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 Page 1


42 | Mechanics-1
10.      x gt
1
2
1
2
=
  x x g t
1 2
2
1
2
2 + = ( )
\     x g t gt
2
2 2
1
2
4
1
2
= × -
      x gt
2
2
3
2
=
   x x gt
2 1
2
- =
Þ      t
x x
g
=
-
2 1
Option (a) is correct.
11.        y x l
2 2 2
+ =
\    2 2 0 y
dy
dt
x
dx
dt
× + × =
or             
dy
dt
x
y
dx
dt
= -
            = -
°
x
x
dx
dt tan30
        = -2 3 m/s
Option (c) is correct.
12. Maximum separation (x
max
) between the
police and the thief will be at time t (shown
in figure)
t
v
a
- = =
-
-
2
90
5
1
2
kmh
ms
    = ´
90
5
1000
3600
 s
i.e.,           t = 7 s
x
max
[ ] = ´ - ´ ´ 90 7
1
2
5 90 kmh s s kmh
–1 –1
       =
´
´
90 1000
3600
9
2
 m
       =112.5 m
Option (a) is correct.
13. If meeting time is t
2
3
30
v
u sin sin q = °
i.e.,           sinq =
3
4
or   q =
-
sin
1
3
4
Option (c) is correct.
14. Distance =
+
æ
è
ç
ö
ø
÷
1
2
2
ab
a b
t
\      0
1
2
1 4
1 4
2
=
´
+
æ
è
ç
ö
ø
÷ t
Þ      t = 500 s
        =22.36 s
Option (a) is correct.
15. Let BC t = ¢
\              
24
4
t¢
=
Þ t¢ =6 s
\              OB =50 s
Let OA t =
\        AB t = - 56
1032
1
2
56 56 24 = + - ´ [ ( )] t
Þ         t =20 s
\ maximum acceleration =
24
20
m / s
s
            =1.2 m/s
2
Option (b) is correct.
16. From DOAB
cos AOB
OA OB AB
OA OB
=
+ -
× ×
2 2 2
2
 
v
time (t)
Thief
Police
108 km/h
90 km/h
2 s 7 s
t
O
24
C B A
v (m/s)
t (s)
60°
O B
(3 – 0.2t) m
Q
– 0.2t
0.2t
P
(4 – 0.2t) m
Page 2


42 | Mechanics-1
10.      x gt
1
2
1
2
=
  x x g t
1 2
2
1
2
2 + = ( )
\     x g t gt
2
2 2
1
2
4
1
2
= × -
      x gt
2
2
3
2
=
   x x gt
2 1
2
- =
Þ      t
x x
g
=
-
2 1
Option (a) is correct.
11.        y x l
2 2 2
+ =
\    2 2 0 y
dy
dt
x
dx
dt
× + × =
or             
dy
dt
x
y
dx
dt
= -
            = -
°
x
x
dx
dt tan30
        = -2 3 m/s
Option (c) is correct.
12. Maximum separation (x
max
) between the
police and the thief will be at time t (shown
in figure)
t
v
a
- = =
-
-
2
90
5
1
2
kmh
ms
    = ´
90
5
1000
3600
 s
i.e.,           t = 7 s
x
max
[ ] = ´ - ´ ´ 90 7
1
2
5 90 kmh s s kmh
–1 –1
       =
´
´
90 1000
3600
9
2
 m
       =112.5 m
Option (a) is correct.
13. If meeting time is t
2
3
30
v
u sin sin q = °
i.e.,           sinq =
3
4
or   q =
-
sin
1
3
4
Option (c) is correct.
14. Distance =
+
æ
è
ç
ö
ø
÷
1
2
2
ab
a b
t
\      0
1
2
1 4
1 4
2
=
´
+
æ
è
ç
ö
ø
÷ t
Þ      t = 500 s
        =22.36 s
Option (a) is correct.
15. Let BC t = ¢
\              
24
4
t¢
=
Þ t¢ =6 s
\              OB =50 s
Let OA t =
\        AB t = - 56
1032
1
2
56 56 24 = + - ´ [ ( )] t
Þ         t =20 s
\ maximum acceleration =
24
20
m / s
s
            =1.2 m/s
2
Option (b) is correct.
16. From DOAB
cos AOB
OA OB AB
OA OB
=
+ -
× ×
2 2 2
2
 
v
time (t)
Thief
Police
108 km/h
90 km/h
2 s 7 s
t
O
24
C B A
v (m/s)
t (s)
60°
O B
(3 – 0.2t) m
Q
– 0.2t
0.2t
P
(4 – 0.2t) m
\ OA OB AB OA OB
2 2 2
+ - = ×
i.e., AB OA OB OA OB
2 2 2
= + - ×
     = - + - ( ) ( ) 4 3
2 2
0.2 0.2 t t
+ - - ( )( ) 4 3 0.2 0.2 t t
or    AB t t t
2 2 2
16 9 = + - + + 0.04 1.6 0.04
- + - + 1.2 1.4 0.04 t t t 12
2
or   AB t t
2 2
37 = - + 0.12 4.2
For AB to be minimum
0.24 4.2 t - = 0
or t = 17.5 s
\   ( ) ( ) ( )
min
AB
2 2
37 = - ´ + 0.12 17.5 4.2 17.5
          = - 36.75 73.5 +37
   ( )
min
AB
2
=0.75 m
2
          = 7500
2
cm
=50 3 cm
Option (d) is correct.
