DC Pandey Solutions (JEE Main): Motion in One Dimension- 1 | DC Pandey Solutions for JEE Physics PDF Download

AIEEE Corner
Subjective Questions (Level 1) 

Basic Definitions
Ques 1: A car moves with 60 km/h in first one hour and with 80 km/h in next half an hour. Find: 
(a) total distance travelled by the car, 
(b) average speed of car in total 1.5 hours.
Ans: (a) 100 km
(b) 66.67 kmh^-1
Sol: (a) D = v₁t₁ + v₂ t₂

(b) Average speed


Ques 2: A particle moves in a straight line with initial velocity 4 m/s and a constant acceleration of 6 m/s². Find the average velocity of the particle in a time interval from 
(a) t = 0 to t = 2s 
(b) t = 2s to t = 4s.
Ans: (a) 10 ms^-1 
(b) 22 ms^-1
Sol: (a) Displacement in first two seconds

= 20 m
∴ Average velocity = 20m/2s = 10m/s
(b) Displacement in first four seconds

∴ Displacement in the time interval
t = 2s to t = 4s
= 64 - 20
= 44 m
∴  Average velocity = 44m/2s = 22 m/s.

Ques 3: A particle is projected upwards from the roof of a tower 60 m high with velocity 20 m/s. Find: 
(a) the average speed and 
(b) average velocity of the particle upto an instant when it strikes the ground. Take g = 10 m/s².
Ans: (a) 16.67 ms^-1 
(b) 10 ms^-1(downwards)
Sol: Let the particle takes t time to reach ground.

∴
i.e.,   5t² - 20t - 64 = 0
t = 6.1 s
If the particle goes h meter above tower before coming down
0 = (20)² + 2 (-10) h
⇒ h = 20 m




= 9.8 m/s (downwards)

Ques 4: A block moves in a straight line with velocity v for time t₀. Then, its velocity becomes 2v for next t₀ time. Finally its velocity becomes 3v for time T. If average velocity during the complete journey was 2.5 v, then find T in terms of t₀.
Ans:  T = 4 t₀ 
Sol: Average velocity

∴  5t₀ + 2.5T = 3t₀ + 3T
or  T = 4t₀

Ques 5: A particle starting from rest has a constant acceleration of 4 m/s² for 4 s. It then retards uniformly for next 8 s and comes to rest. Find during the motion of particle: 
(a) average acceleration, 
(b) average speed, 
(c) average velocity.
Ans: (a) zero
(b) 8 ms^-1 
(c) 8 ms^-1
Sol: (a) Average acceleration

(b) As the particle did not return back distance travelled in 12 s
= Displacement at 12 s
∴ Average speed = 8 m/s.


∴  (∵ a = 4 m/s²)
i.e.,   v_max = 16 m/s
Displacement of particle in 12 seconds
= Area under v-t graph

Average velocity
= +8 m/s

Ques 6: A particle moves in a circle of radius R = 21/22 m with constant speed lm/s. Find:
(a) magnitude of average velocity and 
(b) magnitude of average acceleration in 2 s.
Ans:
Sol: (a) Radius (R) of circle = 21/22 m
∴ Circumference of circle = 2πR

Speed (v) of particle = 1 m/s
∴ Distance moved by particle in 2 s = 2 m
Thus, angle through which the particle moved

Magnitude of Average velocity


(b) Magnitude of average acceleration




Ques 7: A particle is moving in x-y plane. At time t = 0, particle is at (1m, 2m) and has velocity  At t = 4 s, particle reaches at (6m, 4m) and has velocity In the given time interval, find: 
(a) average velocity, 
(b) average acceleration and 
(c) from the given data, can you find average speed?
Ans:
Sol: Position vector at t = 0 s

Position vector at t = 4 s

(a) Displacement from t = 0 s to t = 4 s
 


(b) Average acceleration

(c) We cannot find the average speed as the actual path followed by the particle is not known.

Uniform acceleration
(a) One dimensional motion

Ques 8: Two diamonds begin a free fall from rest from the same height, 1.0 s apart. How long after the first diamond begins to fall will the two diamonds be 10 m apart? Take g = 10 m/s².
Ans: 1.5 s
Sol: If at time t the vertical displacement between A  and B is 10 m

or   t² - (t - 1)² = 2
or   t² - (t² - 2t + 1) = 2
or   2t = 3
t = 1.5 s

Ques 9: Two bodies are projected vertically upwards from one point with the same initial velocity v₀. The second body is projected t₀ s after the first. How long after will the bodies meet?
Ans:
Sol: The two bodies will meet if
Displacement of first after attaining highest point = Displacement of second before attaining highest point

or
or
or  


Ques 10: A stone is dropped from the top of a tower. When it crosses a point 5 m below the top, another stone is let fall from a point 25 m below the top. Both stones reach the bottom of the tower simultaneously. Find the height of the tower. Take g = 10 m/s².
Ans: 45 m

Sol:
⇒ t = 1s

For A

or    …(i)

For B

or  
[Substituting value of H from Eq. (i)]

2t - 1 = 5
⇒ t = 3 s
Substituting t = 3 s in Eq. (i)


Ques 11: A point mass starts moving in a straight line with constant acceleration. After time t₀ the acceleration changes its sign, remaining the same in magnitude. Determine the time t from the beginning of motion in which the point mass returns to the initial position.
Ans: (3.414) t₀
Sol:


∴
or  
or

 (- ive sign being absurd)

From the begining of the motion the point mass will return to the initial position after time 3.141t₀.

