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