JEE PYQ: Motion in a Straight Line - Question ID 16c11659f506 (JEE Main 2023)

Core Formula & Concept:

When an object moves vertically under constant acceleration (gravity), its velocity at any time $t$ is given by the kinematic equation:

$v(t) = u + a t$

Here:

• $v(t)$ = velocity at time $t$,
• $u$ = initial velocity (positive upward),
• $a$ = acceleration (negative for upward motion, $a = -g$),
• $g = 10 \, \mathrm{m/s^2}$ (given).

Since the ball is thrown upward, the acceleration due to gravity acts downward, so $a = -g$.

Step-by-Step Derivation:

Step 1: Write the velocity equation
Given initial velocity $u = 150 \, \mathrm{m/s}$, and $g = 10 \, \mathrm{m/s^2}$, the velocity at time $t$ is: $v(t) = 150 - 10 t$

Step 2: Compute velocity at $t = 3 \, \mathrm{s}$ and $t = 5 \, \mathrm{s}$
At $t = 3 \, \mathrm{s}$: $v(3) = 150 - 10 \times 3 = 150 - 30 = 120 \, \mathrm{m/s}$ At $t = 5 \, \mathrm{s}$: $v(5) = 150 - 10 \times 5 = 150 - 50 = 100 \, \mathrm{m/s}$

Step 3: Form the ratio and set equal to $\frac{x+1}{x}$
The ratio of velocities is: $\frac{v(3)}{v(5)} = \frac{120}{100} = \frac{6}{5}$ We are told this ratio equals $\frac{x+1}{x}$. So: $\frac{x+1}{x} = \frac{6}{5}$

Step 4: Solve for $x$
Cross-multiply: $5(x + 1) = 6x$ $5x + 5 = 6x$ $5 = 6x - 5x$ $5 = x$

Step 5: Match with given options
The value of $x$ is $5$, which corresponds to option C.

Common Traps & Exam Tip:

Students often confuse the sign of acceleration. Since the ball is moving upward, gravity acts downward, so acceleration is $-g$. Using $+g$ would give incorrect velocities and lead to a wrong ratio. Always assign signs based on the chosen coordinate system (upward positive).

Another common mistake is misinterpreting the ratio. Ensure the order of velocities matches the given ratio: $\frac{v(3)}{v(5)}$, not the reverse.