Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

DC Pandey (Questions & Solutions) of Physics: NEET

NEET : Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

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Introductory Exercise 3.2

Ques 1: A ball is thrown vertically upwards. Which quantity remains constant among, speed, kinetic energy, velocity and acceleration?
Ans: acceleration
Sol: Acceleration (due to gravity).

Ques 2: EquationMotion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev does not seem dimensionally correct, why?
Ans: Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Sol: Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev is physically correct as itgives the displacement of the particle in tth second (or any time unit).
st = Displacement in t seconds - displacement in (t - 1) seconds
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Therefore, the given equation is dimensionally incorrect.

Ques 3: Can the speed of a particle increase as its acceleration decreases? If yes give an example.
Ans: Yes, in simple harmonic motion
Sol: Yes. When a particle executing simple harmonic motion returns from maximum amplitude position to its mean position the value of its acceleration decreases while speed increases.

Ques 4: The velocity of a particle moving in a straight line is directly proportional to 3/4th power of time elapsed. How does its displacement and acceleration depend on time?
Ans:  t7/4, t-1/4
Sol: v = t3/4 (given)
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev  …(i)
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
or Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
i.e.,   s ∝ t7/4 
Differentiating Eq. (i) w.r.t. time t,
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
⇒ a ∝ t-1/4

Ques 5: A particle is projected vertically upwards with an initial velocity of 40 m/s. Find the displacement and distance covered by the particle in 6 seconds. Take g = 10 m/s2.
Ans: 60 m, 100 m
Sol: Displacement (s) of the particle
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
= 240 - 180
= 60 m (in the upward direction)
Distance covered (D) by the particle
Time to attain maximum height
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
It implies that particle has come back after attaining maximum height (h) given by
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
∴ D = 80 + (80 - 60)
= 100 m

Ques 6: Velocity of a particle moving along positive x-direction is v = (40 - 101) m/s. Here, t is in seconds. A t time t = 0, the x coordinate of particle is zero. Find the time when the particle is at a distance of 60 m from origin.
Ans: 2 s, 6 s, 2 (2 + √7) s
Sol: v = 40 - 10t
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
or   dx = (40 - 10t) dt
or   x = ∫ (40 - 10t) dt
or    x = 40t - 5t2 + c
As at t = 0 the value of x is zero.
c = 0
∴ x = 40t - 5t2
For x to be 60 m.
60 = 40t - 5t2
or   t2 - 8t + 12 = 0
∴ t = 2 s or 6 s

Ques 7: A particle moves rectilinearly with initial velocity u and a constant acceleration a. Find the average velocity of the particle in a time interval from t = 0 to t = t second of its motion.
Ans:Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

Sol: Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 8: A particle moves in a straight line with uniform acceleration. Its velocity at time t = 0 is v1 and at time t = t is v2. The average velocity of the particle in this time interval is Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev Is this statement true or false?
Ans: True
Sol: v2 = v1 + at
∴ at = v2 - v1 
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 9: Find the average velocity of a particle released from rest from a height of 125 m over a time interval till it strikes the ground, g = 10 m/s2.
Ans: 25 m/s (downwards)
Sol: Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
⇒ t = 25 s
Average velocity = 125 m/5 s  (downwards)
= 25 m/s (downwards)

Ques 10: Velocity of a particle moving along x-axis varies with time as, v = (10 + 5t - t2) At time t = 0, x = 0. 
Find 
(a) acceleration of particle at t = 2 s 
(b) x-coordinate of particle at t = 3 s
Ans: (a) 1 m/s2 
(b) 43 .5 m
Sol: v = 10 + 5t - t2       …  (i)
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
At   t = 2 s
a = 5 - 2 x 2
= 1 m/s2 
From Eq. (i),
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

∴ x = ∫ (10 + 5t - t2) dt
or Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
As, at  t = 0 the value of x is zero
c = 0
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Thus, at t = 3 s
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
= 30 + 22.5 - 9
= 43.5 m

Ques 11: Velocity of a particle at time t = 0 is 2 m/s. A constant acceleration of 2 m/s2 acts on the particle for 2 seconds at an angle of 60° with its initial velocity. Find the magnitude of velocity and displacement of particle at the end of t = 2 s.
Ans:Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Sol: 
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
= 4√3 m

Ques 12: Velocity of a particle at any time t is Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRevFind acceleration and displacement of particle at t = 1 s. Can we apply Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev or not?
Ans: Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Sol: Part I
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev …(i)
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
From Eq. (i),
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Taking initial displacement to be zero.
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Part II  
Yes. As explained below.
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev implies that initial velocity of the particle is Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev and the acceleration is Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
∴  Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 13: The coordinates of a particle moving in x-y plane at any time t are (21, t2).. 
Find: 
(a) the trajectory of the particle, 
(b) velocity of particle at time t and 
(c) acceleration of particle at any time t.
Ans: (a) x2 = 4y
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Sol: x = 2t and y = t2
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
or, x2 = 4y
(The above is the equation to trajectory) x = 2t
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
y = t2
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Thus,
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

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