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Frame of Reference
The frame of reference is a suitable coordinate system involving space and time used as a
reference to study the motion of different bodies. The most common reference frame is
the cartesian frame of reference involving ( x, y, z and t ).
(i) Inertial Frame of Reference A frame of reference which is either at rest or moving
with constant velocity is known as inertial frame of reference. Inertial frame of
reference is one in which Newton’s first law of motion holds good.
(ii) Non-Inertial Frame of Reference A frame of reference moving with some
acceleration is known as non-inertial frame of reference. Non-inertial frame of
reference in one which Newton’s law of motion does not hold good.
Motion in a Straight Line
The motion of a point object in a straight line is one dimensional motion. During such a
motion the point object occupies definite position on the path at each instant of time.
Different terms used to described motion are defined below:
Distance and Displacement
l Distance is the total length of the path travelled by a particle in a given interval of
time. It is a scalar quantity and its SI unit is metre (m).
l Displacement is shortest distance between initial and final
positions of a moving object. It is a vector quantity and its SI
unit is metre.
From the given figure, mathematically it is expressed as,
∆ r = r – r 2 1
l Displacement of motion may be zero or negative
but path length or distance can never be negative.
l For motion between two points displacement is single valued while distance depends
on actual path and so can have many values.
l Magnitude of displacement can never be greater than distance. However, it can be
equal, if the motion is along a straight line without any change in direction.
Kinematics
Learning & Revision for the Day
u (^) Frame of Reference u (^) Motion in a Straight Line u (^) Uniform and Non-uniform Motion u (^) Uniformly Accelerated Motion
u (^) Elementary Concept of Differentiation and Integration for Describing Motion u (^) Graphs
Y
X
r 2
r 1 ∆ r
Distance
Displacement
A
B
O
DAY TWO KINEMATICS
Speed and Velocity
l Speed is defined as the total path length (or actual distance
covered) by time taken by object.
Speed =
Distance
Time taken
It is scalar quantity. Its SI unit is m/s.
l Average Speed, v
av t
Total distance travelled
l When a body travels equal distance with speeds v 1 and v 2 ,
the average speed ( v ) is the harmonic mean of the two
speeds.
v v 1 v 2
l When a body travels for equal time with speeds v 1 and v 2 ,
the average speed v is the arithmetic mean of the two speeds.
v
v v
av =^
l Velocity is defined as ratio of displacement and
corresponding time interval taken by an object.
i.e. velocity =
Displacement
time interval
l Average velocity = Total displacement
Total time taken
x x
t t
x
t
2 1 2 1
Here, x 2 and x 1 are the positions of a particle at the time t 2
and t 1 respectively, with respect to a given frame of
reference.
l For a moving body speed can never be negative or zero
while velocity can be negative and zero.
l The instantaneous speed is average speed for infinitesimal
small time interval (i.e. ∆ t → 0)
i.e. Instantaneous speed, v
s
t
ds
t dt
→
lim
∆
l The instantaneous velocity (or simply velocity) v of a
moving particle is v
x
t
dx
t dt
∆ →
lim
0
It (at a particular time) can be calculated as the slope (at
that particular time) of the graph of x versus t.
Uniform and Non-uniform Motion
l An object is said to be in uniform motion if its velocity is
uniform i.e. it undergoes equal displacement in equal may
be intervals of time, howsoever small these interval.
l An object is said to be in non-uniform motion if its
undergoes equal displacement in unequal intervals of
time., howsoever small these intervals may be.
Acceleration
Acceleration of an object is defined as rate of change
of velocity. It is a vector quantity having unit m/s^2 or ms−^2.
It can be positive, zero or negative.
Average and Instantaneous Acceleration If velocity of a
particle at instant t is v 1 and at instant t 2 is v 2 , then
l Average acceleration, a v^ v
t t
v
av t
2 1 2 1
l Instantaneous acceleration, a v
t
dv
t dt
→
lim
∆
Uniformly Accelerated Motion
l A motion, in which change in velocity in each unit of time
is constant, is called an uniformly accelerated motion. So,
for an uniformly accelerated motion, acceleration is constant.
l For uniformly accelerated motion are given below
Equations of motion, v = u + at …(i)
s = ut + at
2 …(ii)
and v^2 = u^2 + 2 as …(iii)
where, u = initial velocity, v = velocity at time t
and s = displacement of particle at time t.
l Equation of uniformaly accelerated motion under gravity
are
(i) v = u − gt (ii) h = ut − 1 gt
2 (iii) v 2 = u 2 − 2 gh
Elementary Concept of Differentiation and Integration for Describing Motion
l At an instant t , the body is at point P x y z ( , , ).
