An object moves with speed v 1 , v 2 and v 3 along a line segment A B , B C and C D respectively as shown in figure. Where A B = B C and A D = 3 A B , then average speed of the object will be :

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TestHub · Accepted Answer

Speed of an object is defined as the distance travelled by the object per unit time.

Let's assume that, $A B = d$ . So, $B C = d, A D = 3 d$ . Also, assume that the time required by the object to travel $A B, B C, C D$ are $t_{1}, t_{2}, t_{3}$ respectively.

Now,

$\begin{array}{rcl} A D & = & 3 A B \\ A B + B C + C D & = & 3 A B \\ A B + A B + C D & = & 3 A B \\ C D & = & A B \end{array}$

Hence,

$\begin{array}{rcl} t_{1} & = & \frac{A B}{v_{1}} \\ = & \frac{d}{v_{1}} \\ t_{2} & = & \frac{d}{v_{2}} \\ t_{3} & = & \frac{d}{v_{3}} \end{array}$

The formula to calculate the average speed $(v_{a})$ of the object is given by

$v_{a} = \frac{A D}{t_{1} + t_{2} + t_{3}} . . . . . . . . . . . . . . . . . . . . . . . . (1)$

Substitute the expressions for the parameters into equation (1) and simplify to obtain the required average speed of the object.

$\begin{array}{rcl} v_{a} & = & \frac{3 d}{\frac{d}{v_{1}} + \frac{d}{v_{2}} + \frac{d}{v_{3}}} \\ = & \frac{3 v_{1} v_{2} v_{3}}{v_{1} v_{2} + v_{2} v_{3} + v_{3} v_{1}} \end{array}$

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