DC Pandey Solutions: Motion in One Dimension- 1

# DC Pandey Solutions: Motion in One Dimension- 1 | DC Pandey Solutions for NEET Physics PDF Download

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``` Page 1

Introductory Exercise 3.1
1. Suppose a particle is moving with
constant velocity  v (along the axis of x).
Displacement of particle in time t vt
1 1
=
Displacement of particle in time t vt
2 2
=
\ Displacement of the particle in the time
interval Dt t t ( ) = -
2 1
= - vt vt
2 1
= - v t t ( )
2 1
\  Average velocity in the time interval
Dt
v t t
t t
¢ =
-
-
( )
( )
2 1
2 1
= v
Now, as the particle is moving with
constant velocity (i.e., with constant speed
in a given direction) its velocity and speed
at any instant will obviously be v.
Ans.  True.
2. As the stone would be free to acceleration
under earth’s gravity it acceleration will
be g.
3. A second hand takes 1 min i.e., 60s to
complete one rotation (i.e., rotation by an
angle of 2p rad).
\ Angular speed of second hand
=
2
60
s
=
p
30
-1
Linear speed of its tip = radius ´ angular
speed
= ´
-
2.0 cm rad s
p
30
1
=
-
p
15
1
cms
As the tip would be moving with constant
speed.
Average speed =
-
p
15
1
cms
In 15 s the second hand would rotate
through 90° i.e., the displacement of its tip
will be r 2.
\ Modulus of average velocity of the tip of
second hand in 15 s.
=
r 2
15
=
-
2 2
15
1
cms
4. (a) Yes. By changing direction of motion,
there will be change in velocity and so
acceleration.
(b) (i) No. In curved path there will always
be acceleration. (As explained in the
previous answer no. 3)
(ii) Yes. In projectile motion the path of
the particle is a curved one while
acceleration of the particle remains
constant.
(iii) Yes. In curved path the acceleration
will always be there. Even if the path
is circular with constant speed the
direction of the acceleration of the
particle would every time be changing.
3
Motion in One Dimension
Page 2

Introductory Exercise 3.1
1. Suppose a particle is moving with
constant velocity  v (along the axis of x).
Displacement of particle in time t vt
1 1
=
Displacement of particle in time t vt
2 2
=
\ Displacement of the particle in the time
interval Dt t t ( ) = -
2 1
= - vt vt
2 1
= - v t t ( )
2 1
\  Average velocity in the time interval
Dt
v t t
t t
¢ =
-
-
( )
( )
2 1
2 1
= v
Now, as the particle is moving with
constant velocity (i.e., with constant speed
in a given direction) its velocity and speed
at any instant will obviously be v.
Ans.  True.
2. As the stone would be free to acceleration
under earth’s gravity it acceleration will
be g.
3. A second hand takes 1 min i.e., 60s to
complete one rotation (i.e., rotation by an
angle of 2p rad).
\ Angular speed of second hand
=
2
60
s
=
p
30
-1
Linear speed of its tip = radius ´ angular
speed
= ´
-
2.0 cm rad s
p
30
1
=
-
p
15
1
cms
As the tip would be moving with constant
speed.
Average speed =
-
p
15
1
cms
In 15 s the second hand would rotate
through 90° i.e., the displacement of its tip
will be r 2.
\ Modulus of average velocity of the tip of
second hand in 15 s.
=
r 2
15
=
-
2 2
15
1
cms
4. (a) Yes. By changing direction of motion,
there will be change in velocity and so
acceleration.
(b) (i) No. In curved path there will always
be acceleration. (As explained in the
previous answer no. 3)
(ii) Yes. In projectile motion the path of
the particle is a curved one while
acceleration of the particle remains
constant.
(iii) Yes. In curved path the acceleration
will always be there. Even if the path
is circular with constant speed the
direction of the acceleration of the
particle would every time be changing.
3
Motion in One Dimension
5. (a) Time speed =
Circumference
Speed
=
´ 2 4
1
p cm
cm/s
=8p =25.13  s
(b)As particle is moving with constant
speed of 1 cm/s, its average speed in
any time interval will be 1 cm/s.
| |
/
Average velocity =
r
T
2
4
=
æ
è
ç
ö
ø
÷
4 2
2
r
r p
speed
=
2 2
p
speed
=
2 2
p
cms
-1
= 0.9 cms
-1
| |
/
Average acceleration =
v
T
2
4
(where v = speed)
=
4 2
2
v
r
v
p
=
2 2
2
p
v
r
=
2 2 1
4
2
p
( )
=
-
0.23 cm s
2

