
A particle moves in a straight line with velocity \[6\,m{s^{ - 1}}\] and \[3\,m{s^{ - 1}}\] for time intervals which are in the ratio 1:2. Find the average velocity of the particle.
A. \[2\,m{s^{ - 1}}\]
B. \[3\,m{s^{ - 1}}\]
C. \[4\,m{s^{ - 1}}\]
D. \[5\,m{s^{ - 1}}\]
Answer
161.1k+ views
Hint:The average velocity of a particle can be found by the ratio of the total distance travelled to the total time taken. So, we can find the total distance that is travelled by a particle from one point to another as the product of velocity and time taken for a particle.
Formula Used:
To find the average velocity of a particle the formula is,
\[{V_{avg}} = \dfrac{{\Delta x}}{{\Delta t}}\]
Where, \[\Delta x\] is the total distance travelled and \[\Delta t\] is the total time taken.
Complete step by step solution:
Consider a particle that is moving in a straight line with velocity \[{v_1} = 6\,m{s^{ - 1}}\] and \[{v_2} = 3\,m{s^{ - 1}}\] for time intervals that are in the ratio 1:2. That is \[{t_1} = 1t\] and \[{t_2} = 2t\]. We need to find the average velocity of the particle. In order to find the average velocity of a particle we have the formula,
\[{V_{avg}} = \dfrac{{\Delta x}}{{\Delta t}}\]
\[\Rightarrow {V_{avg}} = \dfrac{{{d_1} + {d_2}}}{{{t_1} + {t_2}}}\]
Here, \[{d_1} = {v_1} \times {t_1}\] and \[{d_2} = {v_2} \times {t_2}\]
Then, the above equation becomes,
\[{V_{avg}} = \dfrac{{{v_1}{t_1} + {v_2}{t_2}}}{{{t_1} + {t_2}}}\]
Substitute the value of \[{v_1}\] , \[{v_2}\], \[{t_1}\]and \[{t_2}\]we get,
\[{V_{avg}} = \dfrac{{\left( {6 \times 1t} \right) + \left( {3 \times 2t} \right)}}{{1t + 2t}}\]
\[\Rightarrow {V_{avg}} = \dfrac{{12t}}{{3t}}\]
\[\therefore {V_{avg}} = 4\,m{s^{ - 1}}\]
Therefore, the average velocity of the particle is \[4\,m{s^{ - 1}}\].
Hence, Option C is the correct answer
Note:Remember the formula for the average velocity of a particle which is very important in order to solve these kinds of problems. Here, the total distance travelled by the particle is not given, so in order to calculate the total distance travelled, the total distance is equal to the product of the distance covered or travelled by the particle and the time taken by it.
Formula Used:
To find the average velocity of a particle the formula is,
\[{V_{avg}} = \dfrac{{\Delta x}}{{\Delta t}}\]
Where, \[\Delta x\] is the total distance travelled and \[\Delta t\] is the total time taken.
Complete step by step solution:
Consider a particle that is moving in a straight line with velocity \[{v_1} = 6\,m{s^{ - 1}}\] and \[{v_2} = 3\,m{s^{ - 1}}\] for time intervals that are in the ratio 1:2. That is \[{t_1} = 1t\] and \[{t_2} = 2t\]. We need to find the average velocity of the particle. In order to find the average velocity of a particle we have the formula,
\[{V_{avg}} = \dfrac{{\Delta x}}{{\Delta t}}\]
\[\Rightarrow {V_{avg}} = \dfrac{{{d_1} + {d_2}}}{{{t_1} + {t_2}}}\]
Here, \[{d_1} = {v_1} \times {t_1}\] and \[{d_2} = {v_2} \times {t_2}\]
Then, the above equation becomes,
\[{V_{avg}} = \dfrac{{{v_1}{t_1} + {v_2}{t_2}}}{{{t_1} + {t_2}}}\]
Substitute the value of \[{v_1}\] , \[{v_2}\], \[{t_1}\]and \[{t_2}\]we get,
\[{V_{avg}} = \dfrac{{\left( {6 \times 1t} \right) + \left( {3 \times 2t} \right)}}{{1t + 2t}}\]
\[\Rightarrow {V_{avg}} = \dfrac{{12t}}{{3t}}\]
\[\therefore {V_{avg}} = 4\,m{s^{ - 1}}\]
Therefore, the average velocity of the particle is \[4\,m{s^{ - 1}}\].
Hence, Option C is the correct answer
Note:Remember the formula for the average velocity of a particle which is very important in order to solve these kinds of problems. Here, the total distance travelled by the particle is not given, so in order to calculate the total distance travelled, the total distance is equal to the product of the distance covered or travelled by the particle and the time taken by it.
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