IISER Physics - Description of Motion in One Dimension
Exam Duration: 45 Mins Total Questions : 30
A stone is dropped from an ascending balloon moving with velocity \(\vartheta \) at a height H. The stone will
- (a)
move vertical distance of \(({{\vartheta^{2} \over g}+H})\)
- (b)
cover a total vertical distance of \(({{\vartheta^{2} \over 2g}+H})\)
- (c)
move along a parabolic path and cover a total vertical distance of \(({{\vartheta^{2} \over g}+H})\).
- (d)
mover vertically down and cover a distance H.
At the time of release, the stone has an initial upward velocity v, equal to the speed of the ascending balloon. It moves up and comes to rest after covering a vertical distance upward equal to \(\frac { { v }^{ 2 } }{ 2g } \) . After this instant the stone moves down and covers a vertical distance of \(\left( \frac { { v }^{ 2 } }{ 2g } +H \right) \) From the instant of release, the stone covers a total distance of \(\left( \frac { v^{ 2 } }{ 2g } +\frac { v^{ 2 } }{ 2g } +H \right) =\left( \frac { v^{ 2 } }{ g } +H \right) \)
A particle moves along the X-axis such that its coordinate (x) varies with time (t) according to the expression
x = 2 - 5 t - 6 \(t^{2}\)
where x is in meters and t is in seconds. The initial velocity of the particle is
- (a)
- 5 m \(s^{-1}\)
- (b)
- 3 m \(s^{-1}\)
- (c)
6 m \(s^{-1}\)
- (d)
2 m \(s^{-1}\)
x = 2 - 5t + 6t2
Instantaneous velocity = \(\frac { dx }{ dt } =\frac { d }{ dt } \left( 2-5t+6t^{ 2 } \right) \)
For initial velocity, t = 0. This gives
\(\frac { dx }{ dt } =-5ms^{ -1 }\)
A body travelling with uniform acceleration crosses two points A and B with velocities 20 m \(s^{-1}\) and 30 m \(s^{-1}\) respectively. What is the velocity at mid-point of A and B ?
- (a)
25 \(ms^{-1}\)
- (b)
25.5 \(ms^{-1}\)
- (c)
24 \(ms^{-1}\)
- (d)
None of the above
Use formula: v2 - u2 = 2as
(30)2 - (20)2 = 2as
In the above question the displacement after 4 s from starting point is
- (a)
zero
- (b)
- 40 m
- (c)
+ 20 m
- (d)
+ 40 m
Velocity after 2 s is v = 0 - 10 x 2 = -20 ms-1
Distance travelled is s1 = \(\frac { { v }^{ 2 } }{ 2a } =\frac { \left( -20 \right) ^{ 2 } }{ 2\times \left( -10 \right) } =\frac { 400 }{ -20 } =-20m\)
After 2 s initial velocity u = -20 ms-1
final velocity v = 0
Acceleration a = + 10 ms-2
\({ s }_{ 2 }=\frac { { v }^{ 2 }-{ u }^{ 2 } }{ 2a } =\frac { 0-400 }{ 2\times 10 } =-20m\)
Total distance = -20 + (-20) = -40 m
A car travels first half distance between two places with a speed of 30 \(kmh^{-1}\) and remaining half with a speed of 50 \(kmh^{-1}\). The average speed of the car is
- (a)
37.5 \(kmh^{-1}\)
- (b)
10 \(kmh^{-1}\)
- (c)
42 \(kmh^{-1}\)
- (d)
40 \(kmh^{-1}\)
Let s be the total distance.
For the first half, time t1 = \(\frac { s/2 }{ 30 } \)
For the rest half, t2 = \(\frac { s/2 }{ 50 } \)
Average velocity \({ v }_{ av }=\frac { s }{ { t }_{ 1 }+{ t }_{ 2 } } =\frac { s }{ \frac { s/2 }{ 30 } +\frac { s/2 }{ 50 } } =37.5\quad kmh^{ -1 }\)
A ball is dropped from a bridge 122.5 m above a river. After the ball has been falling for 2s, a second ball is thrown straight down after it. What must its initial velocity be so that both hit the water at the same time?
