DC Pandey Solutions: Motion in One Dimension- 4

# DC Pandey Solutions: Motion in One Dimension- 4 | DC Pandey Solutions for NEET Physics PDF Download

``` Page 1

Graphy
28. OA : slope is + ive and increasing.
\ velocity is + ive and acceleration is + ive.
AB : slope is + ive and constant
\ velocity is +ive and acceleration is zero.
BC : sope is + ive and decreases.
\ velocity is + ive and increasing.
CD : slope is –ive and increasing
\ velocity is - ive and acceleration is - ive.
29.
In M
1
and M
3
0 90 : ° £ < ° q
\ slope is + ive i.e., acceleration is + ive.
In M
2
and M
4
: 90 180 ° < £ ° q
\ slope is - ive i.e., acceleration is - ive.
(a) M
1
: Magnitude of velocity is increasing.
M
2
: Magnitude of velocity is decreasing.
M
3
: Magnitude of velocity is increasing.
M
4
: Magnitude of velocity is decreasing.
Ans : M
1
and M
3
.
(b) P M ®
1
Q M ®
2
R M ®
3
S M ®
4
30. Case I
v mt v = -
0
(st-line)
\   ds mt v dt
ò ò
= - ( )
0
i.e.,  s m
t
v t s = × - +
2
0 0
2
(Parabola)
or   s s v t at = - +
0 0
2
1
2
(Q m a = )
Further,   a
dv
dt
= = tan q
As for 0 90 ° £ < ° q
tan q is + ive, a is + ive.
Case II
v k = (constant) (st line parallel to
time-axis)
Þ ds kdt
ò ò
=
s kt s = +
0
Further,    a
dv
dt
= =0
Motion in One Dimension | 27
O
t
s
A
B
D
C
q
v
M
1
t
q
v
M
2
t O
q
M
4
q
v
M
3
t
O
q
v
0
m = tan q
v
s
0
t
s
Þ
a
Þ
a = tan q
t
s
0
s
t
v
Þ
t
a
Þ
t
Page 2

Graphy
28. OA : slope is + ive and increasing.
\ velocity is + ive and acceleration is + ive.
AB : slope is + ive and constant
\ velocity is +ive and acceleration is zero.
BC : sope is + ive and decreases.
\ velocity is + ive and increasing.
CD : slope is –ive and increasing
\ velocity is - ive and acceleration is - ive.
29.
In M
1
and M
3
0 90 : ° £ < ° q
\ slope is + ive i.e., acceleration is + ive.
In M
2
and M
4
: 90 180 ° < £ ° q
\ slope is - ive i.e., acceleration is - ive.
(a) M
1
: Magnitude of velocity is increasing.
M
2
: Magnitude of velocity is decreasing.
M
3
: Magnitude of velocity is increasing.
M
4
: Magnitude of velocity is decreasing.
Ans : M
1
and M
3
.
(b) P M ®
1
Q M ®
2
R M ®
3
S M ®
4
30. Case I
v mt v = -
0
(st-line)
\   ds mt v dt
ò ò
= - ( )
0
i.e.,  s m
t
v t s = × - +
2
0 0
2
(Parabola)
or   s s v t at = - +
0 0
2
1
2
(Q m a = )
Further,   a
dv
dt
= = tan q
As for 0 90 ° £ < ° q
tan q is + ive, a is + ive.
Case II
v k = (constant) (st line parallel to
time-axis)
Þ ds kdt
ò ò
=
s kt s = +
0
Further,    a
dv
dt
= =0
Motion in One Dimension | 27
O
t
s
A
B
D
C
q
v
M
1
t
q
v
M
2
t O
q
M
4
q
v
M
3
t
O
q
v
0
m = tan q
v
s
0
t
s
Þ
a
Þ
a = tan q
t
s
0
s
t
v
Þ
t
a
Þ
t
Case III.
v v mt = +
0
\    ds v mt dt
ò ò
= + ( )
0
i.e.,    s v t m
t
s = + +
0
2
0
2
or    s s v t at = + -
0 0
2
1
2
(Q m a = - )
Further,  a
dv
dt
= = tan q
As for 90 180 ° < £ ° q
tan q is - ive, a is - ive.