6. Distance = Speed ´ time
D v t
1 1 1
=
D v t
2 2 2
=
Average speed =
+
+
=
+
+
D D
t t
v t v t
t t
1 2
1 2
1 1 2 2
1 2
=
´ + ´
+
( ) ( ) 4 2 6 3
2 3
=
-
5.2ms
1

Introductory Exercise 3.2
1. Acceleration (due to gravity).
2. s u at a
t
= + -
1
2
is physically correct as it
gives the displacement of the particle in t
th
second (or any time unit).
s
t
= Displacement in t seconds
- displacement in ( ) t - 1 seconds
= +
é
ë
ê
ù
û
ú
- - + -
é
ë
ê
ù
û
ú
ut at u t a t
1
2
1
1
2
1
2 2
( ) ( )
Therefore, the given equation is
dimensionally incorrect.
3. 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.
4.        v t =
3 4 /
(given)

ds
dt
t =
3 4 /
…(i)
\         s t dt =
ò
3 4 /
=
+
+
+
t
c
3
4
1
3
4
1
or       s t c = +
4
7
7 4 /
i.e.,    s t µ
74 /
Differentiating Eq. (i) w.r.t. time t,
d s
dt
t
2
2
3
4
1
3
4
=
-
Þ a t µ
-1 4 /
5. Displacement (s) of the particle
s= ´ + - ( ) ( ) 40 6
1
2
106
2
= - 240 180
=60 m (in the upward direction)
Distance covered ( ) D by the particle
Time to attain maximum height
Motion in One Dimension | 15
Displacement
Velocity
st = u + at –
1
2
a
Acceleration
Page 3

Introductory Exercise 3.1
1. Suppose a particle is moving with
constant velocity  v (along the axis of x).
Displacement of particle in time t vt
1 1
=
Displacement of particle in time t vt
2 2
=
\ Displacement of the particle in the time
interval Dt t t ( ) = -
2 1
= - vt vt
2 1
= - v t t ( )
2 1
\  Average velocity in the time interval
Dt
v t t
t t
¢ =
-
-
( )
( )
2 1
2 1
= v
Now, as the particle is moving with
constant velocity (i.e., with constant speed
in a given direction) its velocity and speed
at any instant will obviously be v.
Ans.  True.
2. As the stone would be free to acceleration
under earth’s gravity it acceleration will
be g.
3. A second hand takes 1 min i.e., 60s to
complete one rotation (i.e., rotation by an
angle of 2p rad).
\ Angular speed of second hand
=
2
60
s
=
p
30
-1
Linear speed of its tip = radius ´ angular
speed
= ´
-
2.0 cm rad s
p
30
1
=
-
p
15
1
cms
As the tip would be moving with constant
speed.
Average speed =
-
p
15
1
cms
In 15 s the second hand would rotate
through 90° i.e., the displacement of its tip
will be r 2.
\ Modulus of average velocity of the tip of
second hand in 15 s.
=
r 2
15
=
-
2 2
15
1
cms
4. (a) Yes. By changing direction of motion,
there will be change in velocity and so
acceleration.
(b) (i) No. In curved path there will always
be acceleration. (As explained in the
previous answer no. 3)
(ii) Yes. In projectile motion the path of
the particle is a curved one while
acceleration of the particle remains
constant.
(iii) Yes. In curved path the acceleration
will always be there. Even if the path
is circular with constant speed the
direction of the acceleration of the
particle would every time be changing.
3
Motion in One Dimension
5. (a) Time speed =
Circumference
Speed
=
´ 2 4
1
p cm
cm/s
=8p =25.13  s
(b)As particle is moving with constant
speed of 1 cm/s, its average speed in
any time interval will be 1 cm/s.
| |
/
Average velocity =
r
T
2
4
=
æ
è
ç
ö
ø
÷
4 2
2
r
r p
speed
=
2 2
p
speed
=
2 2
p
cms
-1
= 0.9 cms
-1
| |
/
Average acceleration =
v
T
2
4
(where v = speed)
=
4 2
2
v
r
v
p
=
2 2
2
p
v
r
=
2 2 1
4
2
p
( )
=
-
0.23 cm s
2

