Question

If a car covers \[{\left( {\dfrac{2}{5}} \right)^{th}}\]of the total distance with the speed V and \[{\left( {\dfrac{3}{5}} \right)^{th}}\]distance with\[{V_2}\]. Then find the average speed.
A. \[\dfrac{1}{2}\sqrt {{v_1}{v_2}} \]
B. \[\dfrac{{{v_1} + {v_2}}}{2}\]
C. \[\dfrac{{2{v_1}{v_2}}}{{{v_1} + {v_2}}}\]
D. \[\dfrac{{5{v_1}{v_2}}}{{3{v_1} + 2{v_2}}}\]

Answer 1

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,
\[{V_{avg}} = \dfrac{x}{t}\]
Where, \[x\] is total distance travelled and \[t\] is total time taken.

Complete step by step solution:
Suppose a car covers \[{\left( {\dfrac{2}{5}} \right)^{th}}\] of the total distance with the speed V and \[{\left( {\dfrac{3}{5}} \right)^{th}}\]distance with\[{V_2}\]. Then we need to find the average speed. In order to find the average speed, the formula is,
\[{V_{avg}} = \dfrac{x}{t}\]……………… (1)
Here, the time taken to cover \[{\left( {\dfrac{2}{5}} \right)^{th}}\] distance,
\[{t_1} = \dfrac{{\dfrac{2}{5}x}}{{{v_1}}}\]
\[ \Rightarrow {t_1} = \dfrac{{{\rm{distance}}}}{{{\rm{velocity}}}}\]
Similarly, the time taken to cover \[{\left( {\dfrac{3}{5}} \right)^{th}}\]distance,
\[{t_1} = \dfrac{{\dfrac{3}{5}x}}{{{v_2}}}\]
Now, the equation (1) can be written as,
\[{V_{avg}} = \dfrac{x}{{{t_1} + {t_2}}}\]

Substitute the value of \[{t_1}\]and \[{t_2}\]we get,
\[{V_{avg}} = \dfrac{x}{{\left( {\dfrac{{\dfrac{2}{5}x}}{{{v_1}}}} \right) + \left( {\dfrac{{\dfrac{3}{5}x}}{{{v_2}}}} \right)}} \\ \]
\[\Rightarrow {V_{avg}} = \dfrac{x}{{\left( {\dfrac{{\left( {\dfrac{2}{5}x{v_2}} \right) + \left( {\dfrac{3}{5}x{v_1}} \right)}}{{{v_1}{v_2}}}} \right)}} \\ \]
\[\Rightarrow {V_{avg}} = \dfrac{{x{v_1}{v_2}}}{{\left( {\dfrac{2}{5}x{v_2}} \right) + \left( {\dfrac{3}{5}x{v_1}} \right)}} \\ \]
\[\Rightarrow {V_{avg}} = \dfrac{{x{v_1}{v_2}}}{{\left( {\dfrac{{2x{v_2} + 3x{v_1}}}{5}} \right)}} \\ \]
\[\Rightarrow {V_{avg}} = \dfrac{{5x{v_1}{v_2}}}{{x\left( {2{v_2} + 3{v_1}} \right)}} \\ \]
\[\therefore {V_{avg}} = \dfrac{{5{v_1}{v_2}}}{{3{v_1} + 2{v_2}}}\]
Therefore, the average speed is \[\dfrac{{5{v_1}{v_2}}}{{3{v_1} + 2{v_2}}}\].

Hence, option D is the correct answer.

Note: Remember that the particle's displacement has both magnitude and direction. As a result, it is a vector quantity. Also, don't get the phrases distance and displacement mixed up; they're not the same thing. The particle's displacement is independent of the path and provides no information about the path taken, while the distance changes depending on the path travelled by the particle. A particle's distance travelled can never be 0 or negative; it is always positive. However, when it comes to particle displacement, it might be positive, zero, or negative.