Time-displacement graph
Time Area
under the
graph
Initial
Displacement
Net
displacement
at 0 s 0 m - 10 m - 10 m
2 s 10 m - 10 m 0 m
4 s 30 m - 10 m + 20 m
6 s 40 m - 10 m + 30 m
8 s 30 m - 10 m + 20 m
10 s 20 m - 10 m + 10 m
Time-acceleration graph
time slope acceleration
0 - 2 s 5 5 m/s
2
2 - 4 s zero 0 m/s
2
4 - 6 s - 5 - 5 m/s
2
6 - 8 s - 5 - 5 m/s
2
8 - 10 s + 5 + 5 m/s
2
31. a = slope of v t - graph.
S = area under v t - graph.
Corresponding graphs are drawn in the
32. Average velocity = = =
- s
t
A A
t
Net area
Time
1 2
Average speed = =
+ d
t
A A
t
1 2
33. Average acceleration
=
-
=
- -
=-
v v
f i
time
s
10 20
6
5
2
m/
v v
f i
- = Area under a t - graph.
34. (a) Acceleration =  slope of v t - graph.
35. (b) r r s
f i
- = = area under v t - graph.
(c) Equations are written in answer sheet.
Rel a ti ve Mo ti on
35. (a) Acceleration of 1 w.r.t. 2
= - - - ( ) ( ) g g
=0 m/s
2
(b) Initial velocity of 2 w.r.t. 1
= + - - ( ) ( ) 20 5
=25 m/s
(c) Initial velocity of 1 w.r.t. 2
= - - + ( ) ( ) 5 20
= -25  m/s
28 | Mechanics-1
t 10 8 6 4 2
– 10 m
+ 10 m
+ 20 m
+ 30 m
s
t 10 8 6 4 2
5
a
q
v
v
0
t
Þ
s
s
0
t
Þ
a
t
a = tan q
Page 3

Graphy
28. OA : slope is + ive and increasing.
\ velocity is + ive and acceleration is + ive.
AB : slope is + ive and constant
\ velocity is +ive and acceleration is zero.
BC : sope is + ive and decreases.
\ velocity is + ive and increasing.
CD : slope is –ive and increasing
\ velocity is - ive and acceleration is - ive.
29.
In M
1
and M
3
0 90 : ° £ < ° q
\ slope is + ive i.e., acceleration is + ive.
In M
2
and M
4
: 90 180 ° < £ ° q
\ slope is - ive i.e., acceleration is - ive.
(a) M
1
: Magnitude of velocity is increasing.
M
2
: Magnitude of velocity is decreasing.
M
3
: Magnitude of velocity is increasing.
M
4
: Magnitude of velocity is decreasing.
Ans : M
1
and M
3
.
(b) P M ®
1
Q M ®
2
R M ®
3
S M ®
4
30. Case I
v mt v = -
0
(st-line)
\   ds mt v dt
ò ò
= - ( )
0
i.e.,  s m
t
v t s = × - +
2
0 0
2
(Parabola)
or   s s v t at = - +
0 0
2
1
2
(Q m a = )
Further,   a
dv
dt
= = tan q
As for 0 90 ° £ < ° q
tan q is + ive, a is + ive.
Case II
v k = (constant) (st line parallel to
time-axis)
Þ ds kdt
ò ò
=
s kt s = +
0
Further,    a
dv
dt
= =0
Motion in One Dimension | 27
O
t
s
A
B
D
C
q
v
M
1
t
q
v
M
2
t O
q
M
4
q
v
M
3
t
O
q
v
0
m = tan q
v
s
0
t
s
Þ
a
Þ
a = tan q
t
s
0
s
t
v
Þ
t
a
Þ
t
Case III.
v v mt = +
0
\    ds v mt dt
ò ò
= + ( )
0
i.e.,    s v t m
t
s = + +
0
2
0
2
or    s s v t at = + -
0 0
2
1
2
(Q m a = - )
Further,  a
dv
dt
= = tan q
As for 90 180 ° < £ ° q
tan q is - ive, a is - ive.
Time-displacement graph
Time Area
under the
graph
Initial
Displacement
Net
displacement
at 0 s 0 m - 10 m - 10 m
2 s 10 m - 10 m 0 m
4 s 30 m - 10 m + 20 m
6 s 40 m - 10 m + 30 m
8 s 30 m - 10 m + 20 m
10 s 20 m - 10 m + 10 m
Time-acceleration graph
time slope acceleration
0 - 2 s 5 5 m/s
2
2 - 4 s zero 0 m/s
2
4 - 6 s - 5 - 5 m/s
2
6 - 8 s - 5 - 5 m/s
2
8 - 10 s + 5 + 5 m/s
2
31. a = slope of v t - graph.
