
A particle moving in a straight line covers half the distance with a speed of $3\,m{\text{ }}{s^{ - 1}}$. The other half of the distance is covered in two equal time intervals with uniform speeds of $4.5\,m{\text{ }}{s^{ - 1}}$ and $7.5\,m{\text{ }}{s^{ - 1}}$ respectively.The average speed of the particle during this motion is
A. $4,m{\text{ }}{s^{ - 1}}$
B. $5\,m{\text{ }}{s^{ - 1}}$
C. $5.5\,m{\text{ }}{s^{ - 1}}$
D. $4.8\,m{\text{ }}{s^{ - 1}}$
Answer
434.7k+ views
Hint:To solve this question here we will find time for each interval and then sum up all the intervals all together and we will get the required answer. The average speed of a body in a certain time interval is the distance covered by the body in that time interval divided by time.
Formula used:
${V_{avg}} = \dfrac{{{\text{Total distance covered}}}}{{{\text{Total time}}}}$
Complete step by step answer:
Let us consider, $x$ is the total distance covered by a particle. Half of the distance is covered with speed $3m{\text{ }}{s^{ - 1}}$ , then time required to covered the distance is
${t_1} = \dfrac{x}{6}\sec $
Other half of the distance is covered in two equal time intervals with uniform speed $4.5m{\text{ }}{s^{ - 1}}$ and $7.5m{\text{ }}{s^{ - 1}}$ respectively.
Then time required is,
${t_2} = \dfrac{x}{{4 \times 4.5}} \\
\Rightarrow {t_2} = \dfrac{x}{{18}}\sec \\ $
Similarly,
${t_3} = \dfrac{x}{{4 \times 7.5}} \\
\Rightarrow {t_3} = \dfrac{x}{{30}}\sec $
Therefore, total time required, $t$ is ${t_1} + {t_2} + {t_3}$
The average speed of the particle during the motion is
${V_{avg}} = \dfrac{{{\text{Total distance covered}}}}{{{\text{Total time}}}} \\
\Rightarrow {V_{avg}} = \dfrac{x}{t} \\
\Rightarrow {V_{avg}} = \dfrac{x}{{{t_1} + {t_2} + {t_3}}} \\ $
Now, putting the values of ${t_1},{t_2}$ and ${t_3}$ in the above equation,
${V_{avg}} = \dfrac{x}{{{t_1} + {t_2} + {t_3}}} \\
\Rightarrow {V_{avg}} = \dfrac{x}{{\dfrac{x}{6} + \dfrac{x}{{18}} + \dfrac{x}{{30}}}} \\
\Rightarrow {V_{avg}} = \dfrac{x}{{0.25x}} \\
\therefore {V_{avg}} = 4m{\text{ }}{s^{ - 1}} \\ $
Hence, the correct option is D.
Note:Don’t get confused on average velocity and average speed hence, they both are similar terms but there is slight difference and that is of distance and displacement. In average velocity we take total displacement and in average speed we take total distance.
Formula used:
${V_{avg}} = \dfrac{{{\text{Total distance covered}}}}{{{\text{Total time}}}}$
Complete step by step answer:
Let us consider, $x$ is the total distance covered by a particle. Half of the distance is covered with speed $3m{\text{ }}{s^{ - 1}}$ , then time required to covered the distance is
${t_1} = \dfrac{x}{6}\sec $
Other half of the distance is covered in two equal time intervals with uniform speed $4.5m{\text{ }}{s^{ - 1}}$ and $7.5m{\text{ }}{s^{ - 1}}$ respectively.
Then time required is,
${t_2} = \dfrac{x}{{4 \times 4.5}} \\
\Rightarrow {t_2} = \dfrac{x}{{18}}\sec \\ $
Similarly,
${t_3} = \dfrac{x}{{4 \times 7.5}} \\
\Rightarrow {t_3} = \dfrac{x}{{30}}\sec $
Therefore, total time required, $t$ is ${t_1} + {t_2} + {t_3}$
The average speed of the particle during the motion is
${V_{avg}} = \dfrac{{{\text{Total distance covered}}}}{{{\text{Total time}}}} \\
\Rightarrow {V_{avg}} = \dfrac{x}{t} \\
\Rightarrow {V_{avg}} = \dfrac{x}{{{t_1} + {t_2} + {t_3}}} \\ $
Now, putting the values of ${t_1},{t_2}$ and ${t_3}$ in the above equation,
${V_{avg}} = \dfrac{x}{{{t_1} + {t_2} + {t_3}}} \\
\Rightarrow {V_{avg}} = \dfrac{x}{{\dfrac{x}{6} + \dfrac{x}{{18}} + \dfrac{x}{{30}}}} \\
\Rightarrow {V_{avg}} = \dfrac{x}{{0.25x}} \\
\therefore {V_{avg}} = 4m{\text{ }}{s^{ - 1}} \\ $
Hence, the correct option is D.
Note:Don’t get confused on average velocity and average speed hence, they both are similar terms but there is slight difference and that is of distance and displacement. In average velocity we take total displacement and in average speed we take total distance.
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