hour}\\\ \\ \mathtt{Time\ 2\ =\ \frac{Distance\ 2\ }{Speed\ 2} \ =\ \frac{100}{150} =\frac{2}{3} \ hour}

Now using the average speed formula;

\mathtt{Average\ Speed\ =\ \frac{Total\ Distance\ travelled}{Total\ Time\ Taken}}\\\ \\ \mathtt{Average\ Speed\ =\ \frac{60+\ 100\ }{\frac{1}{2} \ +\ \frac{2}{3} \ }}\\\ \\ \mathtt{Average\ Speed\ =\ \frac{160}{0.5\ +0.67} \ =\ 136.75\ km/hr}

Hence, the average speed of train is 136.75 km/hr

(04) A motorcycle track is in the form of equilateral triangle with each side 20 km. A racer travelled each leg of the triangle with different speed of 80 km/hr, 100 km/hr and 120 km/hr. Find the average speed of the motorcyclist.

Given
Speed 1 (S1) = 80 km/hr
Speed 2 (S2) = 100 km/hr
Speed 3 (S3) = 120 km/hr

For average speed, we will use following formula

\mathtt{Average\ Speed=\ \frac{3.\ S1.\ S2.\ S3}{S1.S2+S2.S3\ +\ S3.S1}}\\\ \\ \mathtt{Putting\ the\ values}\\\ \\ \mathtt{Average\ Speed\ =\ \frac{3\times \ 80\times 100\times 120}{( 80\times 100) +( 100\times 120) +( 120\times 80)}}\\\ \\ \mathtt{Average\ Speed\ =\ \frac{2880000}{8000+12000+9600} \ }\\\ \\ \mathtt{Average\ Speed\ =\frac{2880000}{29600} \ =\ 97.3\ km/hr}

Hence, 97.3 km/hr is the average speed of motorist