Question

Please answer fast It's urgent

Answer 1

Let the distance covered = 1/3 x

Total distance = x

Time taken when velocity is 20 m/s

${t}_{1} = \: \dfrac{x}{60}$

${t}_{2} = \: \dfrac{x}{90}$

${t}_{3} = \: \dfrac{x}{120}$

Now,

<V> = Total distance/Total Time

<V> = x/ x/60+x/90+x/120

<V> = x/ x(6+4+3)/360

<V> = 360/13

<V> = 27.6 m/s

Hence, Average Velocity = <V> = 27.6 m/s Which nearly equal to 28 m/s.

Hence, option 1st is correct

Answer 2

$\huge\boxed{\mathbb{\underline{\underline{\underline{GIVEN}}}}}$

$\textbf{A\: body\: covers first one third of the }$

$\textbf{distance with a velocity 20m/s, }$

$\textbf{the second third with a velocity}$

$\textbf{/speed of 30 m/s and last one }$

$\textbf{third with a velocity of 40 m/s.}$

$\textbf{What is the average velocity}$

$\textbf{of the body ??}$

$\huge\boxed{\mathfrak\pink{Answer}}$

$\textbf{ Option\: A\: is \:the \:correct\: answer}$

$\huge\boxed{\mathbb\red{EXPLANATION}}$

$\texttt{Let the total distance be x}$

$\texttt{Time taken to cover distance}$

$\texttt{ with velocity 20 m/s = \dfrac{x}{60} s }$

$\texttt{Time take to cover distance}$

$\texttt{ with velocity 30 m/s = \dfrac{x}{90} s }$

$\texttt{Time taken to cover distance}$

$\texttt{with velocity 40 m/s = \dfrac{x}{120} s }$

$\boxed{\texttt{Average Velocity = \dfrac{Total \:Distance}{Total \:Time \:Taken} }}$

$\textsf{Total Distance = x }$

$\textsf{Total Time Taken = \dfrac{x}{60}+\dfrac{x}{90}+\dfrac{x}{120} }$

$\Longrightarrow{\textsf{ \dfrac{6x + 4x + 3x }{120} }}$

$\textsf{Average Velocity = \dfrac{x}{\dfrac{6x + 4x + 3x}{360}} }$

$\textsf{Average Velocity = \dfrac{360x}{x(6+4+3)} }$

$\textbf{Average Velocity = \dfrac{360}{13} }$

$\textsf{Average Velocity = 27.692 m/s}$

$\textbf{Therefore, the average velocity of the body is 27.692 m/s or 28 m/s approximately}$

$\mathbb {HOPE \:THE\:ANSWER\:IS \: CORRECT }$

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