17. Velocity of ball w.r.t. elevator = 15 m/s
Acceleration of ball w.r.t. elevator
= - - + ( ) ( ) 10 5 = - 15 m/s
2
Final displacement of ball w.r.t. elevator 
=-2 m
\ - = + - 2 15
1
2
15
2
t t ( )
i.e., 15 30 4 0
2
t t - - =
\ t =
+ + ´ ´
´
30 900 4 4 15
2 15
          =2.13 s
Option (a) is correct.
18. Velocity of ball w.r.t. ground ( ) v
BE
= + ( ) 15 10 m/s
             =25 m/s
Now       v u as
2 2
2 = +
\ 0 25 2 10
2 2
= + - ( ) ( )H
or        H =31.25 m
Maximum height by ball as measured from
ground = + + 31.25 2 50
= 83.25 m
Option (c) is correct.
19. Displacement of ball w.r.t. ground during its
flight = = H 31.25 m
Option (d) is correct.
20. Dis place ment of ball w.r.t. floor of el e va tor
at time t 
s t t = + - + 15
1
2
15 2
2
( )
s will be maximum, when
ds
dt
= 0
i.e., 15 15 0 - = t
i.e., t = 1s
\ s
max
= ´ + - + ( ) ( )( ) 15 1
1
2
10 1 2
2
= 12 m
Option (a) is correct.
21. Let the particles meet at time t 
i.e., displacement of the particles are equal
and which is possible when
Area of D ACB = Area of D A B C ¢ ¢
[Area OBCA O ¢ q being common]
or  Area of D APC = Area A P C ¢ ¢
1
2
1
2
AP PC A P P C ´ = ¢ ¢ ´ ¢
or       AP PC A P P C ´ = ¢ ¢ ´ ¢
or   ( tan ) ( tan ) PC PC P C P C q q = ¢ ¢
or            PC P C = ¢ ¢
Motion in One Dimension | 43
2 m 10 m/s
15 m/s = v
BE
H
2
5 m/s
+
Max. height attained by ball
50 m
v
B
B'
P'
u
B
u
A
C
q
q
P
A
B
v
time (s)
Q R 4 O
v
A
A'
Page 3


42 | Mechanics-1
10.      x gt
1
2
1
2
=
  x x g t
1 2
2
1
2
2 + = ( )
\     x g t gt
2
2 2
1
2
4
1
2
= × -
      x gt
2
2
3
2
=
   x x gt
2 1
2
- =
Þ      t
x x
g
=
-
2 1
Option (a) is correct.
11.        y x l
2 2 2
+ =
\    2 2 0 y
dy
dt
x
dx
dt
× + × =
or             
dy
dt
x
y
dx
dt
= -
            = -
°
x
x
dx
dt tan30
        = -2 3 m/s
Option (c) is correct.
12. Maximum separation (x
max
) between the
police and the thief will be at time t (shown
in figure)
t
v
a
- = =
-
-
2
90
5
1
2
kmh
ms
    = ´
90
5
1000
3600
 s
i.e.,           t = 7 s
x
max
[ ] = ´ - ´ ´ 90 7
1
2
5 90 kmh s s kmh
–1 –1
       =
´
´
90 1000
3600
9
2
 m
       =112.5 m
Option (a) is correct.
13. If meeting time is t
2
3
30
v
u sin sin q = °
i.e.,           sinq =
3
4
or   q =
-
sin
1
3
4
Option (c) is correct.
14. Distance =
+
æ
è
ç
ö
ø
÷
1
2
2
ab
a b
t
\      0
1
2
1 4
1 4
2
=
´
+
æ
è
ç
ö
ø
÷ t
Þ      t = 500 s
        =22.36 s
Option (a) is correct.
15. Let BC t = ¢
\              
24
4
t¢
=
Þ t¢ =6 s
\              OB =50 s
Let OA t =
\        AB t = - 56
1032
1
2
56 56 24 = + - ´ [ ( )] t
Þ         t =20 s
\ maximum acceleration =
24
20
m / s
s
            =1.2 m/s
2
Option (b) is correct.
16. From DOAB
cos AOB
OA OB AB
OA OB
=
+ -
× ×
2 2 2
2
 
v
time (t)
Thief
Police
108 km/h
90 km/h
2 s 7 s
t
O
24
C B A
v (m/s)
t (s)
60°
O B
(3 – 0.2t) m
Q
– 0.2t
0.2t
P
(4 – 0.2t) m
\ OA OB AB OA OB
2 2 2
+ - = ×
i.e., AB OA OB OA OB
2 2 2
= + - ×
     = - + - ( ) ( ) 4 3
2 2
0.2 0.2 t t
+ - - ( )( ) 4 3 0.2 0.2 t t
or    AB t t t
2 2 2
16 9 = + - + + 0.04 1.6 0.04
- + - + 1.2 1.4 0.04 t t t 12
2
or   AB t t
2 2
37 = - + 0.12 4.2
For AB to be minimum
0.24 4.2 t - = 0
or t = 17.5 s
\   ( ) ( ) ( )
min
AB
2 2
37 = - ´ + 0.12 17.5 4.2 17.5
          = - 36.75 73.5 +37
   ( )
min
AB
2
=0.75 m
2
          = 7500
2
cm
=50 3 cm
Option (d) is correct.
