Question

A body moves in a straight line with velocity v1 for 1/4th distance and remaining distance with
velocity V2. Find the average speed.​

Answer 1

Let the total distance be x.

Then,

❖ For first part of the journey:

▪Displacement = (1/4)x = $\sf \dfrac{x}{4}$

▪And, velocity = $\sf v_1$

▪We know that, time = displacement/velocity

$\sf \longrightarrow t_{1} = \Bigg \{ \dfrac{ \: \: \Big( \dfrac{x}{4} \Big) \: \: }{ v_{1} } \Bigg \}$

$\sf \longrightarrow t_{1} = \dfrac{x }{4v_{1}}$

❖ For second part of the journey:-

▪Remaining displacement = (Total distance) - (Displacement in first part) = x - (x/4)

Remaining displacement = $\sf \dfrac{3x}{4}$

▪Velocity = $\sf v_2$

▪We know that, time = displacement/velocity

$\sf \longrightarrow t_{2} = \Bigg \{ \dfrac{ \: \: \Big( \dfrac{3x}{4} \Big) \: \: }{ v_{2} } \Bigg \}$

$\sf \longrightarrow t_{ 2 } = \dfrac{3x }{4v_{2}}$

❖ Finding average speed:

We know that, average velocity = (Total displacement)/(Total time)

$\sf average \: velocity = \dfrac{x}{t_1 + t_2}$

$\sf \ average \: velocity= \Bigg \{\dfrac{x}{ \bigg( \dfrac{x }{ 4 v_{1} } + \dfrac{3x }{4v_{1} } \bigg) } \Bigg \}$

$\sf \longrightarrow average \: velocity= \Bigg \{\dfrac{ \cancel{x}}{ \cancel{x} \bigg( \dfrac{ 1 }{ 4v_{1} } + \dfrac{3 }{4v_{2}} \bigg) } \Bigg \}$

$\sf \longrightarrow average \: velocity= \Bigg \{\dfrac{ 1}{ \bigg( \dfrac{ 1 }{ 4v_{1} } + \dfrac{3 }{4v_{2}} \bigg) } \Bigg \}$

$\sf \longrightarrow average \: velocity= \Bigg \{\dfrac{ 1}{ \bigg( \dfrac{ v_{2} + 3 v_{1} }{ 4 v_1 v_2 } \bigg) } \Bigg \}$

$\sf \longrightarrow average \: velocity = \dfrac{4 v_1 v_2}{ 3v_{1} + v_{2} }$

Hence the answer is,

$\longrightarrow \underline{ \underline{ \sf{ \green{ \dfrac{4 v_1 v_2}{ 3v_{1} + v_{2} } }}}}$