A body covers first 1 3 of the distance


Asked by admin @ in Physics viewed by 315 People


a body covers first 1/3 of distance with a velocity of 20m/s.Second 1/3 with 30m/s and the last one with 40m/s.The average velocity is

Answered by admin @



ANSWER:

  • The average velocity = 27.69 m/s.

GIVEN:

  • A body covers first 1/3 of distance with a velocity of 20m/s.

  • Second 1/3 of distance with a velocity of 30m/s.

  • The last 1/3 of distance with a velocity of 40m/s.

TO FIND:

  • The average velocity.

EXPLANATION:

\boxed{\bold{\gray{Avg\ velocity=\dfrac{ Total\ distance }{Total \ time}}}}

Let the total distance be 3s.

 \sf Total \ time = t_1 + t_2 + t_3

\boxed{\bold{\gray{Time = \dfrac{Distance}{Velocity}}}}

\sf t_1 = \dfrac{s}{20} \ seconds

\sf t_2 = \dfrac{s}{30} \ seconds

\sf t_3 = \dfrac{s}{40} \ seconds

\sf Time = \dfrac{s}{20} + \dfrac{s}{30} + \dfrac{s}{40} \ seconds

\sf Avg\ velocity=\dfrac{ 3s }{\dfrac{s}{20} + \dfrac{s}{30} + \dfrac{s}{40}}

\sf Avg\ velocity=\dfrac{ 3s }{s \left(\dfrac{1}{20} + \dfrac{1}{30} + \dfrac{1}{40} \right)}

\sf Avg\ velocity=\dfrac{ 3}{ \left(\dfrac{1}{20} + \dfrac{1}{30} + \dfrac{1}{40} \right)}

\sf Avg\ velocity=\dfrac{ 3}{ \dfrac{6}{120} + \dfrac{4}{120} + \dfrac{3}{120} }

\sf Avg\ velocity=\dfrac{ 3}{ \dfrac{6 + 4 + 3}{120} }

\sf Avg\ velocity=\dfrac{ 3 \times 120}{13 }

\sf Avg\ velocity=\dfrac{360}{13 }

\sf Avg\ velocity=27.69 \ m {s}^{ - 1}

Hence the average velocity = 27.69 m/s.


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