Question

A car mover from x and y with a constant speed of Va and returned from y to x
with a constant speed Vb. Calculate the average speed for the complete travel.​

Answer 1

Answer:

$\boxed{ \mathfrak{ Average \: speed \ of \ car \ (v_{avg}) = \dfrac{2V_aV_b}{V_a + V_b} }}$

Explanation:

Let distance between x to y be 'd'

So, Total distance covered by car = 2d

Speed of the car from x to y is given as $\rm V_a$

Let the time taken to travel from x to y be $\rm t_{1}$

As, we know

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

$\therefore$

$\rm \implies V_a = \frac{d}{t_{1}} \\ \\ \sf \implies t_{1} = \frac{d}{V_a}$

Speed of the car from y to x is given as $\rm V_b$

Let the time taken to travel from y to x be $\sf t_{2}$

$\therefore$

$\rm \implies V_b = \frac{d}{ t_{2}} \\ \\ \sf \implies t_{2} = \frac{d}{V_b}$

Total time for covering 2d distance (t) = $\sf t_{1} + t_{2}$

$\boxed{ \bold{Average \: speed \ (v_{avg})= \frac{Total \: distance \: travelled}{Total \: time \: taken} }}$

$\rm \implies v_{avg} = \dfrac{2d}{t} \\ \\ \rm \implies v_{avg} = \dfrac{2d}{t_{1} + t_{2}} \\ \\ \rm \implies v_{avg} = \dfrac{2d}{ \dfrac{d}{V_a} + \dfrac{d}{V_b} } \\ \\ \rm \implies v_{avg} = \dfrac{2d}{ \dfrac{dV_b}{V_aV_b} + \dfrac{dV_a}{V_aV_b} } \\ \\ \rm \implies v_{avg} = \dfrac{2 \cancel{d}}{ \dfrac{ \cancel{d}(V_a + V_b)}{V_aV_b} } \\ \\ \rm \implies v_{avg} = \dfrac{2}{ \dfrac{V_a + V_b}{V_aV_b} } \\ \\ \rm \implies v_{avg} = \dfrac{2V_aV_b}{V_a + V_b}$

Answer 2

Answer:

⭐Actually welcome to the concept of the Relative velocity and kinematics

⭐Basically here we can see that, there is similar change in both the ends of the rods,

⭐Thus the time taken travel the distance by both the ends will be same,

⭐so we get as

❄ta = tb

⭐so from this we get as