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:

Question

Earth has mass 8 times and radius 2 times that of a planet. If the escape velocity from the earth is 11.2 km/s, the escape velocity in km/s from the planet will be:

Answer 1

To find the average speed of the object moving along the path $AB \to BC \to CD$, we start by analyzing the given information:

The lengths $AB = BC = x$.
The total length $AD = 3x$.
Velocities along the segments are $v_1, v_2, v_3$ respectively.

The formula for average speed is given by:

\[\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}\]

Calculating Total Distance:

Total Distance $= AD = 3x$

Calculating Total Time:

Time to traverse $AB$: $t_1 = \frac{x}{v_1}$
Time to traverse $BC$: $t_2 = \frac{x}{v_2}$
Time to traverse $CD$: $t_3 = \frac{(AD - AB - BC)}{v_3} = \frac{x}{v_3}$
Total Time $= t_1 + t_2 + t_3 = \frac{x}{v_1} + \frac{x}{v_2} + \frac{x}{v_3}$

Substituting these values into the average speed formula:

\[ \text{Average Speed} = \frac{3x}{\frac{x}{v_1} + \frac{x}{v_2} + \frac{x}{v_3}} = \frac{3x}{x \left( \frac{1}{v_1} + \frac{1}{v_2} + \frac{1}{v_3} \right)} = \frac{3}{\frac{1}{v_1} + \frac{1}{v_2} + \frac{1}{v_3}} \]

Simplifying the expression further:

\[ = \frac{3}{\frac{v_2 v_3 + v_3 v_1 + v_1 v_2}{v_1 v_2 v_3}} = \frac{3 v_1 v_2 v_3}{v_2 v_3 + v_3 v_1 + v_1 v_2} \]

Thus, the average speed of the object is:

$\frac{3 v_1 v_2 v_3}{v_1 v_2 + v_2 v_3 + v_3 v_1}$

The correct answer is: $\frac{3 v_1 v_2 v_3}{\left(v_1 v_2+v_2 v_3+v_3 v_1\right)}$

Answer 2