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Question: A body has speed V, 2V and 3V in first \(\dfrac{1}{3}\) distance S, second \(\dfrac{1}{3}\) of S and...

A body has speed V, 2V and 3V in first 13\dfrac{1}{3} distance S, second 13\dfrac{1}{3} of S and third 13\dfrac{1}{3} of S respectively. Its average speed will be:
A) V
B) 2V
C)1811V\text{C)}\dfrac{18}{11}\text{V}
D)1118V\text{D)}\dfrac{11}{18}\text{V}

Explanation

Solution

To calculate average speed, when time taken is not given, first we need to calculate time taken and then calculate the total time taken which can be used to calculate the average speed of the body. In case the exact value of distance is not given, assume any variable which represents distance.

Formula used:
Average speed = Total DistanceTime taken\dfrac{\text{Total Distance}}{\text{Time taken}}

Complete answer:
In this question, total distance is not given.
So, let’s assume,
Total distance = d
Now, as we know that the formula to calculate average speed is given by:
Average speed=Total DistanceTime Taken\text{Average speed=}\dfrac{\text{Total Distance}}{\text{Time Taken}} -------Equation(1)
From equation (1), Time taken=Total distanceAverage speed\text{Time taken=}\dfrac{\text{Total distance}}{\text{Average speed}}
Now, say t1{{\text{t}}_{1}}= time taken to travel first 13\dfrac{1}{3} distance
t2{{\text{t}}_{2}}= time taken to travel second 13\dfrac{1}{3} distance
t3{{\text{t}}_{3}}= time taken to travel third 13\dfrac{1}{3} distance
From the question and formula we can get the values of t1{{\text{t}}_{1}}, t2{{\text{t}}_{2}} and t3{{\text{t}}_{3}}
t1=d1×3{{\text{t}}_{1}}=\dfrac{\text{d}}{1\times 3}
t2=d2×3{{\text{t}}_{2}}=\dfrac{\text{d}}{2\times 3}
t3=d3×3{{\text{t}}_{3}}=\dfrac{\text{d}}{3\times 3}

Now, as we need to find the average speed, we will use formula from equation no. (1),
Average speed = dt1+t2+t3\Rightarrow \text{Average speed = }\dfrac{\text{d}}{{{\text{t}}_{1}}+{{\text{t}}_{2}}+{{\text{t}}_{3}}}
Average speed = dd1×3+d2×3+d3×3\Rightarrow \text{Average speed = }\dfrac{\text{d}}{\dfrac{\text{d}}{1\times 3}+\dfrac{\text{d}}{2\times 3}+\dfrac{\text{d}}{3\times 3}}
We will take ‘d; common from denominator and numerator,
Average speed = 113+16+19\Rightarrow \text{Average speed = }\dfrac{1}{\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{9}}
Further solving this fraction we will get,
Average speed = 186+3+2\Rightarrow \text{Average speed = }\dfrac{18}{6+3+2}
Average speed = 1811V\Rightarrow \text{Average speed = }\dfrac{18}{11}\text{V}
Hence, we can say that the average speed of the body will be 1811V\dfrac{18}{11}\text{V}.

So, the correct answer is “Option C”.

Note:
It is advisable to use a variable if the actual value is not given in the question, so as to avoid calculation errors.
As, here, in these cases we have assumed, distance as ‘d’.
The average speed is equal to the total distance divided by total time taken.