6. Distance = Speed ´ time
D v t
1 1 1
=
D v t
2 2 2
=
Average speed =
+
+
=
+
+
D D
t t
v t v t
t t
1 2
1 2
1 1 2 2
1 2
=
´ + ´
+
( ) ( ) 4 2 6 3
2 3
=
-
5.2ms
1

Introductory Exercise 3.2
1. Acceleration (due to gravity).
2. s u at a
t
= + -
1
2
is physically correct as it
gives the displacement of the particle in t
th
second (or any time unit).
s
t
= Displacement in t seconds
- displacement in ( ) t - 1 seconds
= +
é
ë
ê
ù
û
ú
- - + -
é
ë
ê
ù
û
ú
ut at u t a t
1
2
1
1
2
1
2 2
( ) ( )
Therefore, the given equation is
dimensionally incorrect.
3. 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.
4.        v t =
3 4 /
(given)

ds
dt
t =
3 4 /
…(i)
\         s t dt =
ò
3 4 /
=
+
+
+
t
c
3
4
1
3
4
1
or       s t c = +
4
7
7 4 /
i.e.,    s t µ
74 /
Differentiating Eq. (i) w.r.t. time t,
d s
dt
t
2
2
3
4
1
3
4
=
-
Þ a t µ
-1 4 /
5. Displacement (s) of the particle
s= ´ + - ( ) ( ) 40 6
1
2
106
2
= - 240 180
=60 m (in the upward direction)
Distance covered ( ) D by the particle
Time to attain maximum height
Motion in One Dimension | 15
Displacement
Velocity
st = u + at –
1
2
a
Acceleration
= =
40
10
4 s < 6 s
It implies that particle has come back
after attaining maximum height (h) given
by
h
u
g
=
2
2

=
´
( ) 40
2 10
2
= 80 m
\ D = + - 80 80 60 ( )
=100 m
6.                      v t = - 40 10
\
dx
dt
t = - 40 10
or        dx t dt = - ( ) 40 10
or x t dt = -
ò
( ) 40 10
or         x t t c = - + 40 5
2
As at t = 0 the value of x is zero.
c =0
\           x t t = - 40 5
2
For x to be 60 m.
60 40 5
2
= - t t
or     t t
2
8 12 0 - + =
\    t =2 s or 6 s
7. Average velocity =
Displacement in time t
t

=
+ ut at
t
1
2
2
= + u at
1
2
8.           v v at
2 1
= +
\          at v v = -
2 1
Average velocity =
Displacement in time t
t
=
+ v t at
t
1
2
1
2
= + v at
1
1
2
= +
-
v
v v
1
2 1
2
=
+ v v
1 2
2
\      Ans.  True.
9.            125 0
1
2
2
= × + t gt
Þ            t =25 s
Average velocity =
125
5
m
s
(downwards)
= 25  m/s (downwards)
10.        v t t = + - 10 5
2
…(i
)
\       a
dv
dt
t = = - 5 2
At       t =2 s
a = - ´ 5 2 2
=1 m/s
2
From Eq. (i),

dx
dt
t t = + - 10 5
2
\ x t t dt = + -
ò
( ) 10 5
2
or x t
t t
c = + - + 10
5
2 3
2 3
As, at        t = 0 the value of x is zero
c =0
\         x t t
t
= + - 10
5
2 3
2
3
Thus, at     t = 3 s
x = ´ + - ( ) ( ) 10 3
5
2
3
3
3
2
3
= + - 30 9 22.5
= 43.5 m
11. u i
®
= 2
^
m/s
a i j
®
= ° + ° ( cos sin )
^ ^
2 60 2 60 m/s
2
= + ( )
^ ^
1 3 i j m/s
2
v u a
® ® ®
= + t
= + + 2 1 3 2 i i j
^ ^ ^
( )
= + 4 2 3 i j
^ ^
16 | Mechanics-1
60°
2
a = 2 m/s
u = 2 m/s
x
y
Page 4

