The displacement x versus time graph is shown below: Figure (A) The average velocity during 0 to 3s is 10 m/s (B) The average velocity during 3 to 5s is 0 m/s (C) The instantaneous velocity at t = 2s is 5 m/s (D) The avberage velocity during 5 to 7s

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Solution

Q.
The displacement x versus time graph is shown below:

(A) The average velocity during 0 to 3s is 10 m/s

(B) The average velocity during 3 to 5s is 0 m/s

(C) The instantaneous velocity at t = 2s is 5 m/s

(D) The avberage velocity during 5 to 7s and instantaneous velocity at t = 6.5s are equal

(E) The average velocity from t = 0 to t = 9s is zero

Choose the correct answer from the options given below: [2025]

1 (A), (D), (E) only

2 (B), (C), (D) only

3 (B), (D), (E) only

4 (B), (C), (E) only

Ans.
(4)

$< \vec{v} > = \frac{△ \vec{s}}{△ t} = \frac{S_{f} - S_{i}}{t_{f} - t_{i}}$

$\vec{v} = \frac{d s}{d t} = slope$

(A) 0 to 3 sec $< \vec{v} > = \frac{5 - 0}{3} = \frac{5}{3} m / s$

(B) 0 to 5 sec, $< \vec{v} > = \frac{5 - 5}{2} = 0$

(C) t = 2 sec, $\vec{v} = \frac{d s}{d t} = Slope = < \vec{v} > = \frac{5 - (- 5)}{2} = 5 m / s$

(D) t = 5 sec to 7 sec, $< \vec{v} > = \frac{0 - 5}{2} = - 2.5 m / s$

At t = 6.5 sec, $\vec{v} = \frac{d s}{d t} = Slope = < \vec{v} > = \frac{0 - (10)}{1} = - 10 m / s$

(E) t = 0 to 9 sec $< \vec{v} > = \frac{0 - 0}{9} = 0$

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