Average Speed

Average speed Formula

Average speed = $\frac{T o t a l d i s t a n c e t r a v e l l e d}{T o t a l t i m e t a k e n i n t r a v e l l i n g t h a t d i s t a n c e}$

Let us consider some scenarios with respect to average speed, and variations of the above formula:

Scenario 1

If a man travels different distances $D_{1}$ , $D_{2}$ , $D_{3}$ , ……. and so on in different times $T_{1}$ , $T_{2}$ , $T_{3}$ , ………. and so on respectively, then

Average speed = $\frac{T o t a l d i s t a n c e t r a v e l l e d}{T o t a l t i m e t a k e n i n t r a v e l l i n g t h a t d i s t a n c e}$ = $\frac{D_{1} + D_{2} + D_{3} + \dots}{T_{1} + T_{2} + T_{3} + \dots}$

Scenario 2

If one part of the distance is covered at a speed of $v_{1}$ in time $t_{1}$ and the other part of the distance is covered at a speed of $v_{2}$ in time $t_{2}$ , then

Average speed = $\frac{T o t a l d i s t a n c e t r a v e l l e d}{T o t a l t i m e t a k e n i n t r a v e l l i n g t h a t d i s t a n c e}$ = $\frac{v_{1} t_{1} + v_{2} t_{2}}{t_{1} + t_{2}}$

Q. The speed of a man for the first 2 hours is 30 km/hr, and for the next 3 hours his speed is 50 km/hr. What is the ratio of average of his speeds to his average speed?

(a) 10 : 11   (b) 21 : 23    (c) 20 : 21   (d) 20 : 23

Explanation:

The speed of a man for the first 2 hours is 30 km/hr, and for the next 3 hours is 50 km/hr.

Distance = Speed × Time
So, Total distance = 2 × 30 + 3 × 50 = 60 + 150 = 210 km
Average speed = Total distance/Total time = 210/(2 + 3) = 210/5 = 42 km/hr

Now, Average of speeds = (30 + 50)/2 = 80/2 = 40 km/hr

So, required ratio = 40 : 42 = 20 : 21

Answer: (c)

Scenario 3

If a man travels different distances $D_{1}$ , $D_{2}$ , $D_{3}$ , …… and so on with different speeds $S_{1}$ , $S_{2}$ , $S_{3}$ , …….. and so on respectively, then

Average speed = $\frac{T o t a l d i s t a n c e t r a v e l l e d}{T o t a l t i m e t a k e n i n t r a v e l l i n g t h a t d i s t a n c e}$ = $\frac{D_{1} + D_{2} + D_{3} + \dots}{(\frac{D_{1}}{S_{1}} + \frac{D_{2}}{S_{2}} + \frac{D_{3}}{S_{3}} + \dots)}$

Special case: If a distance is divided into n equal parts each travelled with different speeds, then $D_{1}$ = $D_{2}$ = $D_{3}$ = … = $D_{n}$

Average speed = $\frac{n D}{(\frac{D}{S_{1}} + \frac{D}{S_{2}} + \frac{D}{S_{3}} + \dots + \frac{D}{S_{n}})}$ = $\frac{n}{(\frac{1}{S_{1}} + \frac{1}{S_{2}} + \frac{1}{S_{3}} + \dots + \frac{1}{S_{n}})}$

where n = number of equal parts and $S_{1}$ , $S_{2}$ , $S_{3}$ , …………. $S_{n}$ are speeds.

Q. Mak covered 275 km of his journey at 25 km/hr and the remaining distance at 35 km/hr. If the total distance covered by him was 590 km, then what must have been his average speed for the whole journey?

(a) 32.5 km/hr   (b) 30 km/hr    (c) 34.5 km/hr   (d) 29.5 km/hr

Explanation:

Speed for first 275 km is 25 km/hr.
Time taken to cover 275 km = 275/25 = 11 hrs.

Speed for next 315 km = 35 km/hr.
Time taken to cover 315 km = 315/35 = 9 hrs.

