Page 1
42  Mechanics1
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  Mechanics1
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  Mechanics1
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  Mechanics1
Page 4
42  Mechanics1
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  Mechanics1
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  Mechanics1
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  Mechanics1
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
\ vt graph will be as shown in (a).
st graph will be as shown in (d).
46  Mechanics1
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