S = area under v t - graph.
Corresponding graphs are drawn in the
32. Average velocity = = =
- s
t
A A
t
Net area
Time
1 2
Average speed = =
+ d
t
A A
t
1 2
33. Average acceleration
=
-
=
- -
=-
v v
f i
time
s
10 20
6
5
2
m/
v v
f i
- = Area under a t - graph.
34. (a) Acceleration =  slope of v t - graph.
35. (b) r r s
f i
- = = area under v t - graph.
(c) Equations are written in answer sheet.
Rel a ti ve Mo ti on
35. (a) Acceleration of 1 w.r.t. 2
= - - - ( ) ( ) g g
=0 m/s
2
(b) Initial velocity of 2 w.r.t. 1
= + - - ( ) ( ) 20 5
=25 m/s
(c) Initial velocity of 1 w.r.t. 2
= - - + ( ) ( ) 5 20
= -25  m/s
28 | Mechanics-1
t 10 8 6 4 2
– 10 m
+ 10 m
+ 20 m
+ 30 m
s
t 10 8 6 4 2
5
a
q
v
v
0
t
Þ
s
s
0
t
Þ
a
t
a = tan q
\ Velocity of 1 w.r.t. 2 at time t =
æ
è
ç
ö
ø
÷
1
2
s
= - +
æ
è
ç
ö
ø
÷ ( ) ( ) 25 0
1
2
= -25 m/s
(d) Initial relative displacement of 2 w.r.t. 1
S
0
20
®
= - m
Using S S u a
® ® ® ®
= + +
0
2
1
2
rel rel
t t ,
as at time t (= time of collision of the
particles) the relative displacement of 2
w.r.t. 1 will be zero (i.e., S O
® ®
= )
O
®
= - + - ×
æ
è
ç
ö
ø
÷ ( ) ( ) 20 25
1
2
Þ        t =0.85 s
36. Let length of escalator L (= 15 m)
walking speed of man =
L
90
Speed of escalator =
L
60
Time taken by man walking on a moving
escalator =
+
L
L L
90 60
=36 s
Required time has been found without
using the actual length of the escalator.
37. S
0
®
= Initial displacement of elevator w.r.t.
ball = + 10 m.
Relative velocity of elevator w.r.t. ball
= - ( ) ( ) 2 18 = -16 m/s
Accelerator of elevator w.r.t. ball
= - - ( ) ( ) 0 10
=10 m/s
2
Using S S
® ®
= + +
0
2
1
2
u t a t
rel rel
0 10 16
1
2
10
2
= + + - + + ( ) ( ) ( ) t t
or 5 16 10 0
2
t t - - =
t =3.65 s
Position of elevator when it meets ball
= + ´ 5 2 ( ) 3.65
= 12.30 m level
38.
(a)
1
2
60
2
( ) 2.2 t =           …(i)
Þ t=7.39 s
(b)
1
2
60
2
( ) 3.5 t x = + …(ii)
Dividing Eq. (ii) by Eq. (i),
60
60
+
=
x 3.5
2.2
or
x
60
=
1.3
2.2
x =35.5 m
(c) At the time of overtaking
Speed of automobile = ( ) ( ) 3.5 7.39
=25.85 m/s
Speed of truck = ( ) ( ) 2.2 7.39
= 16.25 m/s
39. Let, acceleration of lift = a (upward)
\ acceleration of thrown body w.r.t. lift
= - - + ( ) ( ) g a
= - + ( ) g a
If time of flight is t, using
v u a t
rel rel rel
= +
( ) ( ) { ( )} - = + + - + v u a g t
Þ ( ) a g t u + = 2
or       a
u
t
g = -
2
Motion in One Dimension | 29
10 m
12 m
18 m/s
5m
2 m/s
Elevator
T
A
T
x
60 m
2
3.5 m /s
2
2.2 m /s
time = t
s
time = 0
A
Page 4

Graphy
28. OA : slope is + ive and increasing.
\ velocity is + ive and acceleration is + ive.
AB : slope is + ive and constant
\ velocity is +ive and acceleration is zero.
BC : sope is + ive and decreases.