17. Velocity of ball w.r.t. elevator = 15 m/s
Acceleration of ball w.r.t. elevator
= - - + ( ) ( ) 10 5 = - 15 m/s
2
Final displacement of ball w.r.t. elevator 
=-2 m
\ - = + - 2 15
1
2
15
2
t t ( )
i.e., 15 30 4 0
2
t t - - =
\ t =
+ + ´ ´
´
30 900 4 4 15
2 15
          =2.13 s
Option (a) is correct.
18. Velocity of ball w.r.t. ground ( ) v
BE
= + ( ) 15 10 m/s
             =25 m/s
Now       v u as
2 2
2 = +
\ 0 25 2 10
2 2
= + - ( ) ( )H
or        H =31.25 m
Maximum height by ball as measured from
ground = + + 31.25 2 50
= 83.25 m
Option (c) is correct.
19. Displacement of ball w.r.t. ground during its
flight = = H 31.25 m
Option (d) is correct.
20. Dis place ment of ball w.r.t. floor of el e va tor
at time t 
s t t = + - + 15
1
2
15 2
2
( )
s will be maximum, when
ds
dt
= 0
i.e., 15 15 0 - = t
i.e., t = 1s
\ s
max
= ´ + - + ( ) ( )( ) 15 1
1
2
10 1 2
2
= 12 m
Option (a) is correct.
21. Let the particles meet at time t 
i.e., displacement of the particles are equal
and which is possible when
Area of D ACB = Area of D A B C ¢ ¢
[Area OBCA O ¢ q being common]
or  Area of D APC = Area A P C ¢ ¢
1
2
1
2
AP PC A P P C ´ = ¢ ¢ ´ ¢
or       AP PC A P P C ´ = ¢ ¢ ´ ¢
or   ( tan ) ( tan ) PC PC P C P C q q = ¢ ¢
or            PC P C = ¢ ¢
Motion in One Dimension | 43
2 m 10 m/s
15 m/s = v
BE
H
2
5 m/s
+
Max. height attained by ball
50 m
v
B
B'
P'
u
B
u
A
C
q
q
P
A
B
v
time (s)
Q R 4 O
v
A
A'
\    OR PC P C = + ¢ ¢ =2PC = 8 s.
Option (c) is correct.
22. Area of D A B O ¢ ¢ = Area of D ABO
\
1
2
4
1
2
4 ( ) ( ) v v u u
B A A B
- × = - ×
i.e., v v u u
B A A B
- = -
                = - 5 15 = - 10
Þ v v
A B
- =
-
10
1
ms
Option (b) is correct.
23. u
A
=
-
6
1
ms , u
B
=
-
12
1
ms
and at t = 4 s common velocity =
-
8
1
ms
To find velocity of A at t = 10 s
v u
A A
-
-
=
- 8
10 4
8
4
or
v
A
-
=
- 8
6
6 8
4
\     v
A
=
-
5
1
ms
Option (d) is correct.
More than One Cor rect Op tions
1. Q          a v = -a
1 2 /
              …(i)
i.e.,       
dv
dt
v = -a
1 2 /
or  v dv dt
-
ò ò
= -
1 2 /
a
       2
1 2
1
v t C
/
= - + a
Now, at t = 0, v v =
0
\ C v
1 0
1 2
2 =
/
Þ    2 2
1 2
0
1 2
v t v
/ /
= - + a
i.e., the particle will stop at
t
v
=
2
0
1 2 /
a
Option (a) is correct.
Option (b) is incorrect.
From Eq. (i),
   v
dv
dt
v = -a
1 2 /
or v dv dx
1 2 /
ò ò
= -a
or  
2
3
3 2
2
v C
/
= - + a
As at x = 0, v v =
0
, C v
2 0
3 2
2
3
=
/
\    
2
3
2
3
3 2
0
3 2
v x v
/ /
= - + a
i.e., when the particle stops
  x v =
2
3
0
3 2
a
/
Option (d) is correct.
Option (c) is incorrect.
2. a t = - 0.5 (m/s
2
)
dv
dt
t
= -
2
or    dv tdt
ò ò
= -
1
2
       v
t
C = - +
2
1
4
At t =0, v=16 m/s
\ v
t
= - +
2
4
16 …(i)
From above relation v is zero at
t = 8 s
Options (a) is correct.
From relation (i),
ds
dt
t
= - +
2
4
16
\            ds
t
dt
ò ò
= - +
æ
è
ç
ö
ø
÷
2
4
16
i.e.,        s
t
t C = - + +
3
2
12
16
                s
t
t = - +
3
12
16 [As at t = 0, s = 0]
At t = 4 s
s = 58.67 m
Option (b) is correct.
