Average Speed-Time Average Speed, Distance Average Speed, Formulas, Examples


Average Speed

While travelling in a car (or a bus) we have noticed that it is very difficult to keep the speed of the car at a constant or uniform value because at many places the brakes are to be applied to slow down or stop the car due to various reasons. Thus, the speed of a body is usually not constant and the
distance travelled divided by time gives us the average speed during that time.

Mathematically,

$$v_{av} = \frac{\text{Total Distance Travelled}}{\text{Total Time Taken}}$$

Imagine a car travels 120 km in 3 hours and then another 180 km in 2 hours. The total distance traveled is: 120+180=300 km

The total time taken is: 3+2=5 hours

Thus, the average speed is: $$v_{av} = \frac{300}{5} = 60 \text{ km/h}$$

This means the car covered an average of 60 km per hour over the entire journey.


Types of Average Speed

1. Time-Averaged Speed

When a particle moves at different uniform speeds $v_1, v_2, v_3, \dots$ for different time intervals $t_1, t_2, t_3, \dots$, then:

$$v_{av} = \frac{\text{Total Distance Travelled}}{\text{Total Time Taken}}$$

$$v_{av} = \frac{d_1 + d_2 + d_3 + \dots}{t_1 + t_2 + t_3 + \dots}$$

$$v_{av} = \frac{v_1t_1 + v_2t_2 + v_3t_3 + \dots}{t_1 + t_2 + t_3 + \dots}$$

Solution :

The total distance traveled: (40×2)+(60×3)=80+180=260 km

Total time taken: 2+3=5 hours

So, the average speed: $$v_{av} = \frac{260}{5} = 52 \text{ km/h}$$

If a particle moves with speed $v_1$ for half the total time and $v_2$ for the remaining half, then: $$v_{av} = \frac{v_1 + v_2}{2}$$

2. Distance-Averaged Speed

If a particle covers different distances $d_1, d_2, d_3, \dots$ with speeds $v_1, v_2, v_3, \dots$, then:

$$v_{av} = \frac{\text{Total Distance Travelled}}{\text{Total Time Taken}}$$

$$v_{av} = \frac{d_1 + d_2 + d_3 + \dots}{t_1 + t_2 + t_3 + \dots}$$

$$v_{av} = \frac{d_1 + d_2 + d_3 + \dots}{\frac{d_1}{v_1} + \frac{d_2}{v_2} + \frac{d_3}{v_3} + \dots}$$

  1. If a particle moves half the total distance at $v_1$ and the other half at $v_2$, then:

$$v_{av} = \frac{2 v_1 v_2}{v_1 + v_2} $$

  1. If a particle moves one-third of the distance at $v_1$, the next one-third at $v_2$, and the last one-third at $v_3$, then:

$$v_{av} = \frac{3 v_1 v_2 v_3}{v_1 v_2 + v_2 v_3 + v_3 v_1} $$


FAQs

Q1: Is average speed a vector or a scalar?

A: Average speed is a scalar quantity because it depends only on the magnitude of the total distance traveled, not on direction.

Q2: Why is average speed always greater than or equal to average velocity?

A: Since displacement is always ≤ distance, and velocity is based on displacement, the magnitude of average velocity is always ≤ average speed.


Multiple Choice Questions (MCQs) with Answers and Explanations

Q1: A car travels 100 m in 5 seconds and then 150 m in 10 seconds. What is its average speed?

(a) 10 m/s
(b) 12 m/s
(c) 15 m/s
(d) 17 m/s

Answer: (b) 12 m/s
Explanation: $$v_{av} = \frac{\text{Total Distance}}{\text{Total Time}}$$

$$v_{av} = \frac{100 + 150}{5 + 10} = \frac{250}{15} = 12 \text{ m/s}$$


Q2: A bus moves at 40 km/h for 2 hours and then at 60 km/h for 3 hours. What is its average speed?

(a) 48 km/h
(b) 50 km/h
(c) 52 km/h
(d) 54 km/h

Answer: (c) 52 km/h
Explanation:

Total distance traveled: (40×2)+(60×3)=80+180=260 km

Total time taken: 2+3=5 hours

Average speed:

$$v_{av} = \frac{\text{Total Distance}}{\text{Total Time}}$$

$$v_{av} = \frac{260}{5} = 52 \text{ km/h}$$


Q3: A train moves half the total distance at 30 km/h and the other half at 60 km/h. What is its average speed?

(a) 40 km/h
(b) 42 km/h
(c) 45 km/h
(d) 48 km/h

Answer: (a) 40 km/h
Explanation: Using the distance-averaged speed formula:

$$v_{av} = \frac{2 v_1 v_2}{v_1 + v_2} $$

$$v_{av}= \frac{2 \times 30 \times 60}{30 + 60}$$

$$v_{av} = \frac{3600}{90} = 40 \text{ km/h}$$


Q4: A cyclist covers one-third of the total distance at 20 km/h, next one-third at 30 km/h, and the last one-third at 60 km/h. What is the average speed?