Thus, velocity along X -axis, v
dx
x dt
Acceleration along X -axis is a
dv
x dt
= x
Velocity along Y -axis is v
dy
y dt
Acceleration along Y -axis is a
dv
y dt
=^ y
Similarly, v
dz
z dt
= and a
dv
z dt
= z
l For a accelerating body
(i) If ax variable, x = (^) ∫ v dtx , (^) ∫ dv (^) x = (^) ∫ a dtx
(ii) If a (^) y is variable, y = (^) ∫ v (^) ydt , (^) ∫ dv (^) y =∫ a dty
(iii) If az is variable, z = (^) ∫ v (^) zdt , (^) ∫ dv (^) z =∫ a dtz
Also, distance travelled by a particle is s = (^) ∫| | v dt
(i) x -component of displacement is ∆ x = (^) ∫ v (^) xdt
(ii) y -component of displacement is ∆ y = (^) ∫ v (^) ydt
(iii) z -component of displacement is ∆ z = (^) ∫ v (^) zdt
DAY TWO KINEMATICS
1 An aeroplane flies 400 m from North and then flies 300 m
South and then flies 1200 m upwards, then net
displacement is
(a) 1200 m (b) 1300 m (c) 1400 m (d) 1500 m
2 The correct statement from the following is
(a) A body having zero velocity will not necessarily have zero acceleration (b) A body having zero velocity will necessarily have zero acceleration (c) A body having uniform speed can have only uniform acceleration (d) A body having non-uniform velocity will have zero acceleration
3 A vehicle travels half the distance L with speed v 1 and
the other half with speed v 2 , then its average speed is
(a) v^^1 v^2 2
v v v v
(c) 2 1 2 1 2
v v v + v
(d) 2 1 2 1 2
( v v ) v v
4 A particle moves along the sides AB BC , , CD of a square
of side 25 m with a velocity of 15 m/s. Its average
velocity is
(a) 5 m/s (b) 7.5 m/s (c) 10 m/s (d) 15 m/s
5 A body sliding down on a smooth inclined plane slides
down 1/4th of plane’s length in 2 s. It will slide down the
complete plane in
(a) 4 s (b) 5 s (c) 2 s (d) 3 s
6 Three particles P Q , and R are situated at corners of an
equilateral triangle of side length ( d ). At t = 0, they
started to move such that P is moving towards Q , Q is
moving towards R and R is moving towards P at every
instant. After how much time (in second) will they meet
each other?
(a) d u
(b)^2 3
d u
(c) 2 3
d ( ) u
(d) d ( ) 3 u
7 A body is thrown vertically upwards in air, when air
resistance is taken into account, the time of ascent is t 1
and time of descent is t 2 , then which of the following is
true?
(a) t (^) 1 = t 2 (b) t (^) 1 < t 2 (c) t (^) 1 > t 2 (d) t (^) 1 ≥ t 2
8 A stone falls freely from rest and the total distance covered
by it in the last second of its motion equals the distance
covered by it in the first three seconds of its motion. The
stone remains in the air for
(a) 6 s (b) 5 s (c) 7 s (d) 4 s
9 The motor of an electric train can give it an acceleration
of 1 ms– 2^ and brakes can give a negative acceleration of
3 ms –^^2. The shortest time in which the train can make a
trip between the two stations 1215 m apart is
(a) 113.6 s (b) 56.9 s (c) 60 s (d) 55 s
10 A train is moving along a straight path with a uniform
acceleration. Its engine passes a pole with a velocity of
60 kmh−^1 and the end (guard’s van) passes across the
same pole with a velocity of 80 kmh−^1. The middle point
of the train will pass the same pole with a velocity
(a) 70 kmh−^1 (b) 70.7 kmh−^1 (c) 65 kmh−^1 (d) 75 kmh−^1
11 The acceleration experienced by a moving boat after its
engine is cut-off, is given by a = − kv 3 , where k is a
constant. If v 0 is the magnitude of velocity at cut-off,
then the magnitude of the velocity at time t after the
cut-off is
(a) v kt v
0 0 2 2 (b)^
v kt v
0 0
(c) v kv
0 0
(d) v kt v
0 0
12 A body moving with an uniform acceleration describes
12 m in the 3rd second of its motion and 20 m in the
5th second. Find the velocity after the 10th second.
