Graphs of Motion in One Dimension questions | Relative Motion in One Dimension jee questions

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Topic Question Set

Q 1 :
The motion of an airplane is represented by velocity-time graph as shown below. The distance covered by airplane in the first 30.5 second is ______ km. [2025]

9
6
3
12

1

2

3

4

(4)

Distance = area under the graph

$d = 300 \times 2 + 400 \times 28.5$

= 600 + 114000 = 12000 m.

Q 2 :
The velocity-time graph of an object moving along a straight line is shown in figure. What is the distance covered by the object between t = 0 to t = 4 s? [2025]

30 m
10 m
13 m
11 m

1

2

3

4

(1)

Distance travelled = displacement when direction of velocity remains constant

Distance = Area under v/t graph

$\Rightarrow s = \frac{1}{2} (2 s + 4 s) (10 m / s) = 30 m$

Q 3 :
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]

(A), (D), (E) only
(B), (C), (D) only
(B), (D), (E) only
(B), (C), (E) only

1

2

3

4

(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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