\ velocity is + ive and increasing.
CD : slope is –ive and increasing
\ velocity is - ive and acceleration is - ive.
29.
In M
1
and M
3
0 90 : ° £ < ° q
\ slope is + ive i.e., acceleration is + ive.
In M
2
and M
4
: 90 180 ° < £ ° q
\ slope is - ive i.e., acceleration is - ive.
(a) M
1
: Magnitude of velocity is increasing.
M
2
: Magnitude of velocity is decreasing.
M
3
: Magnitude of velocity is increasing.
M
4
: Magnitude of velocity is decreasing.
Ans : M
1
and M
3
.
(b) P M ®
1
Q M ®
2
R M ®
3
S M ®
4
30. Case I
v mt v = -
0
(st-line)
\   ds mt v dt
ò ò
= - ( )
0
i.e.,  s m
t
v t s = × - +
2
0 0
2
(Parabola)
or   s s v t at = - +
0 0
2
1
2
(Q m a = )
Further,   a
dv
dt
= = tan q
As for 0 90 ° £ < ° q
tan q is + ive, a is + ive.
Case II
v k = (constant) (st line parallel to
time-axis)
Þ ds kdt
ò ò
=
s kt s = +
0
Further,    a
dv
dt
= =0
Motion in One Dimension | 27
O
t
s
A
B
D
C
q
v
M
1
t
q
v
M
2
t O
q
M
4
q
v
M
3
t
O
q
v
0
m = tan q
v
s
0
t
s
Þ
a
Þ
a = tan q
t
s
0
s
t
v
Þ
t
a
Þ
t
Case III.
v v mt = +
0
\    ds v mt dt
ò ò
= + ( )
0
i.e.,    s v t m
t
s = + +
0
2
0
2
or    s s v t at = + -
0 0
2
1
2
(Q m a = - )
Further,  a
dv
dt
= = tan q
As for 90 180 ° < £ ° q
tan q is - ive, a is - ive.
Time-displacement graph
Time Area
under the
graph
Initial
Displacement
Net
displacement
at 0 s 0 m - 10 m - 10 m
2 s 10 m - 10 m 0 m
4 s 30 m - 10 m + 20 m
6 s 40 m - 10 m + 30 m
8 s 30 m - 10 m + 20 m
10 s 20 m - 10 m + 10 m
Time-acceleration graph
time slope acceleration
0 - 2 s 5 5 m/s
2
2 - 4 s zero 0 m/s
2
4 - 6 s - 5 - 5 m/s
2
6 - 8 s - 5 - 5 m/s
2
8 - 10 s + 5 + 5 m/s
2
31. a = slope of v t - graph.
S = area under v t - graph.
Corresponding graphs are drawn in the
32. Average velocity = = =
- s
t
A A
t
Net area
Time
1 2
Average speed = =
+ d
t
A A
t
1 2
33. Average acceleration
=
-
=
- -
=-
v v
f i
time
s
10 20
6
5
2
m/
v v
f i
- = Area under a t - graph.
34. (a) Acceleration =  slope of v t - graph.
35. (b) r r s
f i
- = = area under v t - graph.
(c) Equations are written in answer sheet.
Rel a ti ve Mo ti on
35. (a) Acceleration of 1 w.r.t. 2
= - - - ( ) ( ) g g
=0 m/s
2
(b) Initial velocity of 2 w.r.t. 1
= + - - ( ) ( ) 20 5
=25 m/s
(c) Initial velocity of 1 w.r.t. 2
= - - + ( ) ( ) 5 20
= -25  m/s
28 | Mechanics-1
t 10 8 6 4 2
– 10 m
+ 10 m
+ 20 m
+ 30 m
s
t 10 8 6 4 2
5
a
q
v
v
0
t
Þ
s
s
0
t
Þ
a
t
a = tan q
\ Velocity of 1 w.r.t. 2 at time t =
æ
è
ç
ö
ø
÷
1
2
s
= - +
æ
è
ç
ö
ø
÷ ( ) ( ) 25 0
1
2
= -25 m/s
(d) Initial relative displacement of 2 w.r.t. 1
S
0
20
®
= - m
Using S S u a
® ® ® ®
= + +
0
2
1
2
rel rel
t t ,
as at time t (= time of collision of the
particles) the relative displacement of 2
w.r.t. 1 will be zero (i.e., S O
® ®
= )
O
®
= - + - ×
æ
è
ç
ö
ø
÷ ( ) ( ) 20 25
1
2
Þ        t =0.85 s
36. Let length of escalator L (= 15 m)
walking speed of man =
L
90
Speed of escalator =
L
60
Time taken by man walking on a moving
escalator =
+
L
L L
90 60
=36 s
Required time has been found without
using the actual length of the escalator.
37. S
0
®
= Initial displacement of elevator w.r.t.
ball = + 10 m.
Relative velocity of elevator w.r.t. ball
= - ( ) ( ) 2 18 = -16 m/s
Accelerator of elevator w.r.t. ball
= - - ( ) ( ) 0 10
=10 m/s
2
Using S S
® ®
= + +
0
2
1
2
u t a t
rel rel
0 10 16
1
2
10
2
= + + - + + ( ) ( ) ( ) t t
or 5 16 10 0
2
t t - - =
t =3.65 s
Position of elevator when it meets ball
= + ´ 5 2 ( ) 3.65
= 12.30 m level
38.
(a)
1
2
60
2
( ) 2.2 t =           …(i)
Þ t=7.39 s
(b)
1
2
60
2
( ) 3.5 t x = + …(ii)
Dividing Eq. (ii) by Eq. (i),
60
60
+
=
x 3.5
2.2
or
x
60
=
1.3
2.2
x =35.5 m
(c) At the time of overtaking
Speed of automobile = ( ) ( ) 3.5 7.39
=25.85 m/s
Speed of truck = ( ) ( ) 2.2 7.39
= 16.25 m/s
39. Let, acceleration of lift = a (upward)
\ acceleration of thrown body w.r.t. lift
= - - + ( ) ( ) g a
= - + ( ) g a
If time of flight is t, using
v u a t
rel rel rel
= +
( ) ( ) { ( )} - = + + - + v u a g t
Þ ( ) a g t u + = 2
or       a
u
t
g = -
2
Motion in One Dimension | 29
10 m
12 m
18 m/s
5m
2 m/s
Elevator
T
A
T
x
60 m
2
3.5 m /s
2
2.2 m /s
time = t
s
time = 0
A
40.
OB OA AB = + ( ) ( )
2 2
= + ( ) ( ) 20 20
2 2
=20 2 m
Speed along OB is 2 2 m/s
\ Time taken to reach B =
-
20 2
2 2
1
m
ms
=10 s
41. In D OPQ,
OP =
®
| | v
br
OP =
®
| | v
r
PQ
POQ
OP
OPQ sin sin Ð
=
Ð

2
45
4
45 sin( ) sin ° -
=
° q
q=
æ
è
ç
ö
ø
÷- °
é
ë
ê
ù
û
ú
-
sin
1
1
2 2
45
42. Let pilot heads point R to reach point Q
sinq=
RQ
PR
=
200
500
t
t
Þ   q =
-
sin ( )
1
0.4
( ) ( ) ( ) PR RQ PQ
2 2 2
= +
or ( ) ( ) ( ) 500 200 1000
2 2 2
t t - =
or t =
-
1000
500 200
2 2
( ) ( )
=
-
10
25 4
=
10
21
h
Objective Questions (Level 1)
Single Correct Option
1. As the packet is detached from rising
balloon its acceleration will be g in the
downward direction.
Option (b) is correct.
2. While going up :
0
1
= -
+
æ
è
ç
ö
ø
÷
u
mg f
m
T
T
um
mg f
1
=
+
…(i)
Using  v u as
2 2
2 = +
0 2
2 2
= -
+
æ
è
ç
ö
ø
÷
u
mg f
m
s
i.e., S
mu
mg f
=
+
2
2( )
While coming down
s
mg f
m
T =
-
æ
è
ç
ö
ø
÷
1
2
2
2
or
mu
mg f
mg f
m
T
2
2
2
2
1
2 ( ) +
=
-
æ
è
ç
ö
ø
÷
30 | Mechanics-1
f (Air resistance)
S
mg
Stops
u
B A
45°
O
River flow
q
P
Q
20 m
B
A
45°
O
2Ö2 m/s
OAB = OA = 20 m
q
East
P
Q R
North
Wind
200km/h
South
1000 km
Page 5

Graphy
28. OA : slope is + ive and increasing.
\ velocity is + ive and acceleration is + ive.
