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DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET PDF Download

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 | Physics Class 11 - NEET does not seem dimensionally correct, why?
Ans: DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
Sol: DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET 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 | Physics Class 11 - NEET
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 | Physics Class 11 - NEET  …(i)
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
or DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
i.e.,   s ∝ t7/4 
Differentiating Eq. (i) w.r.t. time t,
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
⇒ 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 | Physics Class 11 - NEET
= 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 | Physics Class 11 - NEET
It implies that particle has come back after attaining maximum height (h) given by
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
∴ 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 | Physics Class 11 - NEET
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 | Physics Class 11 - NEET

Sol: DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET

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 | Physics Class 11 - NEET Is this statement true or false?
Ans: True
Sol: v2 = v1 + at
∴ at = v2 - v1 
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET

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 | Physics Class 11 - NEET
⇒ 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 | Physics Class 11 - NEET
At   t = 2 s
a = 5 - 2 x 2
= 1 m/s2 
From Eq. (i),
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET

∴ x = ∫ (10 + 5t - t2) dt
or DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
As, at  t = 0 the value of x is zero
c = 0
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
Thus, at t = 3 s
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
= 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 | Physics Class 11 - NEET
Sol: 
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
= 4√3 m

Ques 12: Velocity of a particle at any time t is DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEETFind acceleration and displacement of particle at t = 1 s. Can we apply DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET or not?
Ans: DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
Sol: Part I
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET …(i)
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
From Eq. (i),
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
Taking initial displacement to be zero.
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
Part II  
Yes. As explained below.
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET implies that initial velocity of the particle is DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET and the acceleration is DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
∴  DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET

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 | Physics Class 11 - NEET
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
Sol: x = 2t and y = t2
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
or, x2 = 4y
(The above is the equation to trajectory) x = 2t
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
y = t2
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
Thus,
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET


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 | Physics Class 11 - NEET
Ans: vt1, at1 and at2 are positive while vt2 is negative
Sol: At t = t1
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
v = tan θ
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET

Corresponding v-t graph will be
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
Acceleration at t = t1 : DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
As α < 90°, a t1 is + ive constant.
Acceleration at t = t2

DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET

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 | Physics Class 11 - NEET
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 | Physics Class 11 - NEET 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 | Physics Class 11 - NEET
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET

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 | Physics Class 11 - NEET
Sol: Speed of ball (just before making first collision with floor) 
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
Time taken to reach ground
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
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 | Physics Class 11 - NEET
∴ Time for the return journey to floor = 2 s.
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
Corresponding velocity-time will be
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET

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 | Physics Class 11 - NEET
Ans: (2 + √3) s
Sol:
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
⇒ h = 2 (t - 2)
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
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 | Physics Class 11 - NEET
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 | Physics Class 11 - NEET
or   s = (vA - vB) t
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET

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 | Physics Class 11 - NEET
Time to cross river
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET

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 | Physics Class 11 - NEET
(b) 50 min
Sol: Let C be the point along which pilot should head the plane.
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
Apply sine formula in Δ ABC
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
= 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 | Physics Class 11 - NEET
or  s = - 10 + 2t - 0.5t2 
For s to be minimum
DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET
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.

The document DC Pandey Solutions: Motion in One Dimension - 2 | Physics Class 11 - NEET is a part of the NEET Course Physics Class 11.
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FAQs on DC Pandey Solutions: Motion in One Dimension - 2 - Physics Class 11 - NEET

1. What is motion in one dimension?
Ans. Motion in one dimension refers to the movement of an object along a straight line. It involves the study of an object's position, velocity, and acceleration in relation to time, considering only one dimension of motion.
2. How is displacement different from distance in one-dimensional motion?
Ans. In one-dimensional motion, displacement refers to the change in an object's position, considering both magnitude and direction. Distance, on the other hand, refers to the total path length traveled by the object, without considering direction. Displacement is a vector quantity, while distance is a scalar quantity.
3. How can we calculate average velocity in one-dimensional motion?
Ans. Average velocity in one-dimensional motion is calculated by dividing the total displacement of an object by the total time taken. The formula for average velocity is: Average velocity = (Change in position) / (Change in time)
4. What is the difference between speed and velocity in one-dimensional motion?
Ans. In one-dimensional motion, speed refers to the rate at which an object covers a distance, without considering direction. It is a scalar quantity and is always positive. Velocity, on the other hand, refers to the rate at which an object changes its position in a particular direction. Velocity is a vector quantity and can be positive or negative, depending on the direction of motion.
5. How can we determine the acceleration in one-dimensional motion?
Ans. Acceleration in one-dimensional motion is determined by calculating the change in velocity of an object over a given time interval. The formula for acceleration is: Acceleration = (Change in velocity) / (Change in time)
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