A car is moving on a straight road covers one third of the distance wi

1.	A car is moving on a straight road covers one third of the distance with a speed of 20 km/h and the resa speed of 60 km/h. The average speed of the car is(1) 40 km/h(2) 50 km/h(3) 36 km/h(4) 55 km/h
Answer» Answer: Given: ⅓rd distance travelled by 20 km/hr and the rest distance travelled by 60 km/hr To find: Average speed of the BODY Concept: Average speed is DEFINED as the RATIO of Total distance to the total time taken. Since distance is involved , average speed can never be ZERO, it's ALWAYS positive. Calculation: Let the total distance travelled be s , Then distance travelled with 20 km/hr is s/3 , the rest 2s/3 is travelled with 60 km/hr. $v \: avg. = \dfrac{total \: distance}{total \: time}$ $= > v \: avg. = \dfrac{s}{( \dfrac{ \frac{s}{3} }{20}) + (\dfrac{ \frac{2s}{3} }{60}) }$ $= > v \: avg. = \dfrac{1}{ \dfrac{1}{60} + \dfrac{1}{90} }$ $= > v \: avg. = \dfrac{1}{ \dfrac{(3 + 2)}{180} }$ $= > v \: avg. = \dfrac{180}{5}$ $= > v \: avg. = 36 \: km \: {hr}^{ - 1}$ So final answer : $\boxed{ \red{ \large{ \sf{ \bold{ v \: avg. = 36 \: km \: {hr}^{ - 1}}}}}}$

A car is moving on a straight road covers one third of the distance with a speed of 20 km/h and the resa speed of 60 km/h. The average speed of the car is(1) 40 km/h(2) 50 km/h(3) 36 km/h(4) 55 km/h

Answer»

⅓rd distance travelled by 20 km/hr and the rest distance travelled by 60 km/hr

Average speed of the BODY

Average speed is DEFINED as the RATIO of Total distance to the total time taken.

Since distance is involved , average speed can never be ZERO, it's ALWAYS positive.

Let the total distance travelled be s , Then distance travelled with 20 km/hr is s/3 , the rest 2s/3 is travelled with

60 km/hr.

$v \: avg. = \dfrac{total \: distance}{total \: time}$

$= > v \: avg. = \dfrac{s}{( \dfrac{ \frac{s}{3} }{20}) + (\dfrac{ \frac{2s}{3} }{60}) }$

$= > v \: avg. = \dfrac{1}{ \dfrac{1}{60} + \dfrac{1}{90} }$

$= > v \: avg. = \dfrac{1}{ \dfrac{(3 + 2)}{180} }$

$= > v \: avg. = \dfrac{180}{5}$

$= > v \: avg. = 36 \: km \: {hr}^{ - 1}$

So final answer :

$\boxed{ \red{ \large{ \sf{ \bold{ v \: avg. = 36 \: km \: {hr}^{ - 1}}}}}}$