Question

A car tavels 1/3 distance with20 km /hr next 1/3 with 20 km/hr and next 1/3 with 60 km/hr find avarage speed

Answer 1

▶ Solution :

Given :

A car travels first 1/3 distance with 20kmph, second 1/3 distance with 20kmph and last 1/3 distance with 60kmph.

To Find :

Average speed of the car.

Concept :

▪ Average speed is defined as ratio of total distance covered to the total time taken.

▪ Average speed is a scalar quantity.

▪ If body covers first 1/3 of total distance with speed X, second 1/3 of total distance with speed Y and final 1/3 of total distance with speed Z then, average speed of body is given by

$\boxed{\bf{\pink{V=\dfrac{3XYZ}{XY+YZ+ZX}}}}$

Calculation :

$\implies\sf\:V=\dfrac{3(20)(20)(60)}{(20\times 20)+(20\times 60)+(20\times 60)}\\ \\ \implies\sf\:V=\dfrac{72000}{400+1200+1200}\\ \\ \implies\sf\:V=\dfrac{72000}{2800}\\ \\ \implies\boxed{\bf{\orange{V=25.71\:kmph}}}$

Answer 2

Given:

Car travels ⅓ rd distance with 20 km/hr , next ⅓rd with 20 km/hr and final ⅓rd with 60 km/hr .

To find:

Average Speed of car

Calculation:

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

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

Cancelling the term d :

$= > avg. \: v = \dfrac{1}{ \bigg(\dfrac{ \frac{1}{3} }{20} + \dfrac{ \frac{1}{3} }{20} + \dfrac{ \frac{1}{3} }{60} \bigg)}$

$= > avg. \: v = \dfrac{1}{ \bigg(\dfrac{ 1 }{60} + \dfrac{ 1 }{60} + \dfrac{ 1 }{180} \bigg)}$

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

$= > avg. \: v = \dfrac{180}{7}$

$= > avg. \: v = 25.71 \: \: km {hr}^{ - 1}$

So final answer :

Average Velocity is 25.71 km/hr