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JEE MainPhysics

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Kinematics: JEE Main Physics Question with Solution

The position-time graphs for two students

\text{A}

and

\text{B}

returning from the school to their homes are shown in figure : (A)

\text{A}

lives closer to the school (B)

\text{B}

lives closer to the school (C)

\text{A}

takes lesser time to reach home (D)

\text{A}

travels faster than

\text{B}

(E)

\text{B}

travels faster than

\text{A}

Choose the correct answer from the options given below :

Position-time graph showing lines for students A and B starting from different positions or times, where B has a steeper slope.

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A(A) and (E) onlyB(B) and (E) onlyC(A), (C) and (E) onlyD(A), (C) and (D) only

Step-by-step solutionView

Correct answer

Statements (A) and (E) are correct because A's home is at a smaller distance from school (origin) and B's graph has a steeper slope, meaning B travels faster.

Option analysis

Why each option works or fails

A · (A) and (E) only

None. The student correctly interprets distances from the vertical axis intercept/final position and speeds from the slope of the x-t graphs. Correct! A's final position $x_A < x_B$ means A lives closer to the origin (school), and B's line has a steeper slope, meaning B has a higher speed.

B · (B) and (E) only

The student confuses the home positions of A and B on the position axis, believing B is closer to the origin. Look at the position axis (vertical axis): A's home corresponds to a smaller position value $x_A$ compared to B's home $x_B$ , so A lives closer to the school.

C · (A), (C) and (E) only

The student assumes that because B starts later or lives farther, A must take less time overall to reach home. Check the time duration $\Delta t = t_{\text{final}} - t_{\text{start}}$ for each student from the time axis. B reaches home in a shorter total time interval than A.

D · (A), (C) and (D) only

The student believes a less steep position-time graph represents a higher speed. The slope of a position-time graph ( $\Delta x / \Delta t$ ) represents velocity/speed. Since B's graph has a greater slope than A's graph, B travels faster than A.

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Solution

StepWorking

01concept

On an $x$ - $t$ graph with the origin $O$ representing the school: 1. Distance of home from school is given by the final position coordinate $x$ . 2. Velocity is the slope of the $x$ - $t$ line: $v = \frac{\Delta x}{\Delta t}$ . 3. Time taken to reach home is the total duration $\Delta t = t_{\text{final}} - t_{\text{start}}$ of travel.

02option_verdict

From the graph, home position $x_A < x_B$ , so A lives closer to the school (Statement A is TRUE). Slope of B's line is steeper than A's line, so $v_B > v_A$ , meaning B travels faster than A (Statement E is TRUE). Hence, (A) and (E) only is correct.

03option_verdict

Statement B claims B lives closer to school, which contradicts the graph where $x_B > x_A$ .

04option_verdict

Statement C claims A takes lesser time to reach home. From the graph, A starts at $t = 0$ and reaches home at $t_A$ , while B starts later at $t_0$ and reaches at $t_B \approx t_A$ , meaning B takes time $t_B - t_0 < t_A$ . Thus, B takes lesser time, making C false.

05option_verdict

Statement D claims A travels faster than B, which is false since slope of A is less than slope of B.

✓discriminator

Directly compare final $x$ ( $x_A < x_B \implies$ A is closer) and slope (B is steeper $\implies$ B is faster), yielding statements (A) and (E).

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Physics
Concepts assessed: Physics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 3 September 2026

Quick checks

Students also ask

Why does B start after A?

B's position-time line starts at a non-zero time on the time axis, meaning B leaves the school later than A.

Answer

Statements (A) and (E) are correct because A's home is at a smaller distance from school (origin) and B's graph has a steeper slope, meaning B travels faster.

Why each option works or fails

A · correct: (A) and (E) only - None. The student correctly interprets distances from the vertical axis intercept/final position and speeds from the slope of the x-t graphs. Correct! A's final position $x_A < x_B$ means A lives closer to the origin (school), and B's line has a steeper slope, meaning B has a higher speed.
B: (B) and (E) only - The student confuses the home positions of A and B on the position axis, believing B is closer to the origin. Look at the position axis (vertical axis): A's home corresponds to a smaller position value $x_A$ compared to B's home $x_B$ , so A lives closer to the school.
C: (A), (C) and (E) only - The student assumes that because B starts later or lives farther, A must take less time overall to reach home. Check the time duration $\Delta t = t_{\text{final}} - t_{\text{start}}$ for each student from the time axis. B reaches home in a shorter total time interval than A.
D: (A), (C) and (D) only - The student believes a less steep position-time graph represents a higher speed. The slope of a position-time graph ( $\Delta x / \Delta t$ ) represents velocity/speed. Since B's graph has a greater slope than A's graph, B travels faster than A.

Step-by-step solution

concept: On an $x$ - $t$ graph with the origin $x = 0$ representing the school: 1. Distance of home from school is given by the final position coordinate $x$ . 2. Velocity is the slope of the $x$ - $t$ line: $v = \frac{\Delta x}{\Delta t}$ . 3. Time taken to reach home is the total duration $\Delta t = t_{\text{final}} - t_{\text{start}}$ of travel.
option_verdict: From the graph, home position $x_A < x_B$ , so A lives closer to the school (Statement A is TRUE). Slope of B's line is steeper than A's line, so $v_B > v_A$ , meaning B travels faster than A (Statement E is TRUE). Hence, (A) and (E) only is correct.
option_verdict: Statement B claims B lives closer to school, which contradicts the graph where $x_B > x_A$ .
option_verdict: Statement C claims A takes lesser time to reach home. From the graph, A starts at $t = 0$ and reaches home at $t_A$ , while B starts later at $t_0$ and reaches at $t_B \approx t_A$ , meaning B takes time $v = \frac{\Delta x}{\Delta t}$ . Thus, B takes lesser time, making C false.
option_verdict: Statement D claims A travels faster than B, which is false since slope of A is less than slope of B.
discriminator: Directly compare final $x$ ( $x_A < x_B \implies$ A is closer) and slope (B is steeper $\implies$ B is faster), yielding statements (A) and (E).