JEE Main 2025 — Motion In One Dimension Question with Solution
JEE Main 2025 (4 Apr Shift 2)
Question
The displacement x versus time graph is shown below.
(A) The average velocity during 0 to 3 s is
(B) The average velocity during 3 to 5 s is
(C) The instantaneous velocity at is
(D) The average velocity during 5 to 7 s and instantaneous velocity at are equal
(E) The average velocity from to is zero
Choose the correct answer from the options given below:
(A) The average velocity during 0 to 3 s is
(B) The average velocity during 3 to 5 s is
(C) The instantaneous velocity at is
(D) The average velocity during 5 to 7 s and instantaneous velocity at are equal
(E) The average velocity from to is zero
Choose the correct answer from the options given below:
Choose an option
Show full solutionCorrect option: D
Correct answer
D(B), (C), (E) only
Step-by-step explanation
$\begin{aligned}
& \langle\overrightarrow{\mathrm{v}}\rangle=\frac{\Delta \overrightarrow{\mathrm{s}}}{\Delta \mathrm{t}}=\frac{\mathrm{S}_{\mathrm{f}}-\mathrm{S}_{\mathrm{i}}}{\mathrm{t}_{\mathrm{f}}-\mathrm{t}_{\mathrm{i}}} \\ & \overrightarrow{\mathrm{v}}=\frac{\mathrm{ds}}{\mathrm{dt}}=\text { slope }
\end{aligned}3 \mathrm{sec} ;\langle\overrightarrow{\mathrm{v}}\rangle=\frac{5-0}{3}=5 / 3 \mathrm{~m} / \mathrm{s}5 \mathrm{sec} ;\langle\vec{v}\rangle=\frac{5-5}{2}=0\mathrm{t}=2 ;=\vec{v}=5 \mathrm{~m} / \mathrm{s}\mathfrak{t}=57 \mathrm{sec} ;\langle\overrightarrow{\mathrm{v}}\rangle=\frac{0-5}{2}=-2.5 \mathrm{~m} / \mathrm{s}t=6.5 \mathrm{sec} ; \vec{v}=10\mathrm{t}=0\mathrm{t}=9 ;\langle\overrightarrow{\mathrm{v}}\rangle=0$
& \langle\overrightarrow{\mathrm{v}}\rangle=\frac{\Delta \overrightarrow{\mathrm{s}}}{\Delta \mathrm{t}}=\frac{\mathrm{S}_{\mathrm{f}}-\mathrm{S}_{\mathrm{i}}}{\mathrm{t}_{\mathrm{f}}-\mathrm{t}_{\mathrm{i}}} \\ & \overrightarrow{\mathrm{v}}=\frac{\mathrm{ds}}{\mathrm{dt}}=\text { slope }
\end{aligned}3 \mathrm{sec} ;\langle\overrightarrow{\mathrm{v}}\rangle=\frac{5-0}{3}=5 / 3 \mathrm{~m} / \mathrm{s}5 \mathrm{sec} ;\langle\vec{v}\rangle=\frac{5-5}{2}=0\mathrm{t}=2 ;=\vec{v}=5 \mathrm{~m} / \mathrm{s}\mathfrak{t}=57 \mathrm{sec} ;\langle\overrightarrow{\mathrm{v}}\rangle=\frac{0-5}{2}=-2.5 \mathrm{~m} / \mathrm{s}t=6.5 \mathrm{sec} ; \vec{v}=10\mathrm{t}=0\mathrm{t}=9 ;\langle\overrightarrow{\mathrm{v}}\rangle=0$
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This is a previous-year question from JEE Main 2025, covering the Motion In One Dimension chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.