The particle returns back at t = 8 s
From relation (ii)
s
8
3
8
12
16 8 = - + ´ = ( ) 85.33 m
  s
10
3
10
12
16 10 = - + ´ = ( ) 76.66 m
Distance travelled in 10 s
= + - s s s
8 8 10
( )
    = ´ - ( ) 2 85.33 76.66
            =94 m 
Option (c) is correct.
Velocity of particle at t = 10 s
44 | Mechanics-1
Page 4


42 | Mechanics-1
10.      x gt
1
2
1
2
=
  x x g t
1 2
2
1
2
2 + = ( )
\     x g t gt
2
2 2
1
2
4
1
2
= × -
      x gt
2
2
3
2
=
   x x gt
2 1
2
- =
Þ      t
x x
g
=
-
2 1
Option (a) is correct.
11.        y x l
2 2 2
+ =
\    2 2 0 y
dy
dt
x
dx
dt
× + × =
or             
dy
dt
x
y
dx
dt
= -
            = -
°
x
x
dx
dt tan30
        = -2 3 m/s
Option (c) is correct.
12. Maximum separation (x
max
) between the
police and the thief will be at time t (shown
in figure)
t
v
a
- = =
-
-
2
90
5
1
2
kmh
ms
    = ´
90
5
1000
3600
 s
i.e.,           t = 7 s
x
max
[ ] = ´ - ´ ´ 90 7
1
2
5 90 kmh s s kmh
–1 –1
       =
´
´
90 1000
3600
9
2
 m
       =112.5 m
Option (a) is correct.
13. If meeting time is t
2
3
30
v
u sin sin q = °
i.e.,           sinq =
3
4
or   q =
-
sin
1
3
4
Option (c) is correct.
14. Distance =
+
æ
è
ç
ö
ø
÷
1
2
2
ab
a b
t
\      0
1
2
1 4
1 4
2
=
´
+
æ
è
ç
ö
ø
÷ t
Þ      t = 500 s
        =22.36 s
Option (a) is correct.
15. Let BC t = ¢
\              
24
4
t¢
=
Þ t¢ =6 s
\              OB =50 s
Let OA t =
\        AB t = - 56
1032
1
2
56 56 24 = + - ´ [ ( )] t
Þ         t =20 s
\ maximum acceleration =
24
20
m / s
s
            =1.2 m/s
2
Option (b) is correct.
16. From DOAB
cos AOB
OA OB AB
OA OB
=
+ -
× ×
2 2 2
2
 
v
time (t)
Thief
Police
108 km/h
90 km/h
2 s 7 s
t
O
24
C B A
v (m/s)
t (s)
60°
O B
(3 – 0.2t) m
Q
– 0.2t
0.2t
P
(4 – 0.2t) m
\ OA OB AB OA OB
2 2 2
+ - = ×
i.e., AB OA OB OA OB
2 2 2
= + - ×
     = - + - ( ) ( ) 4 3
2 2
0.2 0.2 t t
+ - - ( )( ) 4 3 0.2 0.2 t t
or    AB t t t
2 2 2
16 9 = + - + + 0.04 1.6 0.04
- + - + 1.2 1.4 0.04 t t t 12
2
or   AB t t
2 2
37 = - + 0.12 4.2
For AB to be minimum
0.24 4.2 t - = 0
or t = 17.5 s
\   ( ) ( ) ( )
min
AB
2 2
37 = - ´ + 0.12 17.5 4.2 17.5
          = - 36.75 73.5 +37
   ( )
min
AB
2
=0.75 m
2
          = 7500
2
cm
=50 3 cm
Option (d) is correct.
17. Velocity of ball w.r.t. elevator = 15 m/s
Acceleration of ball w.r.t. elevator
= - - + ( ) ( ) 10 5 = - 15 m/s
2
Final displacement of ball w.r.t. elevator 
=-2 m
\ - = + - 2 15
1
2
15
2
t t ( )
i.e., 15 30 4 0
2
t t - - =
\ t =
+ + ´ ´
´
30 900 4 4 15
2 15
          =2.13 s
Option (a) is correct.
18. Velocity of ball w.r.t. ground ( ) v
BE
= + ( ) 15 10 m/s
             =25 m/s
Now       v u as
2 2
2 = +
\ 0 25 2 10
2 2
= + - ( ) ( )H
or        H =31.25 m
Maximum height by ball as measured from
ground = + + 31.25 2 50
= 83.25 m
Option (c) is correct.
19. Displacement of ball w.r.t. ground during its
flight = = H 31.25 m
Option (d) is correct.
20. Dis place ment of ball w.r.t. floor of el e va tor
at time t 
s t t = + - + 15
1
2
15 2
2
( )
s will be maximum, when
ds
dt
= 0
i.e., 15 15 0 - = t
i.e., t = 1s
\ s
max
= ´ + - + ( ) ( )( ) 15 1
1
2
10 1 2
2
= 12 m
Option (a) is correct.
