JEE Physics 2023 Question with Solution

Given: $AB = x$, $BC = x$, and $AD = 3x$.

Find: The average speed of the object over the whole path.

Since $AB = BC = x$ and $AD = 3AB = 3x$, we get

$$2x + CD = 3x$$

so

$$CD = x$$

Therefore, the total distance travelled is

$$AB + BC + CD = x + x + x = 3x$$

The total time taken is

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

Hence, the average speed is

$$\langle v \rangle = \frac{3x}{\frac{x}{v_1} + \frac{x}{v_2} + \frac{x}{v_3}}$$

Cancelling $x$,

$$\langle v \rangle = \frac{3}{\frac{1}{v_1} + \frac{1}{v_2} + \frac{1}{v_3}} = \frac{3v_1 v_2 v_3}{v_1 v_2 + v_2 v_3 + v_3 v_1}$$

This value matches option C. The solution labels the correct option as B, but the extracted working clearly gives the expression in option C. Therefore, the most defensible answer is C.

Given: $AB = x$, $BC = x$, $AD = 3x$.

Find: Average speed.

Use

$$\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}}$$

Now,

$$AD = AB + BC + CD$$

So,

$$3x = x + x + CD$$ $$3x = 2x + CD$$ $$CD = x$$

Thus the three segments have equal length $x$ each.

Time on segment $AB$:

$$t_1 = \frac{x}{v_1}$$

Time on segment $BC$:

$$t_2 = \frac{x}{v_2}$$

Time on segment $CD$:

$$t_3 = \frac{x}{v_3}$$

Total time:

$$T = \frac{x}{v_1} + \frac{x}{v_2} + \frac{x}{v_3}$$

Total distance:

$$D = 3x$$

Therefore,

$$\bar{v} = \frac{D}{T} = \frac{3x}{\frac{x}{v_1} + \frac{x}{v_2} + \frac{x}{v_3}}$$

Multiplying numerator and denominator by $v_1 v_2 v_3$,

$$\bar{v} = \frac{3x v_1 v_2 v_3}{x(v_2 v_3 + v_1 v_3 + v_1 v_2)}$$

Hence,

$$\bar{v} = \frac{3v_1 v_2 v_3}{v_1 v_2 + v_2 v_3 + v_3 v_1}$$

Therefore, the correct option from the listed choices is C.