(a) 40 ms–1^ (b) 42 ms– (c) 52 ms–1^ (d) 4 ms–
13 A train accelerating uniformly from rest attains a
maximum speed of 40 ms–1^ in 20 s. It travels at this
speed for 20 s and is brought to rest with an uniform
retardation in the next 40 s. What is the average velocity
during this period?
(a)^80 3
ms–1^ (b) 25 ms– (c) 40 ms–1^ (d) 30 ms–
15
FOUNDATION QUESTIONS EXERCISE
DAY PRACTICE SESSION 1
P R
Q
u
u
u
B C
v
A D
DAY TWO
14 A frictionless wire AB is fixed on a sphere of radius R. A
very small spherical ball slips on this wire. The time taken
by this ball to slip from A to B is
(a)^2 gR g cosθ
(b) 2 gR g
cosθ (^) (c) 2 R g
(d) gR g cosθ
15 A balloon is going upwards with velocity12 ms–1. It
releases a packet when it is at a height of 65 m from the
ground. How much time the packet will take to reach the
ground if g = 10 ms–2^?
(a) 5 s (b) 6 s (c) 7 s (d) 8 s
16 A ball is dropped from the top of a building. The ball
takes 0.5 s to fall past the 3 m length of window some
distance below from the top of building. With what speed
does the ball pass the top of window?
(a) 6 ms–1^ (b) 12 ms–1^ (c) 7 ms–1^ (d) 3.5 ms–
17 A body starts from the origin and moves along the axis
such that the velocity at any instant is given by
v = 4 t^3 − 2 t where, t is in second and the velocity in
ms −^1. Find the acceleration of the particle when it is at
a distance of 2 m from the origin.
(a) 28 ms–2^ (b) 22 ms– 2^ (c) 12 ms–2^ (d) 10 ms–
18 A point initially at rest moves along the x -axis. Its
acceleration varies with time as a = ( 5 t +6 ms) –2. If it
starts from the origin, the distance covered by it in 2s is
(a) 18.66 m (b) 14.33 m (c) 12.18 m (d) 6.66 m
19 A rod of length l leans by its upper end against a smooth
vertical wall, while its other end leans against the floor.
The end that leans against the wall moves uniformly
downwards. Then,
(a) the other end also move uniformly (b) the speed of other end goes on increasing (c) the speed of other end goes on decreasing (d) the speed of other end first decreases and then increases
20 Which of the following distance-time graphs is not possible?
21 Look at the graphs (i) to (iv) in figure carefully and
choose, which of these can possibly represent
one-dimensional motion of particle?
(a) Both (i) and (ii) (b) Only (iv) (c) Only (iii) (d) Both (iii) and (iv)
22 Figure shows the time-displacement curve of the particles
P and Q. Which of the following statement is correct?
(a) Both P and Q move with uniform equal speed (b) P is accelerated and Q moves with uniform speed but the speed of P is more than the speed of Q (c) Both P and Q moves with uniform speeds but the speed of (^) P is more than the speed of (^) Q (d) Both P and Q moves with uniform speeds but the speed of Q is more than the speed of P
23 The velocity versus time curve of a moving point is shown
in the figure below. The maximum acceleration is
(a) 1 ms−^2 (b) 6 ms−^2 (c) 2 ms−^2 (d) 1 .5 ms−^2
O
R
A
B
C
q
t (a)
x
t (b)
x
t
(c)
t (d)
x x
x
t
(i)
x
t
(ii) x
t
(iii)
x
t
(iv)
O t
x P Q
O
A
F
B C
D E
Velocity (ms –1)
Time (s)
1 The velocity-time plot for a particle moving on a straight
line is as shown in figure, then
(a) the particle has a constant acceleration (b) the particle has never turned around (c) the average speed in the interval 0 to 10 s is the same as the average speed in the interval 10 s to 20 s (d) Both (a) and (c) are correct
2 A body is at rest at x = 0. At t = 0, it starts moving in the
positive x -direction with a constant acceleration. At the
same instant another body passes through x = 0 moving
in the positive x -direction with a constant speed. The
position of the first body is given by x 1 ( ) after time t t and
that of the second body by x 2 ( ) after the same time t
interval. Which of the following graphs correctly
describes ( x 1 − x 2 )as a function of time t?
