JEE PYQ: Motion in a Straight Line - Question ID a9bd9ee36820 (JEE Main 2025)

ID: a9bd9ee36820JEE Main 2025Numerical Value

A person travelling on a straight line moves with a uniform velocity v1v_1 for a distance xx and with a uniform velocity v2v_2 for the next 32x\frac{3}{2} x distance. The average velocity in this motion is 507 m/s\frac{50}{7} \mathrm{~m} / \mathrm{s}. If v1v_1 is 5 m/s5 \mathrm{~m} / \mathrm{s} then v2=v_2= __________ m/s\mathrm{m} / \mathrm{s}.

Your Answer

Step-by-step Explanation

Core Formula & Concept:

In kinematics, the average velocity over a journey is defined as the total displacement divided by the total time taken. Mathematically, Average velocity=Total displacementTotal time.\text{Average velocity} = \frac{\text{Total displacement}}{\text{Total time}}. When an object moves in segments with different constant velocities, we compute the time taken for each segment separately and then sum them to find the total time.

Step-by-Step Derivation:

Given data:
- First segment: distance xx, velocity v1=5 m/sv_1 = 5\ \text{m/s}.
- Second segment: distance 32x\frac{3}{2}x, velocity v2v_2 (to be found).
- Average velocity over the entire journey: 507 m/s\frac{50}{7}\ \text{m/s}.

Step 1: Express the time for each segment.
Time for the first segment: t1=xv1=x5.t_1 = \frac{x}{v_1} = \frac{x}{5}. Time for the second segment: t2=32xv2=3x2v2.t_2 = \frac{\frac{3}{2}x}{v_2} = \frac{3x}{2v_2}.

Step 2: Compute the total displacement and total time.
Total displacement: Δs=x+32x=52x.\Delta s = x + \frac{3}{2}x = \frac{5}{2}x. Total time: T=t1+t2=x5+3x2v2.T = t_1 + t_2 = \frac{x}{5} + \frac{3x}{2v_2}.

Step 3: Write the average velocity formula and substitute.
Average velocity: 507=ΔsT=52xx5+3x2v2.\frac{50}{7} = \frac{\Delta s}{T} = \frac{\frac{5}{2}x}{\frac{x}{5} + \frac{3x}{2v_2}}. Simplify the numerator and denominator by factoring out xx: 507=5215+32v2.\frac{50}{7} = \frac{\frac{5}{2}}{\frac{1}{5} + \frac{3}{2v_2}}.

Step 4: Solve for v2v_2.
Invert both sides: 750=15+32v252.\frac{7}{50} = \frac{\frac{1}{5} + \frac{3}{2v_2}}{\frac{5}{2}}. Multiply both sides by 52\frac{5}{2}: 75052=15+32v2720=15+32v2.\frac{7}{50} \cdot \frac{5}{2} = \frac{1}{5} + \frac{3}{2v_2} \quad\Longrightarrow\quad \frac{7}{20} = \frac{1}{5} + \frac{3}{2v_2}. Subtract 15\frac{1}{5} from both sides: 72015=32v2720420=32v2320=32v2.\frac{7}{20} - \frac{1}{5} = \frac{3}{2v_2} \quad\Longrightarrow\quad \frac{7}{20} - \frac{4}{20} = \frac{3}{2v_2} \quad\Longrightarrow\quad \frac{3}{20} = \frac{3}{2v_2}. Cancel the common factor of 33: 120=12v2.\frac{1}{20} = \frac{1}{2v_2}. Hence, 2v2=20v2=10 m/s.2v_2 = 20 \quad\Longrightarrow\quad v_2 = 10\ \text{m/s}.

Common Traps & Exam Tip:

1. Misidentifying total displacement: Students sometimes mistakenly add the distances incorrectly, e.g., x+32x=52xx + \frac{3}{2}x = \frac{5}{2}x is crucial. 2. Incorrect time expressions: Forgetting to divide the distance by the velocity for each segment leads to wrong total time. 3. Algebraic slips: When solving the equation for v2v_2, a small arithmetic error can yield an incorrect answer. Always double-check each algebraic manipulation.

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