Kinematics

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Question 22 of 308

An object moves with speed v1,v2andv3{v}_{1},{v}_{2}\text{and}{v}_{3} along a line segment AB,BCandCDAB,BC\text{and}CD respectively as shown in figure. Where AB=BCandAD=3ABAB=BC\text{and}AD=3AB, then average speed of the object will be :

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Answer: (C) 3v1v2v3v1v2+v2v3+v3v1\dfrac{3{v}_{1}{v}_{2}{v}_{3}}{{v}_{1}{v}_{2}+{v}_{2}{v}_{3}+{v}_{3}{v}_{1}}

Speed of an object is defined as the distance travelled by the object per unit time.
Let's assume that, AB=dAB=d. So, BC=d,AD=3dBC=d,AD=3d. Also, assume that the time required by the object to travel AB,BC,CDAB,BC,CD are t1,t2,t3{t}_{1},{t}_{2},{t}_{3} respectively.
Now,
AD=3ABAB+BC+CD=3ABAB+AB+CD=3ABCD=AB\begin{matrix}AD & = & 3AB \\ AB+BC+CD & = & 3AB \\ AB+AB+CD & = & 3AB \\ CD & = & AB\end{matrix}
Hence,
t1=ABv1=dv1t2=dv2t3=dv3\begin{matrix}{t}_{1} & = & \dfrac{AB}{{v}_{1}} \\ & = & \dfrac{d}{{v}_{1}} \\ {t}_{2} & = & \dfrac{d}{{v}_{2}} \\ {t}_{3} & = & \dfrac{d}{{v}_{3}}\end{matrix}
The formula to calculate the average speed (va)\left({v}_{a}\right) of the object is given by
va=ADt1+t2+t3........................(1){v}_{a}=\dfrac{AD}{{t}_{1}+{t}_{2}+{t}_{3}}........................(1)
Substitute the expressions for the parameters into equation (1) and simplify to obtain the required average speed of the object.
va=3ddv1+dv2+dv3=3v1v2v3v1v2+v2v3+v3v1\begin{matrix}{v}_{a} & = & \dfrac{3d}{\dfrac{d}{{v}_{1}}+\dfrac{d}{{v}_{2}}+\dfrac{d}{{v}_{3}}} \\ & = & \dfrac{3{v}_{1}{v}_{2}{v}_{3}}{{v}_{1}{v}_{2}+{v}_{2}{v}_{3}+{v}_{3}{v}_{1}}\end{matrix}

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