3 The velocity of a particle is v = v 0 + gt + ft^2. If its position
is x = 0 at t = 0, then its displacement after unit time
( t = 1 is)
(a) v (^) 0 + 2 g + 3 f (b) v (^) 0 g^ f 2 3
(c) v (^) 0 + g + f (d) v (^) 0 g^ f 2
4 A particle located at x = 0 at time t = 0, starts moving
along the positive x -direction with a velocity v that varies
as v = α x. The displacement of the particle varies
with time as
(a) t^2 (b) t (c) t 1 2/^ (d) t^3
5 A stone is dropped from a certain height and reaches the
ground in 5 s. If the stone is stopped after 3 s of its fall
and then allowed to fall again, then the time taken by the
stone to reach the ground after covering the remaining
distance is
(a) 2 s (b) 3 s (c) 4 s (d) None of these
6 A point moves with a uniform acceleration and v 1 , v 2 , v 3
denote the average velocities in three successive
intervals of time t 1 , t 2 , t 3. Which of the following relations is
correct?
(a) ( v (^) 1 − v (^) 2 ) : ( v (^) 2 − v (^) 3 ) = ( t (^) 1 − t (^) 2 ) : ( t (^) 2 + t 3 ) (b) ( v 1 (^) − v (^) 2 ) : ( v (^) 2 − v (^) 3 ) = ( t (^) 1 + t (^) 2 ) : ( t (^) 2 + t 3 ) (c) ( v (^) 1 − v (^) 2 ) : ( v (^) 2 − v (^) 3 ) = ( t (^) 1 − t (^) 2 ) : ( t (^) 1 − t 3 ) (d) ( v 1 (^) − v (^) 2 ) : ( v (^) 2 − v (^) 3 ) = ( t (^) 1 − t (^) 2 ) : ( t (^) 2 − t 3 )
7 From the top of a tower of height 50 m, a ball is thrown
vertically upwards with a certain velocity. It hits the
ground 10 s after it is thrown up. How much time does it
take to cover a distance AB where A and B are two points
20 m and 40 m below the edge of the tower?
( take, g = 10 ms –^2 )
(a) 2.0 s (b) 1 s (c) 0.5 s (d) 0.4 s
8 Car A is moving with a speed of 36 kmh−^1 on a two lane
road. Two cars B and C , each moving with a speed of
54 kmh−^1 in opposite directions on the other lane are
approaching car A. At certain instant of time, when the
distance AB = AC = 1 km, the driver of car B decides to
overtake A before C does. What must be the minimum
acceleration of car B , so as to avoid an accident?
(a) 1 ms−^2 (b) 4 ms−^2 (c) 2 ms−^2 (d) 3 ms−^2
9 The displacement x of a particle varies with time,
according to the relation x a
b
= ( 1 − e − bt ). Then
(a) the particle can not reach a point at a distance x from its starting position, if x > a / b (b) at t = 1/ b , the displacement of the particle is nearly ( /2 3 ) ( / a b ) (c) the velocity and acceleration of the particle at t = 0 are a and − ab respectively (d) the particle will come back its starting point as t → ∞
18 40 DAYS ~ JEE MAIN PHYSICS DAY TWO
v (ms –1) t (s)
PROGRESSIVE QUESTIONS EXERCISE
DAY PRACTICE SESSION 2
O t
( x (^) 1 – x 2 )
(a)
O t
( x (^) 1 – x 2 )
(b)
O t
(c)
O t
(d)
( x (^) 1 – x 2 ) (^ x^ 1 –^ x 2 )
DAY TWO KINEMATICS
10 From the top of a tower, a stone is thrown up which
reaches the ground in time t 1. A second stone thrown
down, with the same speed, reaches the ground in time
t 2. A third stone released from rest, from the same
location, reaches the ground in a time t 3. Then,
(a)^1 1 t (^) 3 t (^) 2 t 1
= − (b) t (^) 32 = t (^) 12 − t 22
