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# Chapter 3 - Motion in One Dimension (Part - 4) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

## DC Pandey (Questions & Solutions) of Physics: NEET

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## NEET : Chapter 3 - Motion in One Dimension (Part - 4) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

``` 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
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