(a) 30 km/h
(b) 32 km/h
(c) 35 km/h
(d) 40 km/h

Answer: (b) 32 km/h
Explanation:

Using the formula for three equal distances:

$$v_{av} = \frac{3 v_1 v_2 v_3}{v_1 v_2 + v_2 v_3 + v_3 v_1}$$

$$v_{av} = \frac{3 \times 20 \times 30 \times 60}{(20 \times 30) + (30 \times 60) + (60 \times 20)}$$

$$v_{av} = \frac{108000}{3360} = 32.14 \approx 32 \text{ km/h}$$


Q5: A person walks 2 km at 5 km/h, then jogs 3 km at 10 km/h. What is the average speed?

(a) 6 km/h
(b) 6.5 km/h
(c) 7 km/h
(d) 7.5 km/h

Answer: (c) 7 km/h
Explanation:

Total distance traveled: 2+3=5 km

Time taken for first part: $$t_1=\frac{2}{5} = 0.4 \text{ hours}$$

Time taken for second part: $$t_2=\frac{3}{10} = 0.3 \text{ hours}$$

Total time taken: 0.4+0.3=0.7 hours

Average speed:

$$v_{av} = \frac{\text{Total Distance}}{\text{Total Time}}$$

$$v_{av} = \frac{5}{0.7} = 7.14 \approx 7 \text{ km/h}$$


Q6: A car travels 100 m in 5 seconds and then 150 m in 10 seconds. What is its average speed?

(a) 10 m/s
(b) 12 m/s
(c) 15 m/s
(d) 17 m/s

Answer: (b) 12 m/s

Explanation: Average speed:

$$v_{av} = \frac{\text{Total Distance}}{\text{Total Time}}$$

$$v_{av} = \frac{(100 + 150)}{(5 + 10)} = \frac{250}{15} = 12 \text{ m/s}$$

These MCQs help in understanding how to calculate average speed under different conditions, which is crucial for JEE, NEET, and CBSE Board Class 11 exams.


Conceptual Questions with Answers

Q1: Can average speed be zero?

A: No, unless the total distance traveled is zero. However, average velocity can be zero if the initial and final positions are the same.


Do You Know?

  • Average speed and velocity are the same only for straight-line motion without turning back.
  • If a body moves at a constant speed in a circular path, its average velocity over one complete revolution is zero.

Worksheet

Solve the following problems:

  1. A train covers half its journey at 40 km/h and the other half at 60 km/h. Find its average speed.
  2. A cyclist moves 100 m in 10 sec and then 150 m in 15 sec. Find the average speed.

Test Paper (Total Marks: 10)

  1. Define average speed. (2 marks)
  2. A car moves at 30 km/h for 2 hours and then at 50 km/h for 3 hours. Find its average speed. (4 marks)
  3. Derive the formula for distance-averaged speed when a particle moves equal distances at different speeds. (4 marks)

Important Points for Quick Revision

  • $v_{av} = \frac{\text{Total Distance}}{\text{Total Time}}$
  • Time-averaged speed considers time intervals; distance-averaged speed considers distances.
  • Use harmonic mean formula for distance-averaged speed when distances are equal.

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Er. Neeraj K.Anand is a freelance mentor and writer who specializes in Engineering & Science subjects. Neeraj Anand received a B.Tech degree in Electronics and Communication Engineering from N.I.T Warangal & M.Tech Post Graduation from IETE, New Delhi. He has over 30 years of teaching experience and serves as the Head of Department of ANAND CLASSES. He concentrated all his energy and experiences in academics and subsequently grew up as one of the best mentors in the country for students aspiring for success in competitive examinations. In parallel, he started a Technical Publication "ANAND TECHNICAL PUBLISHERS" in 2002 and Educational Newspaper "NATIONAL EDUCATION NEWS" in 2014 at Jalandhar. Now he is a Director of leading publication "ANAND TECHNICAL PUBLISHERS", "ANAND CLASSES" and "NATIONAL EDUCATION NEWS". He has published more than hundred books in the field of Physics, Mathematics, Computers and Information Technology. Besides this he has written many books to help students prepare for IIT-JEE and AIPMT entrance exams. He is an executive member of the IEEE (Institute of Electrical & Electronics Engineers. USA) and honorary member of many Indian scientific societies such as Institution of Electronics & Telecommunication Engineers, Aeronautical Society of India, Bioinformatics Institute of India, Institution of Engineers. He has got award from American Biographical Institute Board of International Research in the year 2005.