(c) t t t 3 =^1
(d) t (^) 3 = t t 1 2
11 A bullet moving with a velocity of 100 ms −^1 can just
penetrate two plancks of equal thickness. The number of
such plancks penetrated by the same bullet, when the
velocity is doubled, will be
(a) 4 (b) 6 (c) 8 (d) 10
12 The acceleration in ms−^2 of a particle is given by,
a = 3 t^2 + 2 t + 2 where, t is time. If the particle starts out
with a velocity v = 2 ms −^1 at t = 0, then the velocity at the
end of 2 s is
(a) 36 ms−^1 (b) 18 ms−^1 (c) 12 ms−^1 (d) 27 ms−^1
13 A car, starting from rest, accelerates at the rate f through
a distance s , then continues at constant speed for time t
and then decelerates at the rate
f
to come to rest. If the
total distance travelled is 15 s, then
(a) s = f t (b) s = 1 f t 6
2
(c) s = 1 f t 72
(^2) (d) s = 1 f t 4
2
14 The displacement of a particle is given by x = ( t − 2 ) 2
where, x is in metres and t in seconds. The distance
covered by the particle in first 4 seconds is
(a) 4 m (b) 8 m (c) 12 m (d) 16 m
15 A metro train starts from rest and in five seconds
achieves 108 kmh −^1. 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,
find total time of travelling.
(a) 12.2 s (b) 15.3 s (c) 9 s (d) 17.2 s
16 From a tower of height H , a particle is thrown vertically
upwards with a speed u. The time taken by the particle to
hit the ground, is n times that taken by it to reach the
highest point of its path. The relation between H , u and n
is
(a) 2 gH = n u^2 2 (b) gH = ( n − 2 ) 2 u^2 (c) 2 gH = nu^2 ( n − 2 ) (d) gH = ( n − 2 ) 2 u^2
17 An object, moving with a speed of 6.25 ms −^1 , is
decelerated at a rate given by dv
dt
= − 2 5. v , where, v is
the instantaneous speed. The time taken by the object, to
come to rest, would be
(a) 2 s (b) 4 s (c) 8 s (d) 1 s
18 A ball is released from the top of a tower of height
h metre. It takes T second to reach the ground. What is
the position of the ball in
T
s?
(a) h 9
m from the ground (b)^7 9
h (^) m from the ground
(c)^8 9
h (^) m from the ground (d) 17 18
h (^) m from the ground
19 Two stones are thrown up simultaneously from the edge
of a cliff 240 m high with initial speed at 10 ms −^1 and
40 ms −^1 , respectively. Which of the following graph best
represents the time variation of relative position of the
second stone with respect to the first? (Assume stones
do not rebound after hitting the ground and neglect air
resistance, take g = 10 ms−^2. The figures are schematic
and not drawn to scale
1 (a) 2 (a) 3 (c) 4 (a) 5 (a) 6 (b) 7 (b) 8 (b) 9 (b) 10 (b)
11 (d) 12 (b) 13 (b) 14 (c) 15 (a) 16 (d) 17 (b) 18 (a) 19 (c) 20 (c)
21 (d) 22 (c) 23 (b) 24 (b) 25 (b) 26 (b) 27 (c) 28 (d) 29 (d) 30 (c)
1 (d) 2 (b) 3 (b) 4 (a) 5 (c) 6 (b) 7 (d) 8 (a) 9 (b) 10 (d)
11 (c) 12 (b) 13 (c) 14 (b) 15 (d) 16 (c) 17 (a) 18 (c) 19 (c)
SESSION 1
SESSION 2
ANSWERS
(a)
t 8 12 t (s)
240
(^) ( y (^) 2 – y 1 )m
(b)
t (s)
240
(^) ( y (^) 2 – y 1 )m
(c)
8 12 t (s)
240
(^) ( y (^) 2 – y 1 )m
(d)
8 12 t (s)
240
(^) ( y (^) 2 – y 1 )m
DAY TWO KINEMATICS
12 Using, s n = u + a^ n −
