Question

A particle moving in a straight line covers half the distance with speed of $12m/s$. The other half of the distance is covered in two equal time intervals with speed of $4.5m/s$ and $7.5m/s$ respectively. The average speed of the particle during this motion is:
(A) $8.0m/s$
(B) $12.0m/s$
(C) $10.0m/s$
(D) $9.8m/s$

Answer 1

Hint: To solve this question, we need to use the basic definition of the average speed of a particle in a given interval of time. For that we have to find out the total distance covered, and the total length of the time interval. Substituting these in the formula for the average speed, we will get the final answer.

Complete step-by-step solution:
Let the total distance covered by the particle be $D$ and the total length of the time interval be $T$.
According to the question, the first half of the total distance is covered with a speed of $12m/s$ by the particle. So the time taken for covering this distance is given by
${t_1} = \dfrac{{D/2}}{{12}}$
$ \Rightarrow {t_1} = \dfrac{D}{{24}}$..............(1)
Now, the second half of the total distance is covered in the two equal time intervals of the remaining time. Since the total time is equal to $T$, so the time taken to cover the second half of the distance is given by
${t_2} = T - {t_1}$
From (1)
${t_2} = T - \dfrac{D}{{24}}$..............(2)
Since this time interval is again divided into two equal intervals, so the length of each interval becomes
$t = \dfrac{{{t_2}}}{2}$
From (2)
$t = \dfrac{T}{2} - \dfrac{D}{{48}}$..............................(3)
The distance covered with the speed of $4.5m/s$ is given by
${d_1} = 4.5t$
From (3)
\[{d_1} = 4.5\left( {\dfrac{T}{2} - \dfrac{D}{{48}}} \right)\] (4)
Similarly, the distance covered with the speed of $7.5m/s$ is given by
\[{d_1} = 7.5\left( {\dfrac{T}{2} - \dfrac{D}{{48}}} \right)\] (5)
Since the total distance covered in the second time interval is equal to the half of the total distance covered, so we have
\[{d_1} + {d_2} = \dfrac{D}{2}\]
Putting (4) and (5) above, we get
\[7.5\left( {\dfrac{T}{2} - \dfrac{D}{{48}}} \right) + 4.5\left( {\dfrac{T}{2} - \dfrac{D}{{48}}} \right) = \dfrac{D}{2}\]
$ \Rightarrow 12\left( {\dfrac{T}{2} - \dfrac{D}{{48}}} \right) = \dfrac{D}{2}$
Dividing by $12$ we have
$\dfrac{T}{2} - \dfrac{D}{{48}} = \dfrac{D}{{24}}$
$ \Rightarrow \dfrac{T}{2} = \dfrac{D}{{24}} + \dfrac{D}{{48}}$
Taking the LCM, we have
$\dfrac{T}{2} = \dfrac{{2D + D}}{{48}}$
$ \Rightarrow \dfrac{T}{2} = \dfrac{D}{{16}}$
Multiplying by $2$ we get
$T = \dfrac{D}{8}$.....................(6)
Now, we know that the average speed of a particle is equal to the ratio of the total displacement covered by the particle to the total time taken. So we have
${v_{avg}} = \dfrac{D}{T}$
Putting (6) in the above expression we have
${v_{avg}} = \dfrac{D}{{D/8}}$
$ \Rightarrow {v_{avg}} = 8m{s^{ - 1}}$
Thus, the average speed of the particle is equal to $8.0m/s$.

Hence, the correct answer is option A.

Note: We should not misunderstand the concept of the average speed as being the average of the speeds of the particle in the different intervals of time. Average speed is always calculated by using its basic definition which is, the total distance covered divided by the total time taken.