21. Let the particles meet at time t 
i.e., displacement of the particles are equal
and which is possible when
Area of D ACB = Area of D A B C ¢ ¢
[Area OBCA O ¢ q being common]
or  Area of D APC = Area A P C ¢ ¢
1
2
1
2
AP PC A P P C ´ = ¢ ¢ ´ ¢
or       AP PC A P P C ´ = ¢ ¢ ´ ¢
or   ( tan ) ( tan ) PC PC P C P C q q = ¢ ¢
or            PC P C = ¢ ¢
Motion in One Dimension | 43
2 m 10 m/s
15 m/s = v
BE
H
2
5 m/s
+
Max. height attained by ball
50 m
v
B
B'
P'
u
B
u
A
C
q
q
P
A
B
v
time (s)
Q R 4 O
v
A
A'
\    OR PC P C = + ¢ ¢ =2PC = 8 s.
Option (c) is correct.
22. Area of D A B O ¢ ¢ = Area of D ABO
\
1
2
4
1
2
4 ( ) ( ) v v u u
B A A B
- × = - ×
i.e., v v u u
B A A B
- = -
                = - 5 15 = - 10
Þ v v
A B
- =
-
10
1
ms
Option (b) is correct.
23. u
A
=
-
6
1
ms , u
B
=
-
12
1
ms
and at t = 4 s common velocity =
-
8
1
ms
To find velocity of A at t = 10 s
v u
A A
-
-
=
- 8
10 4
8
4
or
v
A
-
=
- 8
6
6 8
4
\     v
A
=
-
5
1
ms
Option (d) is correct.
More than One Cor rect Op tions
1. Q          a v = -a
1 2 /
              …(i)
i.e.,       
dv
dt
v = -a
1 2 /
or  v dv dt
-
ò ò
= -
1 2 /
a
       2
1 2
1
v t C
/
= - + a
Now, at t = 0, v v =
0
\ C v
1 0
1 2
2 =
/
Þ    2 2
1 2
0
1 2
v t v
/ /
= - + a
i.e., the particle will stop at
t
v
=
2
0
1 2 /
a
Option (a) is correct.
Option (b) is incorrect.
From Eq. (i),
   v
dv
dt
v = -a
1 2 /
or v dv dx
1 2 /
ò ò
= -a
or  
2
3
3 2
2
v C
/
= - + a
As at x = 0, v v =
0
, C v
2 0
3 2
2
3
=
/
\    
2
3
2
3
3 2
0
3 2
v x v
/ /
= - + a
i.e., when the particle stops
  x v =
2
3
0
3 2
a
/
Option (d) is correct.
Option (c) is incorrect.
2. a t = - 0.5 (m/s
2
)
dv
dt
t
= -
2
or    dv tdt
ò ò
= -
1
2
       v
t
C = - +
2
1
4
At t =0, v=16 m/s
\ v
t
= - +
2
4
16 …(i)
From above relation v is zero at
t = 8 s
Options (a) is correct.
From relation (i),
ds
dt
t
= - +
2
4
16
\            ds
t
dt
ò ò
= - +
æ
è
ç
ö
ø
÷
2
4
16
i.e.,        s
t
t C = - + +
3
2
12
16
                s
t
t = - +
3
12
16 [As at t = 0, s = 0]
At t = 4 s
s = 58.67 m
Option (b) is correct.
The particle returns back at t = 8 s
From relation (ii)
s
8
3
8
12
16 8 = - + ´ = ( ) 85.33 m
  s
10
3
10
12
16 10 = - + ´ = ( ) 76.66 m
Distance travelled in 10 s
= + - s s s
8 8 10
( )
    = ´ - ( ) 2 85.33 76.66
            =94 m 
Option (c) is correct.
Velocity of particle at t = 10 s
44 | Mechanics-1
v
10
2
10
4
16 = - +
( )
= - + 25 16
              = -9 m/s
\ Speed of particle at t = 10 s is 9 m/s.
Option (d) is correct.
3. | | v
®
 is scalar. 
\ Option (a) is incorrect.
d
dt
v
a
®
®
= (by definition).
\ Option (b) is correct.
v
2
 is scalar.
\ Option (c) is incorrect.
v
v
®
®
| |
 is v
^
 (unit vector)
\
d
dt
v
a a
^
^
= ¹
®
\ Option (d) is incorrect.
4.
    v j
A
®
-
= -20
1 ^
kmh
    v i j
B
®
= ° + ° ( cos ) ( sin )
^ ^
40 37 40 37  kmh
-1
        = +
-
( )
^ ^
32 24
1
i j kmh
   v v v
AB A B
®
= -
® ®
       = - - + ( ) ( )
^ ^ ^
20 32 24 j i j
       = - - 32i j
^ ^
44
Option (a) is correct.
Option (c) is incorrect.
At any time
    S i j j
A
t
®
= + + - 3 4 20
^ ^ ^
( )
S i j
B
t
®
= + ( )
^ ^
32 24
Position of A relative to B :
S S S
AB A B
® ® ®
= -
    = + - - + ( ) ( )
^ ^ ^ ^ ^
3 4 20 32 24 i j j i j t t t
    = - + - ( ) ( )
^ ^
3 32 4 44 t t i j
Option (b) is correct.
Option (d) is incorrect.