= u + a^ ( 2 × 3 − 1 ) …(i)
20 2
= u + a^ ( 2 × 5 − 1 ) …(ii) On subtracting Eq. (i) from Eq. (ii), we get 8 2
=^ a ( 10 − 6 ) = 2 a a = 4 ms– 2 From Eq. (i), 12 4 2
= u + × 5 u = 2 ms−^1 From v = u + at = 2 + 4 × 10 = 42 ms–
13 As, v = u + at 1 …(i)
40 = 0 + a × 20 a = 2 ms– Now, v^2 − u^2 = 2 as 402 − 0 = 2 × 2 × s 1 s 1 = 400 m s (^) 2 = v × t 2 …(ii) = 40 × 20 = 800 m and v = u + at …(iii) 0 = 40 + a × 40 , a = − 1 ms– 2 Also, v^2 − u^2 = 2 as 0 2 − 402 = 2 ( − 1 ) s 3 s 3 = 800 m ∴Total distance travelled = s (^) 1 + s (^) 2 + s 3 = 400 + 800 + 800 = 2000 m and total time taken = 20 + 20 + 40 = 80 s ∴ Average velocity = 2000 80
= 25ms−^1
14 Acceleration of the body down the
plane (^) = g cos θ Distance travelled by ball in time t second is AB = 1 g t 2
( cos θ) 2 ...(i) From ∆ ABC , AB = 2 R cos θ ...(ii) From Eqs. (i) and (ii), we get 2 1 2
R cos θ = g cosθ t^2
t R g
2 =^4
or t R g
15 a = + g = 10 ms– 2^ ,
s = 65m, t =? As, s = ut + 1 at 2
2
⇒ 65 = − 12 t + 5 t^2 5 t^2 − 12 t − 65 = 0 This gives, t =
= 12 ±^38 =
5s
16 From s = ut + 1 at
x = 0 + 1 × t = t 2 10 2 52 …(i)
Also, x + 3 = 0 + 1 × t + 2
^
t^2 t …(ii)
Subtract Eq. (i) from Eq. (ii), we get 3 5 1 4
^
t + t
− = 5 t
7 4
= 5 t or t = 7 20
s
From v = u + at , v = 0 + 10 ×^7 20 = 3 5. ms−^1
17 v = 4 t^3 − 2 t …(i)
dx dt
= 4 t^3 − 2 t On integration, we get, x = 2 = t^4 − t^2 Let t^2 = α ∴ 2 = α^2 −α …(ii) Let t^2 = α α 2 − α− 2 = 0 ( α − 2 ) ( α + 1 ) = 0 ∴ α = 2, α = − 1, which is not possible t^2 = α = 2 or t = 2, Differentiating Eq. (i) w.r.t. t , dv dt
= 12 t 2 − 2
a = 12 × 2 − 2 =22 ms– 2
18 Acceleration, a dv
dt
= = 5 t + 6 On integrating, we get v = 5 2
t^2 + 6 t = dx dt Integrating again, x = 5 t + t 6
3 2
At t = 2s, x = 5 × + × = 6
8 3 4 18.66 m
19 If ( , x 0 and () y , 0 are the coordinates of)
the end points of the rod at a given location, then x^2 + y^2 = l^2
Differentiating it w.r.t. t , we get 2 x dx 2 0 dt
y dy dt
dx dt
y dy dt x
and v y x x = − vy As, y decreases, x increases, so v (^) x decreases. v (^) x becomes zero when y is zero.
20 The distance travelled can never be
negative in one dimensional motion.
21 In one dimensional motion, there is a
single value of displacement at one particular time.
22 As x - t graph is a straight line in either
case, velocity of both is uniform. As the slope of x - t graph for P is greater, therefore, velocity of P is greater than that of Q.
23 Maximum acceleration is represented by
the maximum slope of the velocity-time graph. Thus, it is the portion CD of the graph, which has a slope = − −
= 6 ms −^2.
24 Displacement is the algebraic sum of area
under velocity-time graph. As, displacement = area of triangles
∆ OAB + ∆ ABC + ∆ CDH + HEFG
= 1 × × + × × +
× 1 × ( − 2 )+ 1 × 1
= 2 + 1 − 1 + 1 =3 m
x
3 m
( , 0) y
( , 0) x
D
t (s)
E F
A
1 B
H
G
O C
v (ms –1)
DAY TWO
25 If velocity versus time graph is a straight
line with negative slope, then acceleration is constant and negative. With a negative slope distance-time graph will be parabolic s = ut − at ^
So, option (b) will be incorrect.
