JEE MAIN - Physics (2025 - 4th April Evening Shift - No. 7)
The displacement x versus time graph is shown below.
_4th_April_Evening_Shift_en_7_1.png)
(A) The average velocity during 0 to 3 s is $10 \mathrm{~m} / \mathrm{s}$
(B) The average velocity during 3 to 5 s is $0 \mathrm{~m} / \mathrm{s}$
(C) The instantaneous velocity at $\mathrm{t}=2 \mathrm{~s}$ is $5 \mathrm{~m} / \mathrm{s}$
(D) The average velocity during 5 to 7 s and instantaneous velocity at $\mathrm{t}=6.5 \mathrm{~s}$ are equal
(E) The average velocity from $t=0$ to $t=9 \mathrm{~s}$ is zero
Choose the correct answer from the options given below :
Explanation
$$\begin{aligned} & <\vec{v}>=\frac{\Delta \overrightarrow{\mathrm{s}}}{\Delta \mathrm{t}}=\frac{\mathrm{S}_{\mathrm{f}}-\mathrm{S}_{\mathrm{i}}}{\mathrm{t}_{\mathrm{f}}-\mathrm{t}_{\mathrm{i}}} \\ & \overrightarrow{\mathrm{v}}=\frac{\mathrm{ds}}{\mathrm{dt}}=\text { slope } \end{aligned}$$
(A) 0 to $3 \mathrm{sec} ;\langle\vec{v}\rangle=\frac{5-0}{3}=5 / 3 \mathrm{~m} / \mathrm{s}$
(B) 0 to $5 \mathrm{sec} ;\langle\overrightarrow{\mathrm{v}}\rangle=\frac{5-5}{2}=0$
(C) $\mathrm{t}=2$; slope $=\overrightarrow{\mathrm{v}}=5 \mathrm{~m} / \mathrm{s}$
(D) $\mathrm{t}=5$ to $7 \mathrm{sec} ;\langle\overrightarrow{\mathrm{v}}\rangle=\frac{0-5}{2}=-2.5 \mathrm{~m} / \mathrm{s}$
At $\mathrm{t}=6.5 \mathrm{sec} ; \overrightarrow{\mathrm{v}}=10$
(E) $\mathrm{t}=0$ to $\mathrm{t}=9 ;\langle\overrightarrow{\mathrm{v}}\rangle=0$
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