Average Speed = Total distance covered / Total time taken = 590/(11 + 9) = 590/20 = 29.5 km/hr.
Thus, his average speed for the whole journey must have been 29.5 km/hr.

Answer: (d)

Finding Average Speed when distance or time are same

We often encounter following scenarios in questions on average speed:

The distances travelled at two different speeds are the same. E.g. A person travels from X to Y with speed A and then comes back with speed B.
The time for which a person travels at the two speeds is the same. E.g. A person travels with speed A for t minutes and then at speed B for t minutes again.

Let us consider both of these scenarios.

Distance travelled is same

If there are two equal parts of journey, it means n = 2 in the following formula:

Average speed = $\frac{n}{(\frac{1}{S_{1}} + \frac{1}{S_{2}} + \frac{1}{S_{3}} + \dots + \frac{1}{S_{n}})}$

In other words, if a person goes from x to y at $S_{1}$ km/hr and comes back from y to x at $S_{2}$ km/hr, then the average speed during the whole journey will be:

Average speed = $\frac{2}{(\frac{1}{S_{1}} + \frac{1}{S_{2}})}$ = $\frac{2 S_{1} S_{2}}{S_{1} + S_{2}}$

Q. A person travels from A to B at a speed of 20 km/hr and then comes back to A at a speed of 30 km/hr. What is his average speed over the entire journey?

Explanations :

Explanation 1: Traditional Method

Let the one side distance be d km.

Time taken to travel from A to B = Distance / Speed = d/20 hrs

Time taken to travel from B to A = Distance / Speed = d/30 hrs

Average speed = $\frac{T o t a l d i s t a n c e t r a v e l l e d}{T o t a l t i m e t a k e n i n t r a v e l l i n g t h a t d i s t a n c e}$ = $\frac{2 d}{(\frac{d}{20} + \frac{d}{30})}$ = $\frac{2 (20 \times 30)}{(20 + 30)}$ = 24 km/hr

Explanation 2: Formula Method

Average speed = $\frac{2 S_{1} S_{2}}{S_{1} + S_{2}}$ = $\frac{2 (20 \times 30)}{(20 + 30)}$ = 24 km/hr

Explanation 3: Allegation Method

The average speed is the weighted average with the time travelled at the different speeds being the weight.

The two speeds are 20 km/hr and 30 km/hr. As the distance is constant, so the time must be in inverse proportion, i.e. 3:2.

So, Average speed = $\frac{(3 \times 20) + (2 \times 30)}{(3 + 2)}$ = 120/5 = 24 km/hr

Q. . A local bus completed a round trip at an average speed of 48 km per hour. The bus covered the first half of the distance at an average speed of 36 km per hour. What would have been its average speed during the second half of the journey?

(a) 48 km/hour   (b) 60 km/hour    (c) 72 km/hour   (d) 84 km/hour

Explanations :

Explanation 1: Traditional Method

Let the total distance covered by bus in the round trip be x km. Let y be the average speed during the second half of the journey.

Hence, the total time taken in complete journey = Total distance/average speed = x/48 hour

Now, total time = time taken during 1st half of the trip + time taken during 2nd half of the trip
Hence, x/48 = 0.5x/36 + 0.5x/y
or x/48 - 0.5x/36 = 0.5x/y
or x/144 = 0.5x/y
or y = 72 km/hour

Explanation 2: Formula Method

Average speed = $\frac{2 S_{1} S_{2}}{S_{1} + S_{2}}$

or 48 = $\frac{2 \times 36 \times S_{2}}{36 + S_{2}}$

or 2 = $\frac{3 \times S_{2}}{36 + S_{2}}$

or $S_{2}$ = 72 km/hour

Explanation 3: Allegation Method

The average speed is the weighted average with the time travelled at the different speeds being the weight.