26 Initially velocity keeps on decreasing at
a constant rate, then it increases in negative direction with same rate.
27 Case I
Relative velocity is v (^) 1 + v 2 = 8 Case II Relative velocity is v (^) 1 − v 2 = 2 On solving, v 1 = 5 ms−^1 , v 2 = 3 ms−^1
28 When a particle moves with constant
velocity, then acceleration of particle is zero and hence particle is not able to change the direction. Hence, statement I is false while statement II is true. Hence, correct answer is (d).
29 When two objects moving in opposite
direction, then their relative velocity becomes ( v (^) 1 + v 2 ), hence statement I is false. When moves in same direction, then relative velocity v = ( v (^) 1 − v 2 ), hence statement II is true. Hence, correct answer is (d).
30 Without changing direction of velocity,
it is possible to change the acceleration of a moving particle, hence statement I is true, while statement II is false. Hence, correct answer is (c).
SESSION 2
1 The slope of velocity-time graph gives
acceleration. Since, the given graph is a straight line and slope of graph is constant. Hence acceleration is constant. Thus, (a) is correct. The area of v - t graph between 0 to 10 s is same as between 10 s to 20 s.
2 As, x 1 t^1 at^2
( ) = and x (^) 2 ( ) t = vt
∴ x 1 (^) x (^) 2 1 at^2 vt 2 − = − (parabola) Clearly, graph (b) represents it correctly.
3 As, v = v 0 + gt + ft^2 or
dx dt
= v (^) 0 + gt + ft^2 ⇒ dx = ( v (^) 0 + gt + ft^2 ) dt So, (^) ∫ 0^ x dx = (^) ∫ 01 ( v (^) 0 + gt + ft^2 ) dt
⇒ x = v (^) 0 + g^ + f 2 3
4 Given, v = α x
or dx dt
= α x Q v dx dt
^
or dx x
= α dt On integration, dx x ∫ 0 x^ =^ ∫ 0^ t α dt [Q at t = 0 , x = 0 and let at any time t , particle is at x ] ⇒ x^ t
1 2^ x (^1 )
/ /
= α
or x 1 2 t 2
/ (^) = α
or x = α × t
(^2 ) 4 ∴ x ∝ t^2
5 From s = ut + 1 at
s = 0 + 1 × × 2 10 52 = 125m Distance covered in 3 s, = 0 + 1 × × 2
10 32 = 45m Distance to be covered = 125 − 45 = 80 m From s = ut + 1 at 2
2
80 0 1 2
= + × 10 t^2
⇒ t^2 5
∴ t = 4 s
6 Suppose velocity at O = zero
As average velocity in interval t 1 is v 1 , ∴ Velocity at A = v 1 As average velocity in interval (^) t 2 is v 2 , ∴ Velocity at B = ( v (^) 2 − v 1 ) As average velocity in interval t 3 is v 3 , Velocity at C = ( v (^) 3 − v (^) 2 + v 1 ) Using v = u + at v (^) 1 = 0 + at 1 …(i) ( v (^) 2 − v (^) 1 ) = 0 + a t ( (^) 1 + t 2 ) …(ii) ( v (^) 3 − v (^) 2 + v (^) 1 ) = 0 + a t ( (^) 1 + t (^) 2 + t 3 ) …(iii) Subtract Eq. (i) from Eq. (iii), we get ( v (^) 3 − v (^) 2 ) = a t ( (^) 2 + t 3 ) …(iv) Divide Eq. (ii) by Eq. (iv), we get ( ) ( )
v v v v
a t t a t t
2 1 3 2
1 2 2 3
v v v v
t t t t
1 2 2 3
1 2 2 3
7 Given, v = − u , a = g = 10 ms–2^ ,
s = 50 m, t = 10 s As, s = ut + 1 at 2
= − u × + × 10 × 102 On solving, u = 45ms−^1 If t 1 and t 2 are the timings taken by the ball to reach the points A and B respectively, then 20 45 1 2
= − t 1 (^) + × 10 × t 12
40 45 1 2 = − t 2 (^) + × 10 × t 22 On solving, we get t 1 = 9 4. s and t 2 = 9 8. s Time taken to cover the distance (^) AB , = ( t (^) 2 − t 1 ) = 9 8. − 9 4. =0 4. s
8 Let us suppose that the cars A and B are
