Question

A car moves a car covers distance S1 with velocity V1 and distance S2 with velocity V2 between two cities P and Q its average velocity will be?​

Answer 1

Given :

A car covers distance s1 with velocity v1 and distance s2 with velocity v2 between two cities P and Q.

To find :

The average velocity.

Solution :

Instantaneous speed of an object is equal to distance travelled per unit time by that object.

So,

$\bf \: v_i = \frac{s}{t}$

(equation 1)

where

s = Distance travelled and

t = Time taken.

For distance s1,

velocity of car = v1,

So, By equation 1,

$\\ \bf \: time \: \: t = \frac{s}{v_i}$

(equation 2)

So, for covering distance s1, time taken:

$\\ \bf \: t_1 = \frac {s_1}{ v_1}$

For covering distance s2, time taken:

$\\ \bf \: t_2 = \frac {s_2}{ v_2}$

Average velocity of an object is equal to the ratio of total displacement travelled and total time taken.

As,

Total distance travelled in this journey = s1 + s2

Total time taken in this journey = t1 + t2,

So,

Putting the values of time t1 and t2,

we get

Average velocity :

$\bf=\frac{ s_1 + s_2 }{ t_1+ t_2} \\ \\ </strong></p><p><strong>[tex] \bf=\frac{ s_1 + s_2 }{ t_1+ t_2} \\ \\ \textbf{average velocity} = \bf\frac{s_1 + s_2}{(\frac{s_1}{v_1} + \frac{s_2}{v_2})}$