Ques 12: A football is kicked vertically upward from the ground and a student gazing out of the window sees it moving upwards past her at 5.00 m/s. The window is 15.0 m above the ground. Air resistance may be ignored. Take g = 10 m/s². 
(a) How high does the football go above ground? 
(b) How much time does it take to go from the ground to its highest point?
Ans: (a) 16.25 m (b) 1.8 s
Sol: 5² = u² + 2 (-10)15
⇒ u² = 325

(a) For H
0² = u² + 2(-10) H
i.e., 20H = 325
or     H = 16.25 m
(b) For t

∴
= 1.8 s

Ques 13: A car moving with constant acceleration covered the distance between two points 60.0 m apart in 6.00 s. Its speed as it passes the second point was 15.0 m/s. 
(a) What is the speed at the first point? 
(b) What is the acceleration? 
(c) At what prior distance from the first was the car at rest? 
(d) Graph s versus t and v versus t for the car, from rest (t = 0).
Ans: (a) 5 ms^-1 
(b)1 .6 7 ms^-2 
(c) 7.5 m
Sol:

(a) 15² = u² + 2a x 60 …(i)
and  15 = u + a x 6 …(ii)
Substituting the value of 6a from Eq. (ii) in Eq. (i)
225 = u² + 20 (15 - u)
i.e., u² - 20 u + 75 = 0
(u - 15)(u - 5) = 0
∴ u = 5 m/s
(15 m/s being not possible)
(b) Using Eq. (ii)

(c)   u² = 0² + 2ax
i.e.,

= 7.5 m
(d)

  …(iii)


Differentiating Eq. (iii) w.r.t. time t




Ques 14: A train stopping at two stations 4 km apart takes 4 min on the journey from one of the station to the other. Assuming that it first accelerates with a uniform acceleration x and then that of uniform retardation y, prove that
Sol: 

Journey A to P
v_max = 0 + xt₁    …(i)
and   …(ii)
Journey P to B
0 = v_max + (- y) t₂    …(iii)
and   …(iv)
⇒
∴
or     …(v)
From Eq. (ii) and Eq. (iv)

or    …(vi)
Dividing Eq. (vi) by Eq. (v)
v_max = 2
Substituting the value of v_max in Eq. (v)
  (Proved)

Ques 15: A particle moves along the x-direction with constant acceleration. The displacement, measured from a convenient position, is 2m at time t = 0 and is zero when t = 10s. If the velocity of the particle is momentary zero when t = 6 s, determine the acceleration a and the velocity v when t = 10s.
Ans: 0.2 ms^-2, 0.8 ms^-1
Sol: Let acceleration of the particle be a using

v = u + at
0 = u + a6
∴
(a) At t = 10 s, s = - 2 m

or   -2 = (-6a) 10 + 50a
or  -10a = - 2
or   a = 0.2 m/s² 
(b) v(at t = 10 s) = u + a 10
= - 6a + 10a
= 4 a
= 0.8 m/s

(b) Two or three dimensional motion

Ques 16: Net force acting on a particle of mass 2 kg is 10 N in north direction. At t = 0, particle was moving eastwards with 10 m/s. Find displacement and velocity of particle after 2 s.
Ans: 10√5 m at cos^-1(2) from east towards north, 10√2 ms-1 at 45° from east towards north.
Sol:






m at cot^-1(2) from east to north.

Ques 17: At time t = 0, a particle is at (2m, 4m). It starts moving towards positive x-axis with constant acceleration 2 m/s² (initial velocity = 0). After 2 s an acceleration of 4 m/s² starts acting on the particle in negative y-direction also. Find after next 2 s: 
(a) velocity and 
(b) coordinates of particle.
Ans:  
Sol: 

(a) Velocity


or  
(b) Co-ordinate of particle




Co-ordinate of the particle [18 m, - 4 m]

Ques 18: A particle moving in x-y plane is at origin at time t = 0. Velocity of the particle is  and acceleration is
(a) velocity of particle and
(b) coordinates of particle.
Ans:
Sol: 


(b) Co-ordinates of the particle


∴ Co-ordinates of particle would be [10 m , - 2 m]

Ques 19: A particle starts from the origin at t = 0 with a velocity of  and moves in the x-y plane with a constant acceleration of  At the instant the particle’s x-coordinate is 29 m, what are: 
(a) its y-coordinate and 
(b) its speed ?
Ans: (a) 45 m (b) 22 ms^-1
Sol:


Comparing the coefficients of
29 = 2t²   …(i)
and  n = 8t + t² …(ii)
⇒
Substituting value of t in Eq. (ii)
n = 8 x 3.807 + (3.807)²
= 44.95



= 24.44 m/s

Ques 20: At time t = 0, the position vector of a particle moving in the x-y plane is  By time t = 0.02 s, its position vector has become  m determine the magnitude v_av of the average velocity during this interval and the angle θ made by the average velocity with the positive x-axis.
Ans: 20.6 ms^-1, tan^-1(4)
Sol: