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 = v_{1}t_{1} + v_{2} t_{2}
(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/s^{2}. 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/s^{2}.
Ans: (a) 16.67 ms^{1}
(b) 10 ms^{1}(downwards)
Sol: Let the particle takes t time to reach ground.
∴
i.e., 5t^{2}  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 t_{0}. Then, its velocity becomes 2v for next t_{0} 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 t_{0}.
Ans: T = 4 t_{0}
Sol: Average velocity
∴ 5t_{0} + 2.5T = 3t_{0} + 3T
or T = 4t_{0}
Ques 5: A particle starting from rest has a constant acceleration of 4 m/s^{2} 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/s^{2})
i.e., v_{max} = 16 m/s
Displacement of particle in 12 seconds
= Area under vt 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 xy 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/s^{2}.
Ans: 1.5 s
Sol: If at time t the vertical displacement between A and B is 10 m
or t^{2}  (t  1)^{2} = 2
or t^{2}  (t^{2}  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 v_{0}. The second body is projected t_{0} 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/s^{2}.
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 t_{0} 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) t_{0}
Sol:
∴
or
or
( ive sign being absurd)
From the begining of the motion the point mass will return to the initial position after time 3.141t_{0}.
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/s^{2}.
(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: 5^{2} = u^{2} + 2 (10)15
⇒ u^{2} = 325
(a) For H
0^{2} = u^{2} + 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) 15^{2} = u^{2} + 2a x 60 …(i)
and 15 = u + a x 6 …(ii)
Substituting the value of 6a from Eq. (ii) in Eq. (i)
225 = u^{2} + 20 (15  u)
i.e., u^{2}  20 u + 75 = 0
(u  15)(u  5) = 0
∴ u = 5 m/s
(15 m/s being not possible)
(b) Using Eq. (ii)
(c) u^{2} = 0^{2} + 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 + xt_{1} …(i)
and …(ii)
Journey P to B
0 = v_{max} + ( y) t_{2} …(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 xdirection 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/s^{2}
(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 ms1 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 xaxis with constant acceleration 2 m/s^{2} (initial velocity = 0). After 2 s an acceleration of 4 m/s^{2} starts acting on the particle in negative ydirection also. Find after next 2 s:
(a) velocity and
(b) coordinates of particle.
Ans:
Sol:
(a) Velocity
or
(b) Coordinate of particle
Coordinate of the particle [18 m,  4 m]
Ques 18: A particle moving in xy 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) Coordinates of the particle
∴ Coordinates 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 xy plane with a constant acceleration of At the instant the particle’s xcoordinate is 29 m, what are:
(a) its ycoordinate and
(b) its speed ?
Ans: (a) 45 m (b) 22 ms^{1}
Sol:
Comparing the coefficients of
29 = 2t^{2} …(i)
and n = 8t + t^{2} …(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 xy 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 xaxis.
Ans: 20.6 ms^{1}, tan^{1}(4)
Sol:
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