Answer
Verified
91.2k+ views
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.
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.
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
Other Pages
The vapour pressure of pure A is 10 torr and at the class 12 chemistry JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main
3 mole of gas X and 2 moles of gas Y enters from the class 11 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Derive an expression for maximum speed of a car on class 11 physics JEE_Main
Velocity of car at t 0 is u moves with a constant acceleration class 11 physics JEE_Main