Question

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

Answer 1

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.