DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev

DC Pandey Solutions for NEET Physics

NEET : DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev

The document DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev is a part of the NEET Course DC Pandey Solutions for NEET Physics.
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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: EquationDC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev does not seem dimensionally correct, why?
Ans: DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
Sol: DC Pandey Solutions: Motion in One Dimension- 2 JEE 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
DC Pandey Solutions: Motion in One Dimension- 2 JEE 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)
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev  …(i)
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
or DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
i.e.,   s ∝ t7/4 
Differentiating Eq. (i) w.r.t. time t,
DC Pandey Solutions: Motion in One Dimension- 2 JEE 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
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
= 240 - 180
= 60 m (in the upward direction)
Distance covered (D) by the particle
Time to attain maximum height
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
It implies that particle has come back after attaining maximum height (h) given by
DC Pandey Solutions: Motion in One Dimension- 2 JEE 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
DC Pandey Solutions: Motion in One Dimension- 2 JEE 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:DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev

Sol: DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
DC Pandey Solutions: Motion in One Dimension- 2 JEE 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 DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev Is this statement true or false?
Ans: True
Sol: v2 = v1 + at
∴ at = v2 - v1 
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
DC Pandey Solutions: Motion in One Dimension- 2 JEE 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: DC Pandey Solutions: Motion in One Dimension- 2 JEE 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)
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
At   t = 2 s
a = 5 - 2 x 2
= 1 m/s2 
From Eq. (i),
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev

∴ x = ∫ (10 + 5t - t2) dt
or DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
As, at  t = 0 the value of x is zero
c = 0
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
Thus, at t = 3 s
DC Pandey Solutions: Motion in One Dimension- 2 JEE 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:DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
Sol: 
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
= 4√3 m

Ques 12: Velocity of a particle at any time t is DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRevFind acceleration and displacement of particle at t = 1 s. Can we apply DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev or not?
Ans: DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
Sol: Part I
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev …(i)
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
From Eq. (i),
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
Taking initial displacement to be zero.
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
Part II  
Yes. As explained below.
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev implies that initial velocity of the particle is DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev and the acceleration is DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
∴  DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
DC Pandey Solutions: Motion in One Dimension- 2 JEE 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
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
Sol: x = 2t and y = t2
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
or, x2 = 4y
(The above is the equation to trajectory) x = 2t
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
y = t2
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
Thus,
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev


Introductory Exercise 3.3

Ques 1: Figure shows the displacement-time graph of a particle moving in a straight line. Find the signs of velocity and acceleration of particle at time t = t1 and t = t2.
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
Ans: vt1, at1 and at2 are positive while vt2 is negative
Sol: At t = t1
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
v = tan θ
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev

Corresponding v-t graph will be
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
Acceleration at t = t1 : DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
As α < 90°, a t1 is + ive constant.
Acceleration at t = t2

DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev

Ques 2: A particle of mass m is released from a certain height h with zero initial velocity. It strikes the ground elastically (direction of its velocity is reversed but magnitude remains the same). Plot the graph between its kinetic energy and time till it returns to its initial position.
Ans:
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
Sol: Let the particle strike ground at time t velocity of particle when it touches ground

would be gt. KE of particle will beDC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev i.e., KE ∝ t2. While going up the velocity will get - ive but the KE will remain. KE will reduce to zero at time 2 t when the particle reaches its initial position.
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev

Ques 3: A ball is dropped from a height of 80 m on a floor. At each collision, the ball loses half of its speed. Plot the speed-time graph and velocity-time graph of its motion till two collisions with the floor. [Take g = 10 m/s2]
Ans: 
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
Sol: Speed of ball (just before making first collision with floor) 
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
Time taken to reach ground
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
Speed of ball (just after first collision with floor)
= 40/2 = 20 m/s
Time to attain maximum height
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
∴ Time for the return journey to floor = 2 s.
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
Corresponding velocity-time will be
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev

Ques 4: Figure shows the acceleration-time graph of a particle moving along a straight line. After what time the particle acquires its initial velocity?
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
Ans: (2 + √3) s
Sol:
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
⇒ h = 2 (t - 2)
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
Particle will attain its initial velocity i.e., net increase in velocity of the particle will be zero when,
area under a-t graph = 0
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
or 3 - (t - 2)2 = 0
or (t - 2)2 = 3
or  t - 2 = ± √3
or   t = 2 ± √3
At time t = 2 + √3 s

Introductory Exercise 3.4

Ques 1: Two balls A and B are projected vertically upwards with different velocities. What is the relative acceleration between them?
Ans:
zero
Sol: Relative acceleration of A w.r.t. B
αAB = (+ g) - (+ g) = 0

Ques 2: In the above problem what is the shape of the graph between distance between the balls and time before either of the two collide with ground?
Ans: straight line passing through origin
Sol: Velocity of A w.r.t. B = vA - vB
∴ Relative displacement (i.e., distance between A  and B) would be
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
or   s = (vA - vB) t
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev

Ques 3: A river 400 m wide is flowing at a rate of 2.0 m/s. A boat is sailing at a velocity of 10.0 m/s with respect to the water in a direction perpendicular to the river. 
(a) Find the time taken by the boat to reach the opposite bank. 
(b) How far from the point directly opposite to the starting point does the boat reach the opposite bank?
Ans: (a) 40 s
(b) 80 m
Sol: In figure, u = speed of boat
v = speed of river flow

DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
Time to cross river
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev

Ques 4: An aeroplane has to go from a point A to another point B, 500 km away due 30° east of north. Wind is blowing due north at a speed of 20 m/s. The air-speed of the plane is 150 m/s. 
(a) Find the direction in which the pilot should head the plane to reach the point B. 
(b) Find the time taken by the plane to go from A to B.
Ans: DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
(b) 50 min
Sol: Let C be the point along which pilot should head the plane.
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
Apply sine formula in Δ ABC
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
= 2989 s
= 50 min

Ques 5: Two particles A and B start moving simultaneously along the line joining them in the same direction with acceleration of 1 m/s2 and 2 m/s2 and speeds 3 m/s and 1 m/s respectively. Initially A is 10 m behind B. What is the minimum distance between them?
Ans: 8 m
Sol: αA = 1 m/s2 ,   αB = 2 m/s2 
vA = 3 m/s,   vB = 1 m/s
Acceleration of A w.r.t. B = 1 - 2 = - 1 m/s2 
Velocity of A w.r.t. B = 3 - 1 = 2 m/s
Initial displacement of A w.r.t. B = - 10 m
At time relative displacement of A w.r.t. B
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
or  s = - 10 + 2t - 0.5t2 
For s to be minimum
DC Pandey Solutions: Motion in One Dimension- 2 JEE Notes | EduRev
or 2 - (0.5 x 2t) = 0
i.e., t = 2 s
∴ smin = - 10 + (2 x 2) - 0.5 x (2)2 
= - 10 + 4 - 2
= - 8 m
Minimum distance between A and B = 8 m.

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