AB : slope is + ive and constant
\ velocity is +ive and acceleration is zero.
BC : sope is + ive and decreases.
\ velocity is + ive and increasing.
CD : slope is –ive and increasing
\ velocity is - ive and acceleration is - ive.
29.
In M
1
and M
3
0 90 : ° £ < ° q
\ slope is + ive i.e., acceleration is + ive.
In M
2
and M
4
: 90 180 ° < £ ° q
\ slope is - ive i.e., acceleration is - ive.
(a) M
1
: Magnitude of velocity is increasing.
M
2
: Magnitude of velocity is decreasing.
M
3
: Magnitude of velocity is increasing.
M
4
: Magnitude of velocity is decreasing.
Ans : M
1
and M
3
.
(b) P M ®
1
Q M ®
2
R M ®
3
S M ®
4
30. Case I
v mt v = -
0
(st-line)
\   ds mt v dt
ò ò
= - ( )
0
i.e.,  s m
t
v t s = × - +
2
0 0
2
(Parabola)
or   s s v t at = - +
0 0
2
1
2
(Q m a = )
Further,   a
dv
dt
= = tan q
As for 0 90 ° £ < ° q
tan q is + ive, a is + ive.
Case II
v k = (constant) (st line parallel to
time-axis)
Þ ds kdt
ò ò
=
s kt s = +
0
Further,    a
dv
dt
= =0
Motion in One Dimension | 27
O
t
s
A
B
D
C
q
v
M
1
t
q
v
M
2
t O
q
M
4
q
v
M
3
t
O
q
v
0
m = tan q
v
s
0
t
s
Þ
a
Þ
a = tan q
t
s
0
s
t
v
Þ
t
a
Þ
t
Case III.
v v mt = +
0
\    ds v mt dt
ò ò
= + ( )
0
i.e.,    s v t m
t
s = + +
0
2
0
2
or    s s v t at = + -
0 0
2
1
2
(Q m a = - )
Further,  a
dv
dt
= = tan q
As for 90 180 ° < £ ° q
tan q is - ive, a is - ive.
Time-displacement graph
Time Area
under the
graph
Initial
Displacement
Net
displacement
at 0 s 0 m - 10 m - 10 m
2 s 10 m - 10 m 0 m
4 s 30 m - 10 m + 20 m
6 s 40 m - 10 m + 30 m
8 s 30 m - 10 m + 20 m
10 s 20 m - 10 m + 10 m
Time-acceleration graph
time slope acceleration
0 - 2 s 5 5 m/s
2
2 - 4 s zero 0 m/s
2
4 - 6 s - 5 - 5 m/s
2
6 - 8 s - 5 - 5 m/s
2
8 - 10 s + 5 + 5 m/s
2
31. a = slope of v t - graph.
S = area under v t - graph.
Corresponding graphs are drawn in the
32. Average velocity = = =
- s
t
A A
t
Net area
Time
1 2
Average speed = =
+ d
t
A A
t
1 2
33. Average acceleration
=
-
=
- -
=-
v v
f i
time
s
10 20
6
5
2
m/
v v
f i
- = Area under a t - graph.
34. (a) Acceleration =  slope of v t - graph.
35. (b) r r s
f i
- = = area under v t - graph.
(c) Equations are written in answer sheet.
Rel a ti ve Mo ti on
35. (a) Acceleration of 1 w.r.t. 2
= - - - ( ) ( ) g g
=0 m/s
2
(b) Initial velocity of 2 w.r.t. 1
= + - - ( ) ( ) 20 5
=25 m/s
(c) Initial velocity of 1 w.r.t. 2
= - - + ( ) ( ) 5 20
= -25  m/s
28 | Mechanics-1
t 10 8 6 4 2
– 10 m
+ 10 m
+ 20 m
+ 30 m
s
t 10 8 6 4 2
5
a
q
v
v
0
t
Þ
s
s
0
t
Þ
a
t
a = tan q
\ Velocity of 1 w.r.t. 2 at time t =
æ
è
ç
ö
ø
÷
1
2
s
= - +
æ
è
ç
ö
ø
÷ ( ) ( ) 25 0
1
2
= -25 m/s
(d) Initial relative displacement of 2 w.r.t. 1
S
0
20
®
= - m
Using S S u a
® ® ® ®
= + +
0
2
1
2
rel rel
t t ,
as at time t (= time of collision of the
particles) the relative displacement of 2
w.r.t. 1 will be zero (i.e., S O
® ®
= )
O
®
= - + - ×
æ
è
ç
ö
ø
÷ ( ) ( ) 20 25
1
2
Þ        t =0.85 s
36. Let length of escalator L (= 15 m)
walking speed of man =
L
90
Speed of escalator =
L
60
Time taken by man walking on a moving
escalator =
+
L
L L
90 60
=36 s
Required time has been found without
using the actual length of the escalator.