Introductory Exercise 3.1
1. Suppose a particle is moving with
constant velocity  v (along the axis of x).
Displacement of particle in time t vt
1 1
=
Displacement of particle in time t vt
2 2
=
\ Displacement of the particle in the time
interval Dt t t ( ) = -
2 1
= - vt vt
2 1
= - v t t ( )
2 1
\  Average velocity in the time interval
Dt
v t t
t t
¢ =
-
-
( )
( )
2 1
2 1
= v
Now, as the particle is moving with
constant velocity (i.e., with constant speed
in a given direction) its velocity and speed
at any instant will obviously be v.
Ans.  True.
2. As the stone would be free to acceleration
under earth’s gravity it acceleration will
be g.
3. A second hand takes 1 min i.e., 60s to
complete one rotation (i.e., rotation by an
angle of 2p rad).
\ Angular speed of second hand
=
2
60
s
=
p
30
-1
Linear speed of its tip = radius ´ angular
speed
= ´
-
2.0 cm rad s
p
30
1
=
-
p
15
1
cms
As the tip would be moving with constant
speed.
Average speed =
-
p
15
1
cms
In 15 s the second hand would rotate
through 90° i.e., the displacement of its tip
will be r 2.
\ Modulus of average velocity of the tip of
second hand in 15 s.
=
r 2
15
=
-
2 2
15
1
cms
4. (a) Yes. By changing direction of motion,
there will be change in velocity and so
acceleration.
(b) (i) No. In curved path there will always
be acceleration. (As explained in the
previous answer no. 3)
(ii) Yes. In projectile motion the path of
the particle is a curved one while
acceleration of the particle remains
constant.
(iii) Yes. In curved path the acceleration
will always be there. Even if the path
is circular with constant speed the
direction of the acceleration of the
particle would every time be changing.
3
Motion in One Dimension
5. (a) Time speed =
Circumference
Speed
=
´ 2 4
1
p cm
cm/s
=8p =25.13  s
(b)As particle is moving with constant
speed of 1 cm/s, its average speed in
any time interval will be 1 cm/s.
| |
/
Average velocity =
r
T
2
4
=
æ
è
ç
ö
ø
÷
4 2
2
r
r p
speed
=
2 2
p
speed
=
2 2
p
cms
-1
= 0.9 cms
-1
| |
/
Average acceleration =
v
T
2
4
(where v = speed)
=
4 2
2
v
r
v
p
=
2 2
2
p
v
r
=
2 2 1
4
2
p
( )
=
-
0.23 cm s
2

6. Distance = Speed ´ time
D v t
1 1 1
=
D v t
2 2 2
=
Average speed =
+
+
=
+
+
D D
t t
v t v t
t t
1 2
1 2
1 1 2 2
1 2
=
´ + ´
+
( ) ( ) 4 2 6 3
2 3
=
-
5.2ms
1

Introductory Exercise 3.2
1. Acceleration (due to gravity).
2. s u at a
t
= + -
1
2
is physically correct as it
gives the displacement of the particle in t
th
second (or any time unit).
s
t
= Displacement in t seconds
- displacement in ( ) t - 1 seconds
= +
é
ë
ê
ù
û
ú
- - + -
é
ë
ê
ù
û
ú
ut at u t a t
1
2
1
1
2
1
2 2
( ) ( )
Therefore, the given equation is
dimensionally incorrect.
3. 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.
4.        v t =
3 4 /
(given)

ds
dt
t =
3 4 /
…(i)
\         s t dt =
ò
3 4 /
=
+
+
+
t
c
3
4
1
3
4
1
or       s t c = +
4
7
7 4 /
i.e.,    s t µ
74 /
Differentiating Eq. (i) w.r.t. time t,
d s
dt
t
2
2
3
4
1
3
4
=
-
Þ a t µ
-1 4 /
5. Displacement (s) of the particle
s= ´ + - ( ) ( ) 40 6
1
2
106
2
= - 240 180
=60 m (in the upward direction)
Distance covered ( ) D by the particle
Time to attain maximum height
Motion in One Dimension | 15
Displacement
Velocity
st = u + at –
1
2
a
Acceleration
= =
40
10
4 s < 6 s
It implies that particle has come back
after attaining maximum height (h) given
by
h
u
g
=
2
2