5. Case I.
                       v a t =
1 1
                       v a t =
2 2
s a t a t
1 1 1
2
2 2
2
1
2
1
2
= +
Case II.
             v a t ¢ =2
1 1
              =2v
            2
2 3
v a t =
         2
2 2 2 3
( ) a t a t =
Þ          t t
3 2
2 =
    s a t at
2 1 1
2
23
2
1
2
2
1
2
= + ( )
         = + a t a t
1 1
2
2 2
2
1
2
2 ( )
         = + a t a t
1 1
2
2 2
2
2
      2
2 1 1
2
2 2
2
s a t a t = +
\      s s
1 1
2 > …(i)
      4 2 2
1 1 1
2
2 2
2
s a t a t = +
\      s s
2 1
4 < …(ii)
Combining Eqs. (i) and (ii),
2 4
1 2 1
s s s < <
Option (d) is correct.
In Case I. v
s
t t
av
=
+
1
1 2
i.e.,        v
a t a t
t t
1
1 1
2
2 2
2
1 2
1
2
1
2
=
+
+
Motion in One Dimension | 45
East
X (km)
20 km/h
A (3, 4)
North
Y (km)
South
B
37°
40 km/h
– a
2
a
1
t
1
t
2
v = (Let)
S
1
Rest
Average
velocity = v
1
Rest
– a
2
2a
1
t
1
t
3
S
2
Rest
Average
velocity = v
2
v' = 2v
Page 5


42 | Mechanics-1
10.      x gt
1
2
1
2
=
  x x g t
1 2
2
1
2
2 + = ( )
\     x g t gt
2
2 2
1
2
4
1
2
= × -
      x gt
2
2
3
2
=
   x x gt
2 1
2
- =
Þ      t
x x
g
=
-
2 1
Option (a) is correct.
11.        y x l
2 2 2
+ =
\    2 2 0 y
dy
dt
x
dx
dt
× + × =
or             
dy
dt
x
y
dx
dt
= -
            = -
°
x
x
dx
dt tan30
        = -2 3 m/s
Option (c) is correct.
12. Maximum separation (x
max
) between the
police and the thief will be at time t (shown
in figure)
t
v
a
- = =
-
-
2
90
5
1
2
kmh
ms
    = ´
90
5
1000
3600
 s
i.e.,           t = 7 s
x
max
[ ] = ´ - ´ ´ 90 7
1
2
5 90 kmh s s kmh
–1 –1
       =
´
´
90 1000
3600
9
2
 m
       =112.5 m
Option (a) is correct.
13. If meeting time is t
2
3
30
v
u sin sin q = °
i.e.,           sinq =
3
4
or   q =
-
sin
1
3
4
Option (c) is correct.
14. Distance =
+
æ
è
ç
ö
ø
÷
1
2
2
ab
a b
t
\      0
1
2
1 4
1 4
2
=
´
+
æ
è
ç
ö
ø
÷ t
Þ      t = 500 s
        =22.36 s
Option (a) is correct.
15. Let BC t = ¢
\              
24
4
t¢
=
Þ t¢ =6 s
\              OB =50 s
Let OA t =
\        AB t = - 56
1032
1
2
56 56 24 = + - ´ [ ( )] t
Þ         t =20 s
\ maximum acceleration =
24
20
m / s
s
            =1.2 m/s
2
Option (b) is correct.
16. From DOAB
cos AOB
OA OB AB
OA OB
=
+ -
× ×
2 2 2
2
 
v
time (t)
Thief
Police
108 km/h
90 km/h
2 s 7 s
t
O
24
C B A
v (m/s)
t (s)
60°
O B
(3 – 0.2t) m
Q
– 0.2t
0.2t
P
(4 – 0.2t) m
\ OA OB AB OA OB
2 2 2
+ - = ×
i.e., AB OA OB OA OB
2 2 2
= + - ×
     = - + - ( ) ( ) 4 3
2 2
0.2 0.2 t t
+ - - ( )( ) 4 3 0.2 0.2 t t
or    AB t t t
2 2 2
16 9 = + - + + 0.04 1.6 0.04
- + - + 1.2 1.4 0.04 t t t 12
2
or   AB t t
2 2
37 = - + 0.12 4.2
For AB to be minimum
0.24 4.2 t - = 0
or t = 17.5 s
\   ( ) ( ) ( )
min
AB
2 2
37 = - ´ + 0.12 17.5 4.2 17.5
          = - 36.75 73.5 +37
   ( )
min
AB
2
=0.75 m
2
          = 7500
2
cm
=50 3 cm
Option (d) is correct.
17. Velocity of ball w.r.t. elevator = 15 m/s
Acceleration of ball w.r.t. elevator
= - - + ( ) ( ) 10 5 = - 15 m/s
2
Final displacement of ball w.r.t. elevator 
=-2 m
\ - = + - 2 15
1
2
15
2
t t ( )
i.e., 15 30 4 0
2
t t - - =
\ t =
+ + ´ ´
´
30 900 4 4 15
2 15
          =2.13 s
Option (a) is correct.
