Question

A point traversed half of the distance with a velocity ${v_0}$. The half of the remaining part of the distance was covered with velocity ${v_1}$ and the second half of the remaining part by ${v_2}$ velocity. The mean velocity of the point, averaged over the whole time of motion is
A. $\dfrac{{{v_0} + {v_1} + {v_2}}}{3}$
B. $\dfrac{{2{v_0} + {v_1} + {v_2}}}{3}$
C. $\dfrac{{{v_0}{v_1}{v_2}}}{{2{v_1}{v_2} + {v_0}{v_1} + {v_0}{v_2}}}$
D. $\dfrac{{4{v_0}{v_1}{v_2}}}{{2{v_1}{v_2} + {v_0}{v_1} + {v_0}{v_2}}}$

Answer 1

Hint: Recall the concept of distance. It is defined as the length that an object has covered during its motion. It is the length of the initial and the final position of an object. It has only magnitude and not direction. Therefore, it is a scalar quantity.

Complete step by step answer:
Step I:
Let the total distance covered by the point is $ = s$
Let the time taken to cover half of the distance is $ = {t_1}$
The formula for speed is written as
$Speed = \dfrac{\text{Distance}}{\text{Time}}$

Step II:
When half of the distance is covered then the value of distance will be $ = \dfrac{s}{2}$
The time taken to cover this distance will be written as
$ \Rightarrow {t_1} = \dfrac{{\dfrac{s}{2}}}{{{v_0}}}$
$ \Rightarrow {t_1} = \dfrac{s}{{2{v_0}}}$

Step III:
Now it is given that the remaining half distance is further divided into two equal distances. Therefore, the distance covered in time ${t_2}$ will be $ = \dfrac{s}{4}$
The time taken to cover this distance will be ${t_2} = \dfrac{{\dfrac{s}{4}}}{{{v_1}}}$
Or $ \Rightarrow {t_2} = \dfrac{s}{{4{v_1}}}$

Step IV:
The last distance is covered in the time ${t_3} = \dfrac{{\dfrac{s}{4}}}{{{v_2}}}$
$ \Rightarrow {t_3} = \dfrac{s}{{4{v_2}}}$

Step V:
Average velocity of an object over the whole time of motion can be calculated by taking the ratio of the total distance covered with the total time taken.
$ \Rightarrow t = \dfrac{s}{{{t_1} + {t_2} + {t_3}}}$
$t = \dfrac{s}{{\dfrac{s}{{2{v_0}}} + \dfrac{s}{{4{v_1}}} + \dfrac{s}{{4{v_2}}}}}$
$t = \dfrac{s}{{s\left(\dfrac{1}{{2{v_0}}} + \dfrac{1}{{4{v_1}}} + \dfrac{1}{{4{v_2}}}\right)}}$
$ \Rightarrow t = \dfrac{{4{v_0}{v_1}{v_2}}}{{2{v_1}{v_2} + {v_0}{v_1} + {v_0}{v_2}}}$

Hence, the correct answer is option (D).

Note: It is important to remember that the terms distance and displacement are not to be confused as they are completely different terms. Displacement is defined as the change in the position of an object. It is usually represented by using an arrow. The arrow points from initial to final position. It has both magnitude and direction. So it is a vector quantity.