Question

body has speed V, 2V and 3V in first 1/3 of distance S, seconds 1/3 of S and third 1/3
of S respectively. average speed Its will be

Answer 1

Answer: 1.6 V  

Explanation:

Average speed = Total distance/Total time taken

First we will calculate the total distance covered = d/3+d/3+d/3 =3 (d/3)=d

Time taken = Distance/ speed

$t_1+t_2+t_3=\frac{d/3}{V}+\frac{d/3}{2V}+\frac{d/3}{3V}=\frac{d}{3V}(11/6)=\frac{11d}{18V}$

Average speed, $S_{avg} = \frac{d}{11d/18V}=\frac{18}{11}V=1.6 V$

Answer 2

Answer:

The average speed  is 1.64 v.

Explanation:

Given that,

A body covered distance of one third with the speed of V.

Next one third distance of S covered with the speed of 2 V.

Next one third distance of S covered with the speed of 3 V.

Let be the total distance 3s.

The time is equal to the distance divided by the speed.

The time taken for first case,

$t = \dfrac{d}{v}$

Put the value of distance and speed

$t=\dfrac{s}{3\times v}$....(I)

The time taken for second case,

$t=\dfrac{s}{3\times 2v}$.....(II)

The time taken for third case,

$t=\dfrac{s}{3\times 3v}$.....(III)

The average speed is equal to the total distance divided by the total time.

The average speed is defined as,

$v_{avg}=\dfrac{D}{T}$

Where, D = total distance

T = total time

Put the value of D and T

$v_{avg}=\dfrac{s}{\dfrac{s}{3v}+\dfrac{s}{6v}+\dfrac{s}{9v}}$

$v_{avg}=\dfrac{18v}{11}$

$v_{avg}=1.64v$

Hence, The average speed  is 1.64 v.