=
´
( ) 40
2 10
2
= 80 m
\ D = + - 80 80 60 ( )
=100 m
6.                      v t = - 40 10
\
dx
dt
t = - 40 10
or        dx t dt = - ( ) 40 10
or x t dt = -
ò
( ) 40 10
or         x t t c = - + 40 5
2
As at t = 0 the value of x is zero.
c =0
\           x t t = - 40 5
2
For x to be 60 m.
60 40 5
2
= - t t
or     t t
2
8 12 0 - + =
\    t =2 s or 6 s
7. Average velocity =
Displacement in time t
t

=
+ ut at
t
1
2
2
= + u at
1
2
8.           v v at
2 1
= +
\          at v v = -
2 1
Average velocity =
Displacement in time t
t
=
+ v t at
t
1
2
1
2
= + v at
1
1
2
= +
-
v
v v
1
2 1
2
=
+ v v
1 2
2
\      Ans.  True.
9.            125 0
1
2
2
= × + t gt
Þ            t =25 s
Average velocity =
125
5
m
s
(downwards)
= 25  m/s (downwards)
10.        v t t = + - 10 5
2
…(i
)
\       a
dv
dt
t = = - 5 2
At       t =2 s
a = - ´ 5 2 2
=1 m/s
2
From Eq. (i),

dx
dt
t t = + - 10 5
2
\ x t t dt = + -
ò
( ) 10 5
2
or x t
t t
c = + - + 10
5
2 3
2 3
As, at        t = 0 the value of x is zero
c =0
\         x t t
t
= + - 10
5
2 3
2
3
Thus, at     t = 3 s
x = ´ + - ( ) ( ) 10 3
5
2
3
3
3
2
3
= + - 30 9 22.5
= 43.5 m
11. u i
®
= 2
^
m/s
a i j
®
= ° + ° ( cos sin )
^ ^
2 60 2 60 m/s
2
= + ( )
^ ^
1 3 i j m/s
2
v u a
® ® ®
= + t
= + + 2 1 3 2 i i j
^ ^ ^
( )
= + 4 2 3 i j
^ ^
16 | Mechanics-1
60°
2
a = 2 m/s
u = 2 m/s
x
y
| | v
®
= + 4 12
2
=2 7 m/s
s u a
® ® ®
= + t t
1
2
2
= + + ( ) ( )
^ ^ ^
2 2
1
2
1 3 2
2
i i j
= + + 4 2 2 3 i i j
^ ^ ^
= + 6 2 3 i j
^ ^
\    | | s
®
= + 36 12
= 4 3 m
12. Part I    v i j
®
= + ( )
^ ^
2 2t m/s …(i)

dv
j
¾®
=
dt
2
^
a j
®
=2
^
m/s
2
From Eq. (i),

d
dt
t
s
i j
®
= + ( )
^ ^
2 2
\       s i j
®
= +
ò
( )
^ ^
2 2t dt
s i j
®
= + + 2
2
t t c
^ ^
Taking initial displacement to be zero.
s
®
(at t = 1 s) = + ( )
^ ^
2 i j m
Part II  Yes. As explained below.
v i j
®
= + 2 2
^ ^
t implies that initial velocity of
the particle is 2 i
^
m/s
2
and the acceleration
is 2
\$
j m/s
2
\ s
®
(at t = 1 s) = ´ + ( ) ( )
^ ^
2 1
1
2
2 1
2
i j
= + ( )
^ ^
2 i j m
13. x t = 2 and y t =
2
\   y
x
=
æ
è
ç
ö
ø
÷
2
2
or, x y
2
4 =
(The above is the equation to trajectory)
x t = 2
\
dx
dt
= 2 i.e., v i
x
®
= 2
^
y t =
2

\
dy
dt
t = 2 i.e., v j
y
t
®
+ 2
^
Thus,           v v v
® ® ®
= +
x y
= + ( )
^ ^
2 2 i j t m/s
a
dv
j
®
¾®
= =
dt
2
^
m/s
2
Introductory Exercise 3.3
1. At t t =
1

v = tan q
As q < ° 90 , v
t
1
is + ive.
At t t =
2

v
t
2
= f tan
As f > ° 90 , v
t
2
is - ive.
Corresponding v-t graph will be
Acceleration at t t =
1
:
a
t
1
= tan a
As a < ° 90 , a t
1
is + ive constant.
Acceleration at t t =
2

a
t
2
= tan b
As b < ° 90 , a
t
2
is + ive constant.
2. Let the particle strike ground at time t
velocity of particle when it touches ground
Motion in One Dimension | 17
s
f
q
0
t
1
t
2 t
v
t
1
t
2 b
t
a
b
Page 5