37. S
0
®
= Initial displacement of elevator w.r.t.
ball = + 10 m.
Relative velocity of elevator w.r.t. ball
= - ( ) ( ) 2 18 = -16 m/s
Accelerator of elevator w.r.t. ball
= - - ( ) ( ) 0 10
=10 m/s
2
Using S S
® ®
= + +
0
2
1
2
u t a t
rel rel
0 10 16
1
2
10
2
= + + - + + ( ) ( ) ( ) t t
or 5 16 10 0
2
t t - - =
t =3.65 s
Position of elevator when it meets ball
= + ´ 5 2 ( ) 3.65
= 12.30 m level
38.
(a)
1
2
60
2
( ) 2.2 t =           …(i)
Þ t=7.39 s
(b)
1
2
60
2
( ) 3.5 t x = + …(ii)
Dividing Eq. (ii) by Eq. (i),
60
60
+
=
x 3.5
2.2
or
x
60
=
1.3
2.2
x =35.5 m
(c) At the time of overtaking
Speed of automobile = ( ) ( ) 3.5 7.39
=25.85 m/s
Speed of truck = ( ) ( ) 2.2 7.39
= 16.25 m/s
39. Let, acceleration of lift = a (upward)
\ acceleration of thrown body w.r.t. lift
= - - + ( ) ( ) g a
= - + ( ) g a
If time of flight is t, using
v u a t
rel rel rel
= +
( ) ( ) { ( )} - = + + - + v u a g t
Þ ( ) a g t u + = 2
or       a
u
t
g = -
2
Motion in One Dimension | 29
10 m
12 m
18 m/s
5m
2 m/s
Elevator
T
A
T
x
60 m
2
3.5 m /s
2
2.2 m /s
time = t
s
time = 0
A
40.
OB OA AB = + ( ) ( )
2 2
= + ( ) ( ) 20 20
2 2
=20 2 m
Speed along OB is 2 2 m/s
\ Time taken to reach B =
-
20 2
2 2
1
m
ms
=10 s
41. In D OPQ,
OP =
®
| | v
br
OP =
®
| | v
r
PQ
POQ
OP
OPQ sin sin Ð
=
Ð

2
45
4
45 sin( ) sin ° -
=
° q
q=
æ
è
ç
ö
ø
÷- °
é
ë
ê
ù
û
ú
-
sin
1
1
2 2
45
42. Let pilot heads point R to reach point Q
sinq=
RQ
PR
=
200
500
t
t
Þ   q =
-
sin ( )
1
0.4
( ) ( ) ( ) PR RQ PQ
2 2 2
= +
or ( ) ( ) ( ) 500 200 1000
2 2 2
t t - =
or t =
-
1000
500 200
2 2
( ) ( )
=
-
10
25 4
=
10
21
h
Objective Questions (Level 1)
Single Correct Option
1. As the packet is detached from rising
balloon its acceleration will be g in the
downward direction.
Option (b) is correct.
2. While going up :
0
1
= -
+
æ
è
ç
ö
ø
÷
u
mg f
m
T
T
um
mg f
1
=
+
…(i)
Using  v u as
2 2
2 = +
0 2
2 2
= -
+
æ
è
ç
ö
ø
÷
u
mg f
m
s
i.e., S
mu
mg f
=
+
2
2( )
While coming down
s
mg f
m
T =
-
æ
è
ç
ö
ø
÷
1
2
2
2
or
mu
mg f
mg f
m
T
2
2
2
2
1
2 ( ) +
=
-
æ
è
ç
ö
ø
÷
30 | Mechanics-1
f (Air resistance)
S
mg
Stops
u
B A
45°
O
River flow
q
P
Q
20 m
B
A
45°
O
2Ö2 m/s
OAB = OA = 20 m
q
East
P
Q R
North
Wind
200km/h
South
1000 km
Þ    T
um
mg f mg f
2
=
+ - ( )( )
…(ii)
Using Eq. (i) and Eq. (ii)
T
T
mg f
mg f
2
1
=
+
-
Þ           T T
2 1
>
Option (c) is correct.