- (a)
\(26.1\ m/s\)
- (b)
\(29\ m/s\)
- (c)
\(32\ m/s\)
- (d)
\(36\ m/s\)
Time taken for the first ball to fall freely 122.5 m will be \(t=\sqrt{2s\over g}=\sqrt{2\times122.5\over 9.8}=5s\)
Thus, the second ball thrown after 2 s with velocity u should cover a distance 122.5 m in 3 s. Taking motion of second ball, we have
\(122.5=u\times 3+{1\over 2}\times 9.8\times (3)^2\Rightarrow u=26.1\ m/s\)
A particle starts from rest and travels a distance l with uniform acceleration, then moves uniform acceleration, then moves uniformly over a further distance 2l and finally come to rest after moving a further distance 3l under uniform retardation. Assuming entire motion to be rectilinear motion, ratio of average speed over the journey to the maximum speed to its way is
- (a)
\({1\over 5}\)
- (b)
\(2\over 5\)
- (c)
\(3\over 5\)
- (d)
\(4\over 5\)
Let \( {v_m}\) be maximum speed, \(v_m=a_1t_1;V_m=\sqrt{2a_1t}\)
\(t_2+{2l\over v_m}; v_m=a_2t_3=\sqrt{2a_3(3l)}\)
Now, average speed, \(v_{av}={l+2l+3l\over t_1+t_2+t_3}\)
\(={6l\over {({v_m\over a_1})+({2l\over v_m})+({v_m\over v_m})}}={6l\over ({10l\over v_m})}={3V_m\over 5};{v_{av}\over v_m}={3\over 5}\)
A body travels a distance of 2 m in 2 s and 2.2 m in next 4 s. What will be velocity of the body at the end of 7th second from the start?
- (a)
0.2 m/s
- (b)
0.1 m/s
- (c)
0.5 m/s
- (d)
0.7 m/s
Case I \(s=ut+{\over 2}at^2\)
\(\Rightarrow\) \(2=-u\times 2+{1\over 2}\times a \times 2^2\)
\(\Rightarrow\) \(1+u+a\) ...(i)
Case II \(4.2 = u\times 6+{1\over 2}\times a\times 6^2\)
\(0.7=u+3a\)
Subtracting Eq. (ii) from Eq. (i), we get
\(0.3=0-2a\ or\ a={-0.3\over 2}=-0.15\ m/s^2\)
From Eq.(i), we get u=1-a=1+0.15 = 1.15 m/s
For the velocity of body at the end of 7th second, we get
\(v=u+at\Rightarrow v=1.15+(-0.15)\times 7=0.1\ m/s\)
The magnitude of the displacement is equal to the distance covered in a given interval of time if the particle:
- (a)
moves with constant acceleration
- (b)
moves with constant speed
- (c)
moves with constant velocity
- (d)
none of the above
To cover the distance equal to magnitude of displacement, the particle has to move with constant velocity.
A stone is dropped from a certain height which can reach the ground in 5 sec. It is stopped after three seconds of its fall and then is again released. The total time taken by the stone to reach the ground will be:
- (a)
6 s
- (b)
6.5 s
- (c)
7 s
- (d)
7.5 s
Here, h=\(\frac{1}{2}\) x 10 x (5)2=125m
In three seconds it falls through
h1=\(\frac{1}{2}\) x 10 x (3)2=45 m
Rest 80 m is covered in 4 sec. Hence, total time taken = 3 sec + 4 see =7 sec.
A ball is dropped from the top of a tower 100 m high. Simultaneously another ball is thrown upward with a speed of 50 ms-1. After what time do they cross each other?
- (a)
1 s
- (b)
2 s
- (c)
3 s
- (d)
4 s
Suppose two balls cross each other after time t.
Then \(\frac{1}{2}\) x 10 x t2-(50t-\(\frac{1}{2}\) x 10 x t2)=100
This gives, t=2 sec
The speed of a boat is 5 km/h in still water. It crosses a river of width 1.0 km along the shortest path in 15 minutes. The velocity (in km/h) of the river water is
- (a)
5
- (b)
1
- (c)
3
- (d)
4
The resultant velocity of the boat and river is 1.0 km/0.25 h=4 km/h
Velocity of the river=\(\sqrt { { 5 }^{ 2 }-4^{ 2 } } \)=3 km/h.
A train is moving at a constant speed V. When its driver observes another train in front of him on the same track and moving in the same direction with constant speed u. If the distance between the trains be x, then what should be the minimum retardation of the train so as to avoid collision?
