Question

28. If a car cover 2/5th of total distance with v, speed
and 3/5th distance with v, speed then the average
speed is :-
(1)
V, V2
V, +V₂
(2)
2
2
5v, V2
(3)
2v, + V2
v + V₂
(4) 3v, +2v

Anyone please help me.......
step by step explanation should needed


spam answers will be reported....​

Answer 1

Answer:

$\boxed{ \mathfrak{Average \: speed \ (v_{avg})= \frac{5v_1v_2}{3v_1 + 2v_2} }}$

Explanation:

Let the total distance be 'd'

Speed of car for covering $\sf {\dfrac{2}{5}}^{th}$ of total distance is given as $\sf v_1$

Let the time taken by car to travel $\sf {\dfrac{2}{5}}^{th}$ of total distance be $\rm t_{1}$

As, we know

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

So,

$\rm \implies v_1 = \dfrac{2d}{5t_{1}} \\ \\ \rm \implies t_{1} = \dfrac{2d}{5 v_1}$

Speed of car for covering $\sf {\dfrac{3}{5}}^{th}$ of total distance is given as $\sf v_2$

Let the time taken by car to travel $\sf {\dfrac{3}{5}}^{th}$ of total distance be $\rm t_{2}$

So,

$\rm \implies v_2 = \frac{3d}{ 5t_{2}} \\ \\ \rm \implies t_{2} = \frac{3d}{5 v_2}$

Total time for covering total distance i.e. 'd' (t) = $\sf t_{1} + t_{2}$

Formula of average speed:

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

By substituting values in the formula we get:

$\rm \implies v_{avg}= \dfrac{d}{t} \\ \\ \rm \implies v_{avg} = \dfrac{d}{t_{1} + t_{2}} \\ \\ \rm \implies v_{avg} = \dfrac{d}{ \dfrac{2d}{5v_1} + \dfrac{3d}{5v_2} } \\ \\ \rm \implies v_{avg}= \dfrac{d}{ \dfrac{2dv_2 + 3dv_1}{5v_1 v_2} } \\ \\ \rm \implies v_{avg}= \dfrac{ \cancel{d} }{\dfrac{ \cancel{d}(2v_2 + 3v_1)}{5v_1 v_2}} \\ \\ \rm \implies v_{avg} = \dfrac{1}{\dfrac{2v_2 + 3v_1}{5v_1 v_2}} \\ \\ \rm \implies v_{avg} = \frac{5v_1v_2}{3v_1 + 2v_2}$

Answer 2

Answer:

(5v1v2)/(2v2 + 3v1)

Explanation:

Answering this Question in the typing way is a bit difficult and confusing,

So please refer the above image,

Hope it helped and you understood it........All the best