Introductory Exercise 3.1
1. Suppose a particle is moving with
constant velocity  v (along the axis of x).
Displacement of particle in time t vt
1 1
=
Displacement of particle in time t vt
2 2
=
\ Displacement of the particle in the time
interval Dt t t ( ) = -
2 1
= - vt vt
2 1
= - v t t ( )
2 1
\  Average velocity in the time interval
Dt
v t t
t t
¢ =
-
-
( )
( )
2 1
2 1
= v
Now, as the particle is moving with
constant velocity (i.e., with constant speed
in a given direction) its velocity and speed
at any instant will obviously be v.
Ans.  True.
2. As the stone would be free to acceleration
under earth’s gravity it acceleration will
be g.
3. A second hand takes 1 min i.e., 60s to
complete one rotation (i.e., rotation by an
angle of 2p rad).
\ Angular speed of second hand
=
2
60
s
=
p
30
-1
Linear speed of its tip = radius ´ angular
speed
= ´
-
2.0 cm rad s
p
30
1
=
-
p
15
1
cms
As the tip would be moving with constant
speed.
Average speed =
-
p
15
1
cms
In 15 s the second hand would rotate
through 90° i.e., the displacement of its tip
will be r 2.
\ Modulus of average velocity of the tip of
second hand in 15 s.
=
r 2
15
=
-
2 2
15
1
cms
4. (a) Yes. By changing direction of motion,
there will be change in velocity and so
acceleration.
(b) (i) No. In curved path there will always
be acceleration. (As explained in the
previous answer no. 3)
(ii) Yes. In projectile motion the path of
the particle is a curved one while
acceleration of the particle remains
constant.
(iii) Yes. In curved path the acceleration
will always be there. Even if the path
is circular with constant speed the
direction of the acceleration of the
particle would every time be changing.
3
Motion in One Dimension
5. (a) Time speed =
Circumference
Speed
=
´ 2 4
1
p cm
cm/s
=8p =25.13  s
(b)As particle is moving with constant
speed of 1 cm/s, its average speed in
any time interval will be 1 cm/s.
| |
/
Average velocity =
r
T
2
4
=
æ
è
ç
ö
ø
÷
4 2
2
r
r p
speed
=
2 2
p
speed
=
2 2
p
cms
-1
= 0.9 cms
-1
| |
/
Average acceleration =
v
T
2
4
(where v = speed)
=
4 2
2
v
r
v
p
=
2 2
2
p
v
r
=
2 2 1
4
2
p
( )
=
-
0.23 cm s
2

6. Distance = Speed ´ time
D v t
1 1 1
=
D v t
2 2 2
=
Average speed =
+
+
=
+
+
D D
t t
v t v t
t t
1 2
1 2
1 1 2 2
1 2
=
´ + ´
+
( ) ( ) 4 2 6 3
2 3
=
-
5.2ms
1

Introductory Exercise 3.2
1. Acceleration (due to gravity).
2. s u at a
t
= + -
1
2
is physically correct as it
gives the displacement of the particle in t
th
second (or any time unit).
s
t
= Displacement in t seconds
- displacement in ( ) t - 1 seconds
= +
é
ë
ê
ù
û
ú
- - + -
é
ë
ê
ù
û
ú
ut at u t a t
1
2
1
1
2
1
2 2
( ) ( )
Therefore, the given equation is
dimensionally incorrect.
3. 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.
4.        v t =
3 4 /
(given)

ds
dt
t =
3 4 /
…(i)
\         s t dt =
ò
3 4 /
=
+
+
+
t
c
3
4
1
3
4
1
or       s t c = +
4
7
7 4 /
i.e.,    s t µ
74 /
Differentiating Eq. (i) w.r.t. time t,
d s
dt
t
2
2
3
4
1
3
4
=
-
Þ a t µ
-1 4 /
5. Displacement (s) of the particle
s= ´ + - ( ) ( ) 40 6
1
2
106
2
= - 240 180
=60 m (in the upward direction)
Distance covered ( ) D by the particle
Time to attain maximum height
Motion in One Dimension | 15
Displacement
Velocity
st = u + at –
1
2
a
Acceleration
= =
40
10
4 s < 6 s
It implies that particle has come back
after attaining maximum height (h) given
by
h
u
g
=
2
2

