Question

06. A car travels one-third distance on a straight road with a velocity of
10km/hr, next one-third with velocity 20km/hr and the last with a velocity
60km/hr. What is the average velocity of the car in the whole journey?​

Answer 1

Average velocity when A body traverses 1/3rd of distance with u1 velocity next 1/3rd of distance with u2 velocity and last 1/3rd distance with u3 velocity is

U.avg.=3u1u2u3/u1u2 +u2u3 +u3u1

u1=10kmph,u2=20kmph,u3=60kmph

U.avg=3(10)(20)(60)/ 10×20 +20×60 +60×10

=36000/200+1200+600

=36000/2000=18kmph

1kmph=5/18m/s

18kmph=5m/s

Answer 2

Answer:

The average velocity of the car in whole journey is 18 km/h.

Explanation:

Given :

$\sf Velocity\;v_{1} = \bf 10\ km/hr\\ \\ \sf Velocity\;v_{2} =\bf20\ km/hr\\ \\ \sf Velocity\;v_{3} = \bf60\ km/hr$

Consider the :

• Total distance as - 3x.
• Total time as - T

We know that :

$\bigstar\;\boxed{\textsf{Velocity} = \bf \dfrac{Distance}{Time}}$

Now,

• The Time taken for the first journey.

$\sf\\ \\\implies t_{1}= \dfrac{3x}{3\times10}\\ \\ \\ \implies t_{1}= \dfrac{x}{10}$

$\rule{300}{1.5}$

• The time for second journey.

$\sf\\ \\\implies t_{2}= \dfrac{3x}{3\times20}\\ \\ \\\implies t_{2}= \dfrac{x}{20}$

$\rule{300}{1.5}$

• The time for third journey.

$\sf\\ \\\implies t_{3}= \dfrac{3x}{3\times60}\\ \\ \\\implies t_{3}= \dfrac{x}{60}$

$\rule{300}{1.5}$

Now, We know that :

$\bigstar\;\boxed{\textsf{Average Velocity} = \sf \dfrac{Total\;Distance}{Total\;Time}}$

• Putting the values,

$\sf\\ \\\implies v = \dfrac{3x}{\dfrac{x}{10}+\dfrac{x}{20}+\dfrac{x}{60}}\\ \\ \\ \implies v = \dfrac{3x\times60}{10x}\\ \\ \\ \implies \bf v = 18\ km/hr$

∴ The average velocity of the car in whole journey is 18 km/h.