Question

A man travels one third length of a straight road with a constant speed
of 40 m/sec. and the remaining portion of the road with a constant
speed of 60 m/sec., then calculate its average speed.​

Answer 1

Answer:

36/7 m per second

..................

Answer 2

Answer:

$\boxed{ \mathfrak{Average \: speed \ (v_{avg})= 51.4 \: m/s}}$

Explanation:

Let the total length of straight road be 'd'

Speed of man for first one third distance is given as 40 m/s

Let the time taken by man to travel one third of length be $\rm t_{1}$

As, we know

$\rm speed = \dfrac{distance}{time}$

So,

$\rm \implies 40 = \frac{d}{3t_{1}} \\ \\ \rm \implies t_{1} = \frac{d}{3 \times 40} \\ \\ \rm \implies t_{1} = \frac{d}{120}$

Speed of man for the remaining distance i.e. two third length of road is given as 60 m/s

Let the time taken by man to travel two third of length be $\rm t_{2}$

So,

$\rm \implies 60 = \frac{2d}{ 3t_{2}} \\ \\ \rm \implies t_{2} = \frac{2d}{3 \times 60} \\ \\ \rm \implies t_{2} = \frac{2d}{180} \\ \\ \rm \implies t_{2} = \frac{d}{90}$

Total time for covering covering complete length of road i.e. 'd' (t) = $\sf t_{1} + t_{2}$

$\rm \implies Average \: speed \ (v_{avg})= \frac{Total \: distance}{Total \: time} \\ \\ \rm \implies v_{avg}= \dfrac{d}{t} \\ \\ \rm \implies v_{avg} = \dfrac{d}{t_{1} + t_{2}} \\ \\ \rm \implies v_{avg} = \dfrac{d}{ \dfrac{d}{120} + \dfrac{d}{90} } \\ \\ \rm \implies v_{avg}= \dfrac{d}{ \dfrac{3d}{360} + \dfrac{4d}{360} } \\ \\ \rm \implies v_{avg}= \dfrac{d}{ \dfrac{3d + 4d}{360} } \\ \\ \rm \implies v_{avg}= \dfrac{d}{ \dfrac{7d}{360} } \\ \\ \rm \implies v_{avg}= \frac{ \cancel{d} \times 360}{7 \cancel{d}} \\ \\ \rm \implies v_{avg} = \frac{360}{7} \\ \\ \rm \implies v_{avg} = 51.4 \: m {s}^{ - 1}$