- (a)
\({ \left( V+\upsilon \right) }^{ 2 }/x\)
- (b)
\({ \left( V-\upsilon \right) }^{ 2 }/x\)
- (c)
\({ \left( V+\upsilon \right) }^{ 2 }/2x\)
- (d)
\({ \left( V-\upsilon \right) }^{ 2 }/2x\)
Here relative velocity of the train W.r.t. other train =V-\(\upsilon \)
\(Hence,\quad 0-{ \left( V-\upsilon \right) }^{ 2 }=2ax\quad or\quad a=-\frac { { \left( V-\upsilon \right) }^{ 2 } }{ 2x } \\ \therefore \quad Minimum\quad retardation=\frac { { \left( V-\upsilon \right) }^{ 2 } }{ 2x } .\)
A particle returns to the starting point after 10 s. If the rate of change of velocity during the motion is constant in magnitude, then its location after 7 seconds will be same as that after:
- (a)
1 sec
- (b)
2 sec
- (c)
3 sec
- (d)
3.5 sec
Because the particle returns to the starting point after 10 sec and the rate of change of velocity during motion remains constant in magnitude, hence time taken in upward motion = time taken in backward motion = 5 sec (Example: A stone thrown upwards). Hence, the location of particle after 7 sec, i. e. ,2 sec after the start of downward motion will be same as its location 2 see before the end of upward motion, i. e., after 3 sec.
A car starting from rest and moving with uniform acceleration possesses average velocities 5ms-1 , 10 ms-1 and 15 ms -1 in the first, second and third seconds. What is
the total distance covered by the car in these three seconds?
- (a)
15 m
- (b)
30m
- (c)
55 m
- (d)
None of these
A particle in uniform motion can possess:
- (a)
radial acceleration
- (b)
tangential acceleration
- (c)
both radial and tangential accelerations
- (d)
neither radial nor tangential acceleration
A thief is running away on a straight road in a jeep moving with a speed of 9 ms-1 A policeman chases him on a motor cycle moving at a speed of 10 ms-1. If the instantaneous separation of the jeep from the motor cycle is 100 m, how long will it take for the policeman to catch the thief?
- (a)
1s
- (b)
19s
- (c)
90s
- (d)
100s
Relative velocity of police man w.r.t. the thief is 10ms-1- 9ms-1= 1 m s-1. Since, the separation between them is 100 m, hence the time taken will be 100 s.
Two particles P1 and P2 are moving with velocities V1 and V2 respectively. Which of the following statements about their relative velocity vr12 is true?
- (a)
vr12 cannot be greater than v1 + v2
- (b)
vr12 cannot be greater than v1 - v2
- (c)
vr12 > (v1 + v2)
- (d)
vr12 < (v1 + v2)
A 2 m wide truck is moving with a uniform speed Vo = 8m/s along a straight horizontal road. A pedestrian starts to cross the road with a uniform speed v when the truck is 4 m away from him. The minimum value of v so that he can cross the road safely is:
- (a)
2.62 m/s
- (b)
4.6 m/s
- (c)
3.57 m/s
- (d)
1.414 m/s
Let the man starts crossing the road at an angle θ with the roadside. For safe crossing, the condition is that the man must cross the road by the time truck describes the distance (4+2 cotθ)
So, \(\frac{4+2cot\theta}{8}=\frac{2lsin\theta}{v}\)
or v=\(\frac{8}{2sin\theta+cos\theta}\)
For minimum v, \(\frac{dv}{d\theta}=0\)
or \(\frac{-8(2cos\theta-sin\theta)}{(2sin\theta+cos\theta)^2}=0\)
or 2cosθ-sinθ=0
or tanθ=2
So, sinθ=\(\frac{2}{\sqrt{5}}; cos\theta=\frac{1}{\sqrt{5}}\)
ஃ vmin=\(\frac{8}{2(\frac{2}{\sqrt{5}})+\frac{1}{\sqrt{5}}}=\frac{8}{\sqrt{5}}=3.57m/s\)
A body starts from rest with a uniform acceleration of 2m/ s2 for 10 sec, it moves with constant speed for 30 sec then decelerates by 4 m/s2 to zero. What is the distance covered by the body?
- (a)
750 m
- (b)
850 m
- (c)
600 m
- (d)
None of these
u=0, a=2m/s2, t=10 sec
ஃ s1=ut+\(\frac{1}{2}at^2=0+\frac{1}{2}\times2\times100=100m\)
Velocity after 10 sec, v =u + at = 0 + 2 x 10 = 20 m/s
ஃ s2 = v x 30 = 20 x 30 = 600 m
Final velocity = 0, a = -4 m/s2
ஃ 0 = v2 + 2as3
0 = (20)2 - 2 x 4 x s3
ஃ \(s_3=\frac{400}{8}=50m\)
s = s1 + s2 + s3
= 100+ 600+ 50 = 750 m.
Water drops fall from a tap on the floor 5 m below at regular intervals of time, the first drop striking the floor when the fifth drop begins to fall. The height at which the third drop will be, from ground, at the instant when the first drop strikes the ground, will be:(g = 10m s-2)
- (a)
1.25 m
- (b)
2.15 m
- (c)
2.75 m
- (d)
3.75 m
By the time 5th water drop starts falling, the first water drop reaches the ground.