=
´
( ) 40
2 10
2
= 80 m
\ D = + - 80 80 60 ( )
=100 m
6.                      v t = - 40 10
\
dx
dt
t = - 40 10
or        dx t dt = - ( ) 40 10
or x t dt = -
ò
( ) 40 10
or         x t t c = - + 40 5
2
As at t = 0 the value of x is zero.
c =0
\           x t t = - 40 5
2
For x to be 60 m.
60 40 5
2
= - t t
or     t t
2
8 12 0 - + =
\    t =2 s or 6 s
7. Average velocity =
Displacement in time t
t

=
+ ut at
t
1
2
2
= + u at
1
2
8.           v v at
2 1
= +
\          at v v = -
2 1
Average velocity =
Displacement in time t
t
=
+ v t at
t
1
2
1
2
= + v at
1
1
2
= +
-
v
v v
1
2 1
2
=
+ v v
1 2
2
\      Ans.  True.
9.            125 0
1
2
2
= × + t gt
Þ            t =25 s
Average velocity =
125
5
m
s
(downwards)
= 25  m/s (downwards)
10.        v t t = + - 10 5
2
…(i
)
\       a
dv
dt
t = = - 5 2
At       t =2 s
a = - ´ 5 2 2
=1 m/s
2
From Eq. (i),

dx
dt
t t = + - 10 5
2
\ x t t dt = + -
ò
( ) 10 5
2
or x t
t t
c = + - + 10
5
2 3
2 3
As, at        t = 0 the value of x is zero
c =0
\         x t t
t
= + - 10
5
2 3
2
3
Thus, at     t = 3 s
x = ´ + - ( ) ( ) 10 3
5
2
3
3
3
2
3
= + - 30 9 22.5
= 43.5 m
11. u i
®
= 2
^
m/s
a i j
®
= ° + ° ( cos sin )
^ ^
2 60 2 60 m/s
2
= + ( )
^ ^
1 3 i j m/s
2
v u a
® ® ®
= + t
= + + 2 1 3 2 i i j
^ ^ ^
( )
= + 4 2 3 i j
^ ^
16 | Mechanics-1
60°
2
a = 2 m/s
u = 2 m/s
x
y
| | v
®
= + 4 12
2
=2 7 m/s
s u a
® ® ®
= + t t
1
2
2
= + + ( ) ( )
^ ^ ^
2 2
1
2
1 3 2
2
i i j
= + + 4 2 2 3 i i j
^ ^ ^
= + 6 2 3 i j
^ ^
\    | | s
®
= + 36 12
= 4 3 m
12. Part I    v i j
®
= + ( )
^ ^
2 2t m/s …(i)

dv
j
¾®
=
dt
2
^
a j
®
=2
^
m/s
2
From Eq. (i),

d
dt
t
s
i j
®
= + ( )
^ ^
2 2
\       s i j
®
= +
ò
( )
^ ^
2 2t dt
s i j
®
= + + 2
2
t t c
^ ^
Taking initial displacement to be zero.
s
®
(at t = 1 s) = + ( )
^ ^
2 i j m
Part II  Yes. As explained below.
v i j
®
= + 2 2
^ ^
t implies that initial velocity of
the particle is 2 i
^
m/s
2
and the acceleration
is 2
\$
j m/s
2
\ s
®
(at t = 1 s) = ´ + ( ) ( )
^ ^
2 1
1
2
2 1
2
i j
= + ( )
^ ^
2 i j m
13. x t = 2 and y t =
2
\   y
x
=
æ
è
ç
ö
ø
÷
2
2
or, x y
2
4 =
(The above is the equation to trajectory)
x t = 2
\
dx
dt
= 2 i.e., v i
x
®
= 2
^
y t =
2

