Question:

An object moves with speed $v_1, v_2$ and $v_3$ along a line segment $AB , BC$ and $C D$ respectively as shown in figure Where $AB = BC$ and $AD =3 AB$, then average speed of the object will be:

Updated On: Mar 19, 2025
  • $\frac{\left(v_1+v_2+v_3\right)}{3}$
  • $\frac{3 v_1 v_2 v_3}{\left(v_1 v_2+v_2 v_3+v_3 v_1\right)}$
  • $\frac{\left(v_1+v_2+v_3\right)}{3 v_1 v_2 v_3}$
  • $\frac{v_1 v_2 v_3}{3\left(v_1 v_2+v_2 v_3+v_3 v_1\right)}$
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The Correct Option is B

Approach Solution - 1

\(AB=x\)
\(BC=x\)
\(2x+CD=3x\)
\(CD=x\)
\(⟨v⟩= \frac{3x}{\frac{x}{v_1}+\frac{x}{v_2}+\frac{x}{v_3}}=\frac{3 v_1 v_2 v_3}{\left(v_1 v_2+v_2 v_3+v_1 v_3\right)}\)
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Approach Solution -2

Given: \( AB = x \), \( BC = x \), \( AD = 3x \) \[ \text{Total Distance} = 3x \] \[ \text{Total Time} = \frac{x}{v_1} + \frac{x}{v_2} + \frac{x}{v_3} \] \[ \text{Average Speed} = \frac{3x}{\frac{x}{v_1} + \frac{x}{v_2} + \frac{x}{v_3}} = \frac{3v_1 v_2 v_3}{v_1 v_2 + v_2 v_3 + v_3 v_1} \] 

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Concepts Used:

Motion in a Plane

It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions. 

Equations of Plane Motion

The equations of motion in a straight line are:

v=u+at

s=ut+½ at2

v2-u2=2as

Where,

  • v = final velocity of the particle
  • u = initial velocity of the particle
  • s = displacement of the particle
  • a = acceleration of the particle
  • t = the time interval in which the particle is in consideration