As the distance is constant, so the time must be in inverse proportion.
So,

\frac{36}{x}

\frac{12}{(x - 48)}

\frac{3}{x}

\frac{1}{(x - 48)}

or 3x - 3 × 48 = x
or x = 72 km/hour

Time taken is same

When time taken is constant: The average speed of travelling at two different speeds for the same time span is just the simple average of two speeds.

Average speed = $\frac{S_{1} + S_{2}}{2}$

(Where, $S_{1}$ and $S_{2}$ are the two speeds at which we traveled for the same time.)

Note: The average speed is the weighted average with the time travelled at the different speeds being the weight and not the distance travelled at the different speeds.

Q. A person travels from A to B. For half the time he travelled at a speed of 20 km/hr and then at a speed of 30 km/hr for the other half. What is his average speed over the entire journey?

Explanations :

Explanation 1: Traditional Method

Let the distance travelled be d meters and the total time be 2 hours.

So, d = $d_{1}$ + $d_{2}$ = (20 × 1) + (30 × 1) = 50 kms

Average speed = $\frac{T o t a l d i s t a n c e t r a v e l l e d}{T o t a l t i m e t a k e n i n t r a v e l l i n g t h a t d i s t a n c e}$ = 50/2 = 25 km/hr

Explanation 2: Formula Method

Time spent in the two parts of the journey is the same.

So, Average speed = $\frac{S_{1} + S_{2}}{2}$ = (20 + 30)/2 = 25 km/hr

Explanation 3: Allegation Method

The average speed is the weighted average with the time travelled at the different speeds being the weight.

So, Average speed = $\frac{(1 \times 20) + (1 \times 30)}{(1 + 1)}$ = 50/2 = 25 km/hr

Special Cases of Average Speed

Case 1

A person goes a certain distance at a speed of $S_{1}$ km/hr and returns back to the same place at a speed of $S_{2}$ km/hr. If he takes ‘T’ hours in all (i.e. addition of times taken is given), then the distance between first place and the second place is –

Distance = $\frac{A v e r a g e s p e e d \times T o t a l t i m e t a k e n i n t w o s i d e j o u r n e y}{2}$ = $\frac{(\frac{2 S_{1} S_{2}}{S_{1} + S_{2}}) \times T}{2}$
= $\frac{S_{1} S_{2}}{S_{1} + S_{2}}$ × T km

Q. A person travels from A to B at a speed of 20 km/hr and then comes back to A at a speed of 30 km/hr. What is the distance between A and B if the total time taken in the two way journey is 5 hours?

Explanations :

Explanation 1: Traditional Method

Let the one side distance be d km.

Time taken to travel from A to B = Distance / Speed = d/20 hrs

Time taken to travel from B to A = Distance / Speed = d/30 hrs

Total time taken = (d/20) + (d/30) = 5
Or (5d/60) = 5
Or d = 60 km

Explanation 2: Formula Method

Distance = $\frac{S_{1} S_{2}}{S_{1} + S_{2}}$ × T km = $\frac{(20 \times 30)}{(20 + 30)}$ × 5 = 60 km

Case 2

If an object travels a distance ’d’ with speed $S_{1}$ in time $t_{1}$ and travels back the same distance with $\frac{m}{n}$ of the usual speed $S_{1}$ , then:

Change in time taken to cover the same distance = $[(\frac{m}{n}) - 1] t_{1}$ … (if n>m)

Or Change in time taken to cover the same distance = $[1 - (\frac{n}{m})] t_{1}$ … (if m>n)

Q. A person travels from A to B at a speed of 20 km/hr in 3 hours and then comes back to A at 3/2 times the previous speed. What is the difference in time taken in the two journeys?

