Question

Leela travels from x to y on straight line path with a speed of 20 m/s. On the return journey from y to x on the same path, she travels with 30 m/s. The average speed of the journey is-​

Answer 1

Answer:

24 m/s

Explanation:

As per the provided information in the given question, we have :

• Speed taken by Leela to cover the distance from X to Y = 20 m/s
• Speed taken by Leela to cover the distance from Y to X = 30 m/s

We are asked to calculate the average speed.

$\bigstar \; \boxed {\bf {Speed_{(avg)} = \dfrac{Total \; distance}{Total \; time} } }$

Finding total distance :

Let us suppose the distance from X to Y be d m. Therefore,

$\longmapsto \rm { Distance_{(Total)} = Distance_{(XY)} + Distance_{(YZ)} }$

$\longmapsto \rm { Distance_{(Total)} = (d + d) \; m }$

$\longmapsto \bf { Distance_{(Total)} = 2d \; m }$

∴ Total distance covered is 2d metres.

Finding total time :

$\longmapsto \rm { Time_{(Total)} = Time_{(XY)} + Time_{(YZ)} }$

$\longmapsto \rm { Time_{(Total)} = \dfrac{Distance_{(XY)}}{Speed_{(XY)}} + \dfrac{Distance_{(YX)}}{Speed_{(YX)}} }$

$\longmapsto \rm { Time_{(Total)} =\Bigg ( \dfrac{d}{20} + \dfrac{d}{30} \Bigg ) \; s}$

$\longmapsto \rm { Time_{(Total)} =\Bigg ( \dfrac{3d + 2d}{60} \Bigg ) \; s}$

$\longmapsto \rm { Time_{(Total)} =\Bigg ( \dfrac{5d}{60} \Bigg ) \; s}$

$\longmapsto \bf { Time_{(Total)} = \dfrac{1d}{12} \; s}$

$\rule{200}2$

Now, by using the formula of average speed,

$\bigstar \; \boxed {\bf {Speed_{(avg)} = \dfrac{Total \; distance}{Total \; time} } }$

$\longmapsto \rm {Speed_{(avg)} =\left ( \dfrac{2d}{\cfrac{1d}{12} }\right ) \; ms^{-1}}$

$\longmapsto \rm {Speed_{(avg)} = \Bigg ( 2d \times \dfrac{12}{d} \Bigg) \; ms^{-1} }$

$\longmapsto \rm {Speed_{(avg)} = (2 \times 12)\; ms^{-1} }$

$\longmapsto \bf {Speed_{(avg)} = 24 \; ms^{-1} }$

∴ Average speed of the journey is 24 m/s.