As u = 0, h=\(\frac{1}{2}at^2=\frac{1}{2}\times10\times t^2\)
or 5=\(\frac{1}{2}\times10\times t^2\) or t=1 sec
Hence, the interval of each water drop=\(\frac{1 sec}{4}=0.25sec\)
When the 5th drop starts its journey towards ground, the third drop travels in air for 0.25 + 0.25 = 0.5 sec.
ஃ Height (distance) covered by 3rd drop in air is,
h1=\(\frac{1}{2}gt^2=\frac{1}{2}\times10\times(0.5)^2\)
= 5 x 0.25 = 1.25 m
ஃ The third water drop will be at a height of= 5 - 1.25 = 3.75 m.
Two trains A and B initially 120 km apart, start moving towards each other on the same track with a velocity of 60 km/h each. At the moment of start A blows a whistle,
which reflects on B and subsequently reflects from A and so on. Take the velocity of sound waves in air as 1200 km/hr. The distance travelled by sound waves before the trains crash will be:
- (a)
2400 km
- (b)
1200 km
- (c)
240 km
- (d)
120 km
A body moves for a total of nine second starting from rest with uniform acceleration and then with uniform retardation, which is twice the value of uniform acceleration. The time taken during retardation is:
- (a)
3 s
- (b)
4.5 s
- (c)
5 s
- (d)
6 s
A boat travels 50 km east, then 120 km north and finally it comes back to the starting point through the shortest distance. The total time of journey is 3 hours. What is the average velocity, over the entire trip?
- (a)
0 km h-1
- (b)
100 km h-1
- (c)
17 km h-1
- (d)
33.33 km h-1
- (e)
86.7 km h-1
Average velocity=\(\frac{displacement}{total\quad time\quad taken}\)
Displacement = 0, as boat cOip.es back at the starting point.
ஃ Average velocity = 0.
A metro train starts from rest and in five seconds achieves 108 km/h. After that it moves with constant velocity and comes to rest after travelling 45 m with uniform retardation. If total distance travelled is 395 m, then total time of travelling is:
- (a)
12.2 sec
- (b)
15.3 sec
- (c)
9.0 sec
- (d)
17.2 sec
A car moves from X to Y with a uniform speed vu and returns to Y with a uniform speed vd. The average speed for this round trip is:
- (a)
\(\sqrt{v_uv_d}\)
- (b)
\(\frac{v_dv_u}{v_d+v_u}\)
- (c)
\(\frac{v_u+v_d}{2}\)
- (d)
\(\frac{2v_dv_u}{v_d+v_u}\)
Average speed=\(\frac{Total\quad distance\quad travelled}{Total\quad time\quad taken}\)
=\(\frac{s+s}{t_1+t_2}=\frac{2s}{\frac{s}{v_u}+\frac{s}{v_d}}=\frac{2v_d v_u}{vd+v_u}\)
A particle is thrown above, then the correct v-t graph will be:
- (a)
- (b)
- (c)
- (d)
From an elevated point P, a stone is projected vertically upwards. When the stone reaches a distance h below P, its velocity is double of its velocity at a height h above P. The greatest height attained by the stone from the point of projection P is:
- (a)
\(\frac{3}{5}h\)
- (b)
\(\frac{5}{3}h\)
- (c)
\(\frac{7}{5}h\)
- (d)
\(\frac{5}{7}h\)
- (e)
\(\frac{2}{3}h\)
From equation of motion, at a point above point P
v2 = u2 - 2gh...(i)
At a point below point P
(2v)2 = v2 + 4gh
= u2 - 2gh + 4gh
= u2 + 2gh...(ii)
Adding, equations (i) and (ii), we get;
5v2 = 2u2
or u2=\(\frac{5v^2}{2}\)...(iii)
Subtracting eqn. (i) from eqn. (ii), we get;
\(v^2=\frac{4}{3}gh\)...(iv)
Maximum height attained by stone is
H=\(\frac{u^2}{2g}=\frac{\frac{5v^2}{2}}{2g}=\frac{\frac{5}{2}[\frac{4}{3}gh]}{2g}=\frac{5}{3}h\)
Hence, correct answer is (b).
A particle moves along x-axis and its displacement at any time is given by x( t) = 2t3 - 3t2 + 4t in Sl units. The velocity of the particle when its acceleration is zero, is :
- (a)
2.5 ms-1
- (b)
3.5ms-1
- (c)
4.5ms-1
- (d)
8.5ms-1
Which of the following velocity-time graphs shows a realistic situation for a body in motion?
- (a)
- (b)
- (c)
- (d)