Explanations :

Explanation 1: Traditional Method

Let the one side distance be d km.

d = Speed × Time = 20 × 3 = 60 km

Speed while going from B to A = (3/2) × 20 = 30 km/hr

Time taken to travel from B to A = Distance / Speed = 60/30 = 2 hours

So, the difference in time taken in the two journeys = 3 – 2 = 1 hour

Explanation 2: Formula Method

m/n = 3/2

So, m = 3 and n = 2

Change in time taken to cover the same distance = $[1 - (\frac{n}{m})] t_{1}$ = [1 – ( $\frac{2}{3}$ )] × 3 = (1/3) × 3 = 1 hour

Case 3

If a body covers part of the journey at speed $S_{1}$ and the remaining part of the journey at speed $S_{2}$ and the distances of the two parts of the journey are in the ratio m : n, then:

Average speed for the entire journey = $\frac{(m + n) S_{1} S_{2}}{(m S_{2} + n S_{1})}$

Q. A person travels one part of the journey at a speed of 20 km/hr and the other part at a speed of 30 km/hr. If the distance travelled in two parts is in the ratio 1:3, then what is his average speed in the entire journey?

Explanations :

Explanation 1: Traditional Method

Let first part of the journey be x km and the second part be 3x km.

Time taken to travel the first part of the journey = Distance / Speed = x/20 hrs

Time taken to travel the second part of the journey = Distance / Speed = 3x/30 hrs

Average speed = $\frac{T o t a l d i s t a n c e t r a v e l l e d}{T o t a l t i m e t a k e n i n t r a v e l l i n g t h a t d i s t a n c e}$ = $\frac{4 x}{(\frac{x}{20}) + (\frac{3 x}{30})}$ = $\frac{(4 \times 60)}{(3 + 6)}$ = 80/3 km/hr

Explanation 2:

Let first part of the journey be 20 km and the second part be 60 km.

Time taken to travel the first part of the journey = Distance / Speed = 20/20 = 1 hr

Time taken to travel the second part of the journey = Distance / Speed = 60/30 = 2 hrs

Average speed = $\frac{T o t a l d i s t a n c e t r a v e l l e d}{T o t a l t i m e t a k e n i n t r a v e l l i n g t h a t d i s t a n c e}$ = $\frac{(20 + 60)}{(1 + 2)}$ = 80/3 km/hr

Explanation 3: Formula Method

Average speed for the entire journey = $\frac{(m + n) S_{1} S_{2}}{(m S_{2} + n S_{1})}$ = $\frac{(1 + 3) (20 \times 30)}{(1 \times 30 + 3 \times 20)}$ = $\frac{(4 \times 20 \times 30)}{(30 + 60)}$ = 80/3 km/hr

Case 4

Suppose a person takes T hours in covering a distance of x km. If he covers some of the distance at a speed of $S_{1}$ km/hr on foot and rest of the distance on cycle at a speed of $S_{2}$ km/hr, then:

Time taken to cover the distance on foot = $\frac{(S_{2} T - x)}{(S_{2} - S_{1})}$ hours

Q. A person travels one part of the journey on foot at a speed of 10 km/hr and the other part on cycle at a speed of 20 km/hr. If the person travels 80 km in 5 hours, then what is the time taken to cover the distance on foot ?

Explanations :

Explanation 1: Traditional Method

Let the distance travelled on foot be x km.

The person travels 80 km in 5 hours, i.e. a speed of 80/5 = 16 km/hr

Time taken to travel the first part of the journey = Distance / Speed = x/10 hrs

Time taken to travel the second part of the journey = Distance / Speed = (80 - x)/20 hrs

Average speed = $\frac{T o t a l d i s t a n c e t r a v e l l e d}{T o t a l t i m e t a k e n i n t r a v e l l i n g t h a t d i s t a n c e}$
Or 16 = $\frac{80}{(\frac{x}{10}) + (\frac{80 - 𝑥}{20})}$
Or x + 80 = 100
Or x = 20 km
So, the time taken to cover the distance on foot = Distance/Speed = 20/10 = 2 hours

Explanation 2: Formula Method

Time taken to cover the distance on foot = $\frac{(S_{2} T - x)}{(S_{2} - S_{1})}$ hours
= $\frac{(20 \times 5 - 80)}{(20 - 10)}$ = 20/10 = 2 hours

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Last updated on Nov 13, 2021