Question

a particle travels 1/3 of its distance with the velocity of 30m/s,next 1/3 distance with the velocity of 40m/s and the remaining distance with the velocity of 50m/s calcutale the averaga speed of the particle​

Answer 1

Let the total distance be $\sf{x.}$

The first $\sf{\dfrac{x}{3}}$ distance is travelled with the velocity of $\sf{30\ m\,s^{-1}.}$ The time taken for this,

$\sf{\longrightarrow t_1=\dfrac{\left(\frac{x}{3}\right)}{30}}$

$\sf{\longrightarrow t_1=\dfrac{x}{90}}$

The next $\sf{\dfrac{x}{3}}$ distance is travelled with the velocity of $\sf{40\ m\,s^{-1}.}$ The time taken for this,

$\sf{\longrightarrow t_2=\dfrac{\left(\frac{x}{3}\right)}{40}}$

$\sf{\longrightarrow t_2=\dfrac{x}{120}}$

The final $\sf{\dfrac{x}{3}}$ distance is travelled with the velocity of $\sf{50\ m\,s^{-1}.}$ The time taken for this,

$\sf{\longrightarrow t_3=\dfrac{\left(\frac{x}{3}\right)}{50}}$

$\sf{\longrightarrow t_3=\dfrac{x}{150}}$

Hence average velocity,

$\sf{\longrightarrow v=\dfrac{x}{t_1+t_2+t_3}}$

$\sf{\longrightarrow v=\dfrac{x}{\dfrac{x}{90}+\dfrac{x}{120}+\dfrac{x}{150}}}$

$\sf{\longrightarrow v=\dfrac{90\times120\times150}{90\times120+90\times150+120\times150}}$

$\sf{\longrightarrow\underline{\underline{v=38.3\ m\,s^{-1}}}}$