Question

please explain fast please explain fast i need it rn​

Answer 1

Answer:

option (4)

Formulae :

• Speed = distance/time
• Average speed = total distance/total time taken

Explanation:

Let the total distance be 'x'

The car covers 2/5 th of the total distance with speed v₁

Let the time taken be t₁

$\tt v_1 = \dfrac{\dfrac{2x}{5}}{t_1} \\ \tt v_1 = \dfrac{2x}{5t_1} \\ \longrightarrow \tt t_1 = \dfrac{2x}{5v_1}$

The car covers 3/5 th of the total distance with speed v₂

Let the time taken be t₂

$\tt v_2 = \dfrac{\dfrac{3x}{5}}{t_2} \\ \tt v_2 = \dfrac{3x}{5t_2} \\ \longrightarrow \tt t_2 = \dfrac{3x}{5v_2}$

Average speed :

total distance covered = x

total time taken = t₁ + t₂

$\tt = \dfrac{2x}{5v_1} + \dfrac{3x}{5v_2} \\ \tt = \dfrac{x}{5} \bigg ( \dfrac{2}{v_1} + \dfrac{3}{v_2} \bigg ) \\ \tt = \dfrac{x}{5} \bigg ( \dfrac{2v_2 + 3v_1}{v_1 v_2} \bigg )$

Average speed = total distance covered/total time taken

$\tt = \dfrac{x}{t_1+t_2} \\ \tt = \dfrac{x}{\dfrac{x}{5} \bigg(\dfrac{2v_2 +3v_1}{v_1 v_2} \bigg)} \\ \tt = \dfrac{1}{\dfrac{1}{5} \bigg( \dfrac{2v_2 +3v_1}{v_1 v_2} \bigg)} \\ \tt = \dfrac{5}{\bigg( \dfrac{2v_2 +3v_1}{v_1 v_2} \bigg)} \\ \tt = \dfrac{5v_1 v_2}{2v_2 + 3v_1}$