Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev

DC Pandey (Questions & Solutions) of Physics: NEET

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NEET : Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev

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Graphs

Ques 28: Displacement-time graph of a particle moving in a straight line is as shown in figure.
Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev
(a) Find the sign of velocity in regions oa, ab, be and cd. 
(b) Find the sign of acceleration in the above region.
Ans: (a) positive, positive, positive, negative (b) positive, zero, negative, negative
Sol: OA : slope is + ive and increasing.
Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev

∴ velocity is + ive and acceleration is + ive.
AB : slope is + ive and constant
∴ velocity is +ive and acceleration is zero.
BC : sope is + ive and decreases.
∴ velocity is + ive and increasing.
CD : slope is –ive and increasing
∴ velocity is - ive and acceleration is - ive.

Ques 29: Let us call a motion as:
M1 → if velocity and acceleration both are positive.
M2 → if velocity is positive but acceleration is negative.
M→ if velocity and acceleration both are negative.
M4 → if velocity is negative but acceleration is positive.

(a) State, in which of the above four motions, magnitude of velocity is increasing and in which it is decreasing.
Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev
(b) v-t graph of a particle moving in a straight line is as shown in figure. The whole graph is made up of four straight lines P, Q, R and S. These four straight lines indicate four type of motions (M1.. , M4) discussed above. State, which straight line corresponds to which type of motion.
Ans: (a) In M1 and M3 magnitude is increasing, in M2 and M4 magnitude is decreasing
(b) P → M1; Q → M2; R → M3; S → M4
Sol: 
Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev

In M1 and M3 : 0° < θ < 90°
∴ slope is + ive i.e., acceleration is + ive.
In M2 and M4 : 90° < θ < 180°
∴ slope is - ive i.e., acceleration is - ive.
(a) M1 : Magnitude of velocity is increasing.
M2 : Magnitude of velocity is decreasing.
M3 : Magnitude of velocity is increasing.
M4 : Magnitude of velocity is decreasing.
M1 and M3.
(b) P → M1
Q → M2
R → M3
S → M4

Ques 30: Velocity-time graph of a particle moving in a straight line is shown in figure. At time t = 0, s = - 10 m. Plot corresponding a-t and s-t graphs.
Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev
Ans: 
Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev
Sol: Case I
Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev
v = mt - v0 (st-line)
∴ ∫ ds = ∫ (mt - v0) dt

Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev
As for 0° < q < 90°
tan θ is + ive, a is + ive.
Case II
Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev
v = k (constant) (st line parallel to time-axis)
⇒  ∫ ds = ∫ k dt
s = kt + s0
Further, a = dv/dt = 0
Case III.

Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev

Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev

Time-displacement graph

Time

Area

under the graph

Initial

Displacement

Net

displacement

at 0 s

0 m

- 10 m

- 10 m

2 s

10 m

- 10 m

0 m

4 s

30 m

- 10 m

+ 20 m

6 s

40 m

- 10 m

+ 30 m

8 s

30 m

- 10 m

+ 20 m

10 s

20 m

- 10 m

+ 10 m

Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev

Time-acceleration graph

time

slope

acceleration

0 - 2 s

5

5 m/s2

2 - 4 s

zero

0 m/s2

4 - 6 s

- 5

- 5 m/s2

6 - 8 s

- 5

- 5 m/s2

8 - 10 s

+ 5

+ 5 m/s2

Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 31: Velocity-time graph of a particle moving in a straight line is shown in figure. At time t = 0, s = 20 m. Plot a-t and s-t graphs of the particle.
Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev
Ans: 
Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev
Sol: a = slope of v-t graph.
S = area under v-t graph.
Corresponding graphs are drawn in the answer sheet.

Ques 32: Velocity-time graph of a particle moving in a straight line is shown in figure. In the time interval from t = 0 to t = 14s, find:
Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev
(a) average velocity and 
(b) average speed of the particle
Ans: (a) (50/7) ms-1 
(b) 10 ms-1
Sol: Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 33: Acceleration-time graph of a particle moving in a straight line is as shown in figure. At time t = 0, velocity of the particle is zero. Find:
Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev
(a) average acceleration in a time interval from t = 6s to t = 12s, 
(b) velocity of the particle at t = 14s.
Ans: (a) 5 ms-2 
(b) 90 ms-1
Sol: Average acceleration
Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev
vf - vi = Area under a-t graph.

Ques 34: Velocity-time graph of a particle moving in a straight line is shown in figure. At time t = 0, displacement of the particle from mean position is 10 m. Find:
Motion in One Dimension: JEE Main(Part- 3) - Physics, Solution by DC Pandey NEET Notes | EduRev
(a) acceleration of particle at t = Is, 3s and 9 s. 
(b) position of particle from mean position at t = 10s. 
(c) write down s-t equation for time interval: 
(i) 0 < / < 2s, 
(ii) 4s < t < 8s
Ans: (a) 5 ms-2, zero, 5 ms-2 
(b) s = 30 m
(c) (i) s = 10 + 2.5 t2 
(ii) s = 40 + 10 (t - 4) - 2.5 (t - 4)2
Sol: (a) Acceleration = slope of v-t graph.
(b) rf - ri = s = area under v-t graph.
(c) Equations are written in answer sheet.

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