\
dy
dt
t = 2 i.e., v j
y
t
®
+ 2
^
Thus,           v v v
® ® ®
= +
x y
= + ( )
^ ^
2 2 i j t m/s
a
dv
j
®
¾®
= =
dt
2
^
m/s
2
Introductory Exercise 3.3
1. At t t =
1

v = tan q
As q < ° 90 , v
t
1
is + ive.
At t t =
2

v
t
2
= f tan
As f > ° 90 , v
t
2
is - ive.
Corresponding v-t graph will be
Acceleration at t t =
1
:
a
t
1
= tan a
As a < ° 90 , a t
1
is + ive constant.
Acceleration at t t =
2

a
t
2
= tan b
As b < ° 90 , a
t
2
is + ive constant.
2. Let the particle strike ground at time t
velocity of particle when it touches ground
Motion in One Dimension | 17
s
f
q
0
t
1
t
2 t
v
t
1
t
2 b
t
a
b
would be gt. KE of particle will be
1
2
2 2
mg t
i.e., KE µ t
2
. While going up the velocity
will get - ive but the KE will remain. KE
will reduce to zero at time 2t when the
particle reaches its initial position.
KE =
1
2
2 2
mg t =
1
2
2
2
mg
h
g
= mgh
3. Speed of ball (just before making first
collision with floor)
= 2gh = ´ ´ 2 10 80
= 40 m/s
Time taken to reach ground
=
2h
g
=
´ 2 80
10
= 4 s
Speed of ball (just after first collision with
floor)
= =
40
2
20 m/s
Time to attain maximum height
t =
-
-
=
20
10
2 s
\ Time for the return journey to floor = 2 s.
Corresponding velocity-time will be
4.
h
t ( )
tan
( ) -
= =
- 2
2
2 1
q
Þ h t = - 2 2 ( )
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

( ) ( ) ( ) 1 2 2
2
2
2
0
+ ´
+
- -
=
h t
or 3 2 0
2
- - = ( ) t
or   ( ) t - = 2 3
2
or       t - = ± 2 3
or             t = ± 2 3
Ans . At time t = + 2 3 s
(t = - 2 3 not possible).
Introductory Exercise 3.4
1. Relative acceleration of A w.r.t. B
a g g
AB
= + - + ( ) ( ) = 0
2. Velocity of A w.r.t. B v v
A B
= -
\ Relative displacement (i.e., distance
between A and B) would be
s v v t a t
A B AB
= - + ( )
1
2
2
18 | Mechanics-1
KE
2t time
2h
g
= t
8 4
t (s)
Speed (m/s)
8 4
t (s)
Velocity (m/s)
2
a (m/s )
2
1 2
(t–2)
t
h
q
q
```

122 docs

## FAQs on DC Pandey Solutions: Motion in One Dimension- 1 - DC Pandey Solutions for NEET Physics

 1. What is motion in one dimension?
Ans. Motion in one dimension refers to the movement of an object along a straight line, where the object can only move in either the positive or negative direction. It involves the study of the object's displacement, velocity, and acceleration in relation to time.
 2. How do you calculate displacement in one dimension motion?
Ans. Displacement in one dimension motion can be calculated by subtracting the initial position from the final position of the object. It is represented by the symbol Δx and can be positive or negative depending on the direction of motion.
 3. What is the difference between speed and velocity in one dimension motion?
Ans. Speed in one dimension motion is a scalar quantity that represents the rate at which an object covers distance, irrespective of the direction. On the other hand, velocity in one dimension motion is a vector quantity that includes both the speed and direction of motion.
 4. How is acceleration calculated in one dimension motion?
Ans. Acceleration in one dimension motion can be calculated by dividing the change in velocity (∆v) by the time taken (∆t). It is represented by the symbol 'a' and its unit is meters per second squared (m/s^2).
 5. What are the different types of motion in one dimension?
Ans. In one dimension, there are three types of motion: uniform motion, uniformly accelerated motion, and non-uniform motion. Uniform motion occurs when an object covers equal distances in equal intervals of time. Uniformly accelerated motion refers to the motion where the object's acceleration remains constant. Non-uniform motion is when an object's velocity changes irregularly over time.

## DC Pandey Solutions for NEET Physics

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