3. Angular speed (w) of seconds hand
=
2
60
p
-1
Speed of the tip of seconds hand
v = ´
p
30
1 cm/s    [Q v r = w and
r=1cm]
As in 15s the seconds hand rotates
through 90°, the change in velocity of its
tip in 15 s will be
=v 2 =
-
p 2
30
1
cms
Option (d) is correct.
4. Average speed =
+
+
=
-
5 5
5
30
5
60
40
1
ms
Option (c) is correct.
5. Relative velocity of boat w.r.t. water
= + - - - ( ) ( )
^ ^ ^ ^
3 4 3 4 i j i j = + 6 8 i j
^ ^
Option (b) is correct.
6. 21
18 11 42
18 42
=
´ + ´
+
( ) ( ) v
Þ v = 25.29 m/s
7.     x t
t
= - 32
8
3
3
v
dx
dt
t = = - 32 8
2
…(i)
\ Particle is at rest when
32 8 0
2
- = t
i.e.,        t =2 s
Differentiating Eq. (i) w.r.t. time t
a
dv
dt
t = = - 16
\  a at time t = 2 s (when particle is at rest)
= - ´ ( ) ( ) 16 2 = -32 m/s
2
Option (b) is correct.
8. For first one second
2
1
2
1
2
= ´ a
Þ a = 4 m/s
2
Velocity at the end of next second
v = ´ ( ) ( ) 4 2
= 8 m/s
Option (b) is correct.
9. x t t = - + 3
3
\ displacement at time t (= 1 s)
= - + = - 3 1 1 2
3
( ) ( ) m
and displacement at time t (= 3 s)
= - + = 3 3 3 18
3
( ) ( ) m
And as such displacement in the time
interval (t = 1 s to t = 3 s)
= - - ( ) ( ) 18 2 m m
= +20 m
Option (c) is correct.
10. Acceleration a bt =
\
dv
dt
bt =
or       dv btdt
ò ò
=
or         v bt C = +
1
2
2
Now, at t = 0, v v =
0
\          C v =
0
i.e., v bt v = +
1
2
2
0
or
ds
dt
v bt = +
0
2
1
2

or   ds v bt dt
ò ò
= +
æ
è
ç
ö
ø
÷
0
2
1
2
or   s v t bt k = + +
0
3
1
6
At t =0, s =0,
\            k=0
Þ s v t bt = +
0
3
1
6
Option (a) is correct.
Motion in One Dimension | 31
```

122 docs

## FAQs on DC Pandey Solutions: Motion in One Dimension- 4 - 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. It involves the analysis of the object's position, velocity, and acceleration in relation to time. This type of motion occurs when the object's motion is restricted to only one direction, either forward or backward.
 2. What are the basic principles of motion in one dimension?
Ans. The basic principles of motion in one dimension include the concepts of displacement, velocity, and acceleration. Displacement refers to the change in position of an object with respect to a reference point. Velocity is the rate at which an object changes its position, while acceleration is the rate at which an object changes its velocity.
 3. How can we calculate displacement in motion in one dimension?
Ans. Displacement in motion in one dimension can be calculated by subtracting the initial position of an object from its final position. It is represented by the symbol Δx and is a vector quantity, meaning it has both magnitude and direction. The formula for displacement is Δx = xf - xi, where xf is the final position and xi is the initial position.
 4. What is the difference between speed and velocity in motion in one dimension?
Ans. Speed and velocity are both measures of how fast an object is moving, but they differ in their definitions. Speed is a scalar quantity that only considers the magnitude of the object's motion, while velocity is a vector quantity that takes into account both the magnitude and direction of the object's motion. Therefore, velocity includes information about the object's speed and the direction it is moving in.
 5. How is acceleration calculated in motion in one dimension?
Ans. Acceleration in motion in one dimension can be calculated by dividing the change in velocity by the change in time. It is represented by the symbol a and is a vector quantity. The formula for acceleration is a = Δv / Δt, where Δv is the change in velocity and Δt is the change in time. Positive acceleration indicates an increase in velocity, while negative acceleration indicates a decrease in velocity.

## DC Pandey Solutions for NEET Physics

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