JEE PYQ: Motion in a Straight Line - Question ID 4221ac2b6548 (JEE Main 2023)

Core Formula & Concept:

In problems involving mean speed (also called average speed), the key concept is:

• The mean speed over an entire journey is defined as the total distance traveled divided by the total time taken.
• Mathematically, this is expressed as: $\text{Mean speed} = \frac{\text{Total distance}}{\text{Total time}}$
• This is not the arithmetic mean of speeds unless the time intervals for each speed are equal.

In this problem, the journey is split into two parts:

1. First half of the distance is covered at a constant speed of $5 \mathrm{~m/s}$.
2. The second half of the distance is covered in two sub-parts: for half the time at $10 \mathrm{~m/s}$, and the other half of the time at $15 \mathrm{~m/s}$.

We must carefully compute the time taken for each segment and then use the definition of mean speed to find the value of $x$.

--- Step-by-Step Derivation:

Let the total distance be $D$ meters.

1. First half of the distance:

   Distance covered = $\frac{D}{2}$ m
   Speed = $5 \mathrm{~m/s}$
   Time taken, $t_1 = \frac{\text{Distance}}{\text{Speed}} = \frac{D/2}{5} = \frac{D}{10}$ seconds.

2. Second half of the distance:

   Distance to cover = $\frac{D}{2}$ m
   This part is traveled in two time intervals, each of duration $t_2/2$, where $t_2$ is the total time taken for the second half.
   Let $t_2$ be the total time for the second half. Then:

   • For the first half of $t_2$: speed = $10 \mathrm{~m/s}$
   • For the second half of $t_2$: speed = $15 \mathrm{~m/s}$
   The total distance covered in the second half is: $\text{Distance} = \left(10 \cdot \frac{t_2}{2}\right) + \left(15 \cdot \frac{t_2}{2}\right) = 5t_2 + 7.5t_2 = 12.5 t_2$ But this distance must equal $\frac{D}{2}$: $12.5 t_2 = \frac{D}{2} \implies t_2 = \frac{D}{25} \text{ seconds}$

3. Total time for the entire journey: $T = t_1 + t_2 = \frac{D}{10} + \frac{D}{25} = \frac{5D + 2D}{50} = \frac{7D}{50} \text{ seconds}$
4. Mean speed: $\text{Mean speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{D}{T} = \frac{D}{\frac{7D}{50}} = \frac{50}{7} \mathrm{~m/s}$ This matches the given form $\frac{x}{7} \mathrm{~m/s}$. Hence: $\frac{x}{7} = \frac{50}{7} \implies x = 50$

--- Common Traps & Exam Tip:

Students often make the following mistakes:

• Confusing distance and time splits: The first half is a distance split, while the second half is a time split. Mixing these leads to incorrect time calculations.
• Assuming arithmetic mean: Some students incorrectly average the speeds $5$, $10$, and $15$ without considering the actual time or distance proportions.
• Algebraic errors in time calculation: Misapplying the relation between distance, speed, and time in the second half can lead to wrong values for $t_2$.

Exam Tip: Always define the total distance as a variable (e.g., $D$) and express all times in terms of $D$. This ensures consistency and avoids confusion between distance and time splits.