18. Velocity of ball w.r.t. ground ( ) v
BE
= + ( ) 15 10 m/s
             =25 m/s
Now       v u as
2 2
2 = +
\ 0 25 2 10
2 2
= + - ( ) ( )H
or        H =31.25 m
Maximum height by ball as measured from
ground = + + 31.25 2 50
= 83.25 m
Option (c) is correct.
19. Displacement of ball w.r.t. ground during its
flight = = H 31.25 m
Option (d) is correct.
20. Dis place ment of ball w.r.t. floor of el e va tor
at time t 
s t t = + - + 15
1
2
15 2
2
( )
s will be maximum, when
ds
dt
= 0
i.e., 15 15 0 - = t
i.e., t = 1s
\ s
max
= ´ + - + ( ) ( )( ) 15 1
1
2
10 1 2
2
= 12 m
Option (a) is correct.
21. Let the particles meet at time t 
i.e., displacement of the particles are equal
and which is possible when
Area of D ACB = Area of D A B C ¢ ¢
[Area OBCA O ¢ q being common]
or  Area of D APC = Area A P C ¢ ¢
1
2
1
2
AP PC A P P C ´ = ¢ ¢ ´ ¢
or       AP PC A P P C ´ = ¢ ¢ ´ ¢
or   ( tan ) ( tan ) PC PC P C P C q q = ¢ ¢
or            PC P C = ¢ ¢
Motion in One Dimension | 43
2 m 10 m/s
15 m/s = v
BE
H
2
5 m/s
+
Max. height attained by ball
50 m
v
B
B'
P'
u
B
u
A
C
q
q
P
A
B
v
time (s)
Q R 4 O
v
A
A'
\    OR PC P C = + ¢ ¢ =2PC = 8 s.
Option (c) is correct.
22. Area of D A B O ¢ ¢ = Area of D ABO
\
1
2
4
1
2
4 ( ) ( ) v v u u
B A A B
- × = - ×
i.e., v v u u
B A A B
- = -
                = - 5 15 = - 10
Þ v v
A B
- =
-
10
1
ms
Option (b) is correct.
23. u
A
=
-
6
1
ms , u
B
=
-
12
1
ms
and at t = 4 s common velocity =
-
8
1
ms
To find velocity of A at t = 10 s
v u
A A
-
-
=
- 8
10 4
8
4
or
v
A
-
=
- 8
6
6 8
4
\     v
A
=
-
5
1
ms
Option (d) is correct.
More than One Cor rect Op tions
1. Q          a v = -a
1 2 /
              …(i)
i.e.,       
dv
dt
v = -a
1 2 /
or  v dv dt
-
ò ò
= -
1 2 /
a
       2
1 2
1
v t C
/
= - + a
Now, at t = 0, v v =
0
\ C v
1 0
1 2
2 =
/
Þ    2 2
1 2
0
1 2
v t v
/ /
= - + a
i.e., the particle will stop at
t
v
=
2
0
1 2 /
a
Option (a) is correct.
Option (b) is incorrect.
From Eq. (i),
   v
dv
dt
v = -a
1 2 /
or v dv dx
1 2 /
ò ò
= -a
or  
2
3
3 2
2
v C
/
= - + a
As at x = 0, v v =
0
, C v
2 0
3 2
2
3
=
/
\    
2
3
2
3
3 2
0
3 2
v x v
/ /
= - + a
i.e., when the particle stops
  x v =
2
3
0
3 2
a
/
Option (d) is correct.
Option (c) is incorrect.
2. a t = - 0.5 (m/s
2
)
dv
dt
t
= -
2
or    dv tdt
ò ò
= -
1
2
       v
t
C = - +
2
1
4
At t =0, v=16 m/s
\ v
t
= - +
2
4
16 …(i)
From above relation v is zero at
t = 8 s
Options (a) is correct.
From relation (i),
ds
dt
t
= - +
2
4
16
\            ds
t
dt
ò ò
= - +
æ
è
ç
ö
ø
÷
2
4
16
i.e.,        s
t
t C = - + +
3
2
12
16
                s
t
t = - +
3
12
16 [As at t = 0, s = 0]
At t = 4 s
s = 58.67 m
Option (b) is correct.
The particle returns back at t = 8 s
From relation (ii)
s
8
3
8
12
16 8 = - + ´ = ( ) 85.33 m
  s
10
3
10
12
16 10 = - + ´ = ( ) 76.66 m
Distance travelled in 10 s
= + - s s s
8 8 10
( )
    = ´ - ( ) 2 85.33 76.66
            =94 m 
Option (c) is correct.
Velocity of particle at t = 10 s
44 | Mechanics-1
v
10
2
10
4
16 = - +
( )
= - + 25 16
              = -9 m/s
\ Speed of particle at t = 10 s is 9 m/s.
Option (d) is correct.
3. | | v
®
 is scalar. 
\ Option (a) is incorrect.
d
dt
v
a
®
®
= (by definition).
\ Option (b) is correct.
v
2
 is scalar.
\ Option (c) is incorrect.
v
v
®
®
| |
 is v
^
 (unit vector)
\
d
dt
v
a a
^
^
= ¹
®
\ Option (d) is incorrect.
