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Average Speed

The motion of objects is described in the branch of physics that is Kinematics which comes under mechanics. This is studied with terms like scalar and vector quantities, displacement and distance, speed, acceleration, and velocity which are manly used for the motion of objects. Vector quantities are explained by their magnitude with direction while scalar are used only their numerical value without the explanation of direction. The scalar quantity speed shows the fastness of any object that how fast the object can be moved.

The value of speed is zero when there is no movement is shown by object. This is basically a distance which is covered by the moving object. When an object is moved it undergoes many changes in speed. So the needle of speedometer constantly moves up or down to show the correct speed at a particular time. But the average of all speed shows the whole motion of object at a given period of time. Let’s discuss average speed and its problem solving formula.


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Average Speed Definition

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The Average speed, as evident from the name itself, is the average of the speed of a moving object for the overall distance it has covered.

The average speed is related to the distance traveled by the object and is a scalar quantity, which means, it is only represented by the magnitude and direction of travel is not important.
The formula for average speed is calculated by finding the ratio of the total distance covered by the object to the time taken to cover that distance. It is not the average of the speed.

The equation for average speed is given by:

SAVG = $\frac{Total\ Distance\ Traveled}{Total\ Time\ taken}$ .............(1)

SAVG = $\frac{D_{total}}{T_{total}}$ ................(2)

The average speed and average velocity are also related like the speed and the velocity. The average velocity is the ratio of total displacement of the object over a given time. While average velocity is related to displacement of the object, the average speed is related to the total distance traveled by the object.

The equation (2) represents the average speed formula of an object moving with a varying speed.

The Average speed is sometimes misunderstood for instantaneous speed. Both are different from each other, in average speed the total time is large while in instantaneous speed limiting case of the speed where time approaches zero.
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Average Speed Problems

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The following examples will help us understand how to calculate average speed.

Solved Examples

Question 1: A runner runs at a track meet. He completes 800 meter lap in 80 seconds. After the finish he is at the starting point. Calculate average speed of the runner during this lap?
Solution:
 
For finding the Average speed of the runner, we must find the total distance covered by him and the total time taken to complete that distance.

In this case the distance covered by him is equal to 800 meters and he has completed it in 80 seconds.
So, applying formula for the average speed
we have
                SAVG = $\frac{800} {80}$,
                SAVG = 10 m/s,
So, the Average speed of the runner on the track is 10 m/s.
 

Question 2: A Man is traveling in his car from city A to city B and back. In the journey from city A to city B, he is traveling with the constant speed of 40 kmph, and he is traveling with the 45 kmph while he is coming back. The total journey took 3 hours to complete. Find the average speed of the car for the whole journey?

Solution:
 
As you can see that we are provided with the speed in both the direction, one can directly calculate the average speed by averaging the two speeds, but it is the wrong approach.

Let us assume that the distance between two cities is D.
Time taken is equal is 3 hours to complete the round trip journey. Also assume that the time taken from A to b is t hours so the time taken from B to A is 3 – t hours.

Now, the correct approach for finding average speed is as follows, first find the distance in both the direction.
DAB = 40 $\times$ t
DBA = 45 $\times$ (3 - t)

Since both the distance $D_{AB}$ and $D_{BA}$ are same (from city A to B and from city B to A), so we can say that;
$D_{AB}$ = $D_{BA}$
40 $\times$ t = 45 $\times$ (3 - t)
40t = 135 - 45t
85t = 135
t = $\frac{135}{85}$
t = 1.59 hours
So, the time from city A to B is 1.59 hours and the time from city A to B is 1.41 hours.

Now we will find the distance between the city A to B is
$D_{AB}$ = $S_{AB}$ $\times$ t
$D_{AB}$ = 40 $\times$ 1.59 = 63.53 kms
So, the average speed of the round trip journey is
$S_{AVG}$ = $\frac{(D_{AB} + D_{BA})}{(T_{AB} + T_{BA})}$

Since $D_{AB}$ = $D_{BA}$, we will take it D.
So, the total distance is 2D = 127.05 km, putting these values in the above equation for finding average speed
$S_{AVG}$ = $\frac{(127.05)}{(3)}$ 
$S_{AVG}$ = 42.35 kmph.

 

Question 3: Vikram drove his car for 3 hours at the rate of 60 miles per hour and for 4 hours at 50 miles per hour. Find his average speed for the journey?
Solution:
 
For calculating average speed we need to find the total distance traveled by Vikram.
D1 = 60 $\times$ 3 = 180 miles
D2 = 50 $\times$ 4 = 200 miles
Therefore, the total distance traveled is
D = D1 + D2
D = 180 + 200
D = 380 miles

So, the average speed is
SAVG = $\frac{(380)}{(3 + 4)}$
SAVG = $\frac{(380)}{(7)}$
SAVG = 54.29 miles per hour.
So, the average speed of the vikram’s journey by car is 54.29 miles per hour.


 

Question 4: Mr. B and Mr. A ride their bikes from their house to school which is 14.4 kilometers away from their house. It takes Mr. A 40 minutes to arrive at school. Mr. B arrives 20 minutes after Mr. A. Find how much faster Mr. A is moving with respect to Mr. B?
Solution:
 
The distance to be covered by both of them is equal to 14.4 kms.
Mr. A completes it in 40 minutes and Mr. B takes 20 minutes more than Mr. A, so Mr. B completes it in 60 minutes.

Average speed of A = $\frac{14.4}{(\frac{40}{60})}$
SA = $\frac{(14.4 \times  60)}{40}$
SA = 21.6 kmph

Average speed of B = $\frac{14.4}{(\frac{60}{60})}$
SB = $\frac{(14.4 \times 60)}{60}$
SB = 14.4 kmph

So, difference of speed of Mr. A and Mr. B is:
SA - SB = 21.6 - 14.4 = 7.2
So, Mr. A is 7.2 kmph faster than Mr. B.
 

Question 5: A car is travelling with the speed of 30 mph from city A to B and back from city B to A with the speed of 40 mph. Find its average speed?
Solution:
 
For finding the average speed of the car, we need to first identify total distance which is equal to two times the distance between cities A and B.
Time taken from A to B is = $\frac{D}{S_{1}}$
Time taken from B to A is = $\frac{D}{S_{2}}$

Total distance travelled is = 2D
So,
$S_{(avg)}$ = $\frac{(2D)}{((\frac{D}{S_{1}}) + (\frac{D}{S_{2}}))}$

Putting the value and solving we have
$S_{(avg)}$ = 2 $\times$ $\frac{(S_{1} \times S_{2})}{(S_{1} + S_{2})}$
$S_{(avg)}$ = 2 $\times$ $\frac{(30 \times 40)}{70}$
$S_{(avg)}$ = 34.29 mph.


 

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