moving in the positive x -direction. Then, car C is moving in the negative x -direction. Therefore, v (^) A = 36 kmh −^1 = 10 ms−^1 v (^) B = 54 kmh −^1 = 15 ms−^1 and vC = − 54 kmh− 1 = − 15 ms−^1 Thus, the relative speed of B with respect to A is, v (^) BA = v (^) B − vA = 15 − 10 =5 ms −^1 and the relative speed of C with respect to A is, v (^) CA = v (^) C − vA = − 15 − 10 = − 25ms −^1 At time t = 0, the distance between A and B = distance between A and C = 1 km = 1000 m. The car (^) C covers a distance (^) AC = 1000 m and reaches car A at a time t given by t AC vCA
m 1 40 ms
s
Car B will overtake car A just before car C does and the accident can be avoided if it acquires a minimum acceleration a such that it covers a distance, s = AB = 1000 m in time t = 40 s travelling with a relative speed of u = v (^) BA = 5ms −^1. This gives, from s = ut + 1 at a = 2
(^2) , 1 ms − 2
Time taken in travelling 45 m is t 3 30 10
= = 3 s Now, total distance = 395 m i.e. 75 + s ′ + 45 = 395 m or s ′ = 395 − ( 75 + 45 ) = 275 m ∴ t 2 = 275 = 30
9.2 s Hence, total time taken in whole journey = t (^) 1 + t (^) 2 + t 3 = 5 + 9 2. + 3 = 17 2. s
16 Time taken to reach the maximum
height, t u (^1) g
If t 2 is the time taken to hit the ground, i.e. − H = ut (^) 2 −^1 gt 22 2 But t (^) 2 = nt 1 [Given] So, − H = u nu − g
g n u g
2 2 2
− H = nu − g
n u g
H n u g
nu g
n u nu g
2 2 2 2 2 2
2 gH = n u^2 2^ − 2 nu^2 2 gH = nu^2 ( n − 2 )
17 Given, dv
dt
= − 2 5. v
⇒ dv v
= − 2 5. dt
⇒ v dv dt −^ t ∫ 1 2 = − ∫
0 2 5 0
⇒ − 2 5. [ ] t 0 t =[ 2 v 1 2/^ ] 6250.
= 2 ( − 6 25. ) = 2 ×2. ⇒ t = 2 s
18 From equation law of motion gives,
s = ut + 1 gT 2
2
or h = 0 +^1 gT 2
(^2) (Q u = 0)
⇒ T h g
At, t = T 3
s,
s = + g T ^
2
or s = 1 g ⋅ T 2 9
2
⇒ s g^ h g
= ×
T h g
or s = h 9
m Hence, the position of ball from the ground = h − h^ = h 9
m
19 Central idea concept of relative motion
can be applied to predict the nature of motion of one particle with respect to the other.
Consider the stones thrown up simultaneously as shown in the diagram below. As motion of the second particle with respect to the first we have relative acceleration | a 21 (^) | |= a 2 (^) − a 1 (^) |= g − g = 0. Thus, motion of first particle is straight line with respect to second particle till the first particle strikes ground at a time is given by − 240 = 10 − 1 × × 2
t 10 t^2
or t^2 − 2 t − 48 = 0 or t^2 − 8 t + 6 t − 48 = 0 or (^) t = 8 ,− 6 [As, t = − 6 s is not possible] i.e., t = 8 s Thus, distance covered by second particle with respect to first particle in 8 s is s 12 = ( v 21 ) t = (40 −10) (8s) = 30 × 8 = 240 m Similarly, time taken by second particle to strike the ground is given by − 240 = 40 − 1 × × 2
t 10 t
or − 240 = 40 t − 5 t 2 or 5 t^2 − 40 t − 240 = 0 or t^2 − 8 t − 48 = 0 t^2 − 12 t + 4 t − 48 = 0 or t ( t − 12 ) + 4 ( t − 12 ) = 0 or t = 12 , − 4 (As, t = − 4 s is not possible) i.e. t = 12 s Thus, after 8 s, magnitude of relative velocity will increase upto 12 s when second particle strikes the ground. Hence, graph (c) is the correct description.
DAY TWO
240 m
10 ms
40 ms-^1
Cliff
H (^) t 2
t 1 u
s h
u =
t =
t = 3^ T
O t = T Ground