4.
    v j
A
®
-
= -20
1 ^
kmh
    v i j
B
®
= ° + ° ( cos ) ( sin )
^ ^
40 37 40 37  kmh
-1
        = +
-
( )
^ ^
32 24
1
i j kmh
   v v v
AB A B
®
= -
® ®
       = - - + ( ) ( )
^ ^ ^
20 32 24 j i j
       = - - 32i j
^ ^
44
Option (a) is correct.
Option (c) is incorrect.
At any time
    S i j j
A
t
®
= + + - 3 4 20
^ ^ ^
( )
S i j
B
t
®
= + ( )
^ ^
32 24
Position of A relative to B :
S S S
AB A B
® ® ®
= -
    = + - - + ( ) ( )
^ ^ ^ ^ ^
3 4 20 32 24 i j j i j t t t
    = - + - ( ) ( )
^ ^
3 32 4 44 t t i j
Option (b) is correct.
Option (d) is incorrect.
5. Case I.
                       v a t =
1 1
                       v a t =
2 2
s a t a t
1 1 1
2
2 2
2
1
2
1
2
= +
Case II.
             v a t ¢ =2
1 1
              =2v
            2
2 3
v a t =
         2
2 2 2 3
( ) a t a t =
Þ          t t
3 2
2 =
    s a t at
2 1 1
2
23
2
1
2
2
1
2
= + ( )
         = + a t a t
1 1
2
2 2
2
1
2
2 ( )
         = + a t a t
1 1
2
2 2
2
2
      2
2 1 1
2
2 2
2
s a t a t = +
\      s s
1 1
2 > …(i)
      4 2 2
1 1 1
2
2 2
2
s a t a t = +
\      s s
2 1
4 < …(ii)
Combining Eqs. (i) and (ii),
2 4
1 2 1
s s s < <
Option (d) is correct.
In Case I. v
s
t t
av
=
+
1
1 2
i.e.,        v
a t a t
t t
1
1 1
2
2 2
2
1 2
1
2
1
2
=
+
+
Motion in One Dimension | 45
East
X (km)
20 km/h
A (3, 4)
North
Y (km)
South
B
37°
40 km/h
– a
2
a
1
t
1
t
2
v = (Let)
S
1
Rest
Average
velocity = v
1
Rest
– a
2
2a
1
t
1
t
3
S
2
Rest
Average
velocity = v
2
v' = 2v
   =
+
+
( ) ( )
( )
a t t a t t
t t
1 1 1 2 2 2
1 2
2
            =
+
+
vt vt
t t
1 2
1 2
2( )
 =
v
2
In Case II.
v
s
t t
av
=
+
2
1 3
i.e., v
a t a t
t t
2
1 1
2
2 2
2
1 3
2
=
+
+
              =
+
+
( ) ( ) a t t a t t
t t
1 1 1 2 2 2
1 2
2
2
              =
+
+
vt v t
t t
1 2
1 2
2
2
( )
              =v =2
1
v
Option (a) is correct.
Option (b) is incorrect.
6. If the particle’s initial velocity is + ive and
has some constant - ive acceleration the
particle will stop somewhere and then
return back to have zero displacement at
same time t ( ) > 0 .
Also if particle’s initial velocity is - ive and
has some constant + ive acceleration the
particle will stop somewhere and then
return back to have zero displacement at
same time t ( ) > 0 .
\ Options (b) and (c) are correct.
and options (a) and (d) are incorrect.
7. F t = a
Þ ma t = a
or   a
m
t =
a
…(i)
\ Graph between a (acceleration) and time ( ) t
will be as curve 1.
\ Option (a) is correct.
From equation
dv
dt m
t = ×
a
\  dv
m
tdt
ò ò
=
a
i.e.,      v
m
t
c = × +
a
2
2
\ Graph between velocity ( ) v and time ( ) t  will 
be as curve 2.
Option (b) is correct.
8.         a s = - +
6
30
6
        5 30 a s = - +
      5 30 × = - + v
dv
ds
s
or 5 30 vdv s ds
ò ò
= - + ( )
or    
5
2 2
30
2 2
1
v s
s c = - + +
or
5
2 2
30
2
2
v
S
s = - + …(i)
[C
1
0 = as at s = 0 particle is at rest]
Substituting s = 10 m in Eq. (i),
5
2
10
2
300
2
2
v = - +
( )
i.e.,          v=10 m/s
\ Option (b) is correct.
From Eq. (i) v to be maximum
- + = s 30 0
       s =30
\     
5
2
30
2
30 30
2
2
v
max
( )
( ) = - + ´
or   
5
2
450
2
v
max
=
or     v
max
= 180 m/s
Option (c) is correct.
9. If particle’s path is
(i) straight with backward motion
(ii) not straight somewhere.
Distance moved will be greater than the
modulus of displacement
\ | | v
av av
®
< v
Option (a) is correct.
If particle returns to its initial position, the
value of v
av
®
 will be zero while its average
speed (v
av
) will not be zero.
\ Option (c) is correct.
10. If u = 0, v at = and s at =
1
2
2
\ v-t graph will be as shown in (a).
s-t graph will be as shown in (d).
46 | Mechanics-1
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