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Page 1
AIEEE Corner
Subjective Questions (Level 1)
1. (a) D v t v t = +
1 1 2 2
= ´
æ
è
ç
ö
ø
÷ + ´
æ
è
ç
ö
ø
÷ 60 1 80
1
2
km
h
h
km
h
h
=100 km
(b) Average speed
= =
100km
1.5h
66.67 km/h.
2. (a) Displacement in first two seconds
= + ( )( ) ( ) 4 2
1
2
6 2
2
=20 m
\ Average velocity = =
20
2
10
m
s
m/s
(b) Displacement in first four seconds
= + ( )( ) ( ) 4 4
1
2
6 4
2
=64 m
\ Displacement in the time interval
t =2 s to t = 4 s
= - 64 20
= 44 m
\ Average velocity = =
44
2
22
m
s
m/s.
3. Let the particle takes t time to reach
ground.
\ - = + - 64 20
1
2
10
2
t t ( )
i.e., 5 20 64 0
2
t t - - =
t = 6.1 s
If the particle goes h meter above tower
before coming down
0 20 2 10
2
= + - ( ) ( ) h
Þ h = 20 m
(a) Average speed =
Total distance moved
Time taken
=
+ + ( ) 20 20 60
6.1
=16.4 m/s
(b) Average velocity =
Net displacement
Time taken
=
-60m
6.1s
= -9.8 m/s
=9.8 m/s (downwards)
4. Average velocity =
S
S
vt
t
2.5v
vt vt vT
t t T
=
+ +
+ +
0 0
0 0
2 3
\ 5 3 3
0 0
t T t T + = + 2.5
or T t = 4
0
5. (a) Average acceleration
=
-
+
Final velocity initial velocity
s ( ) 4 8
=
- 0 0
12
=0 m/s
2
(c) a
v
= = tan
max
q
4
\ 4
4
=
v
max
(Q a = 4 m/s
2
)
i.e., v
max
=16 m/s
Displacement of particle in 12 seconds
= Area under v-t graph
= ´ 12
2
v
max
=
´ 12 16
2
=96 m
Average velocity
=
Displacement m) at
Time
(
( )
96 12
12
s
s
= +8 m/s
20 | Mechanics-1
h
+20 m/s
–
60 m
v
time
12s
8s 4s
v
max
q
Page 2
AIEEE Corner
Subjective Questions (Level 1)
1. (a) D v t v t = +
1 1 2 2
= ´
æ
è
ç
ö
ø
÷ + ´
æ
è
ç
ö
ø
÷ 60 1 80
1
2
km
h
h
km
h
h
=100 km
(b) Average speed
= =
100km
1.5h
66.67 km/h.
2. (a) Displacement in first two seconds
= + ( )( ) ( ) 4 2
1
2
6 2
2
=20 m
\ Average velocity = =
20
2
10
m
s
m/s
(b) Displacement in first four seconds
= + ( )( ) ( ) 4 4
1
2
6 4
2
=64 m
\ Displacement in the time interval
t =2 s to t = 4 s
= - 64 20
= 44 m
\ Average velocity = =
44
2
22
m
s
m/s.
3. Let the particle takes t time to reach
ground.
\ - = + - 64 20
1
2
10
2
t t ( )
i.e., 5 20 64 0
2
t t - - =
t = 6.1 s
If the particle goes h meter above tower
before coming down
0 20 2 10
2
= + - ( ) ( ) h
Þ h = 20 m
(a) Average speed =
Total distance moved
Time taken
=
+ + ( ) 20 20 60
6.1
=16.4 m/s
(b) Average velocity =
Net displacement
Time taken
=
-60m
6.1s
= -9.8 m/s
=9.8 m/s (downwards)
4. Average velocity =
S
S
vt
t
2.5v
vt vt vT
t t T
=
+ +
+ +
0 0
0 0
2 3
\ 5 3 3
0 0
t T t T + = + 2.5
or T t = 4
0
5. (a) Average acceleration
=
-
+
Final velocity initial velocity
s ( ) 4 8
=
- 0 0
12
=0 m/s
2
(c) a
v
= = tan
max
q
4
\ 4
4
=
v
max
(Q a = 4 m/s
2
)
i.e., v
max
=16 m/s
Displacement of particle in 12 seconds
= Area under v-t graph
= ´ 12
2
v
max
=
´ 12 16
2
=96 m
Average velocity
=
Displacement m) at
Time
(
( )
96 12
12
s
s
= +8 m/s
20 | Mechanics-1
h
+20 m/s
–
60 m
v
time
12s
8s 4s
v
max
q
(b)As the particle did not return back
distance travelled in 12 s
= Displacement at 12 s
\ Average speed = 8 m/s.
6. (a) Radius ( ) R of circle =
21
22
m
\ Circumference of circle = 2pR
= ´ ´ 2
22
7
21
22
m
=6 m
Speed ( ) v of particle = 1 m/s
\ Distance moved by particle in 2 s = 2 m
Thus, angle through which the particle
moved
= ´ = = °
2
6
2
2
3
120 p
p
Magnitude of Average velocity
=
=
Magnitude of displacement
Time s ( ) 2
= =
° AB R
2
2 60
2
sin
=R
3
2
=
21
22
3
2
=
21 3
44
m/s
(b) Magnitude of average acceleration
=
-
½
½
½
½
½
½
½
½
® ®
v v
B A
2s
=
°
2
120
2
2
vsin
[Q | | | | v v
A B
v
® ®
= = = 1 m/s)
= ° vsin 60
=
3
2
m/s
2
7. Position vector at t = 0 s
r i j
1
1 2
®
= + ( )
^ ^
m
Position vector at t = 4 s
r i j
2
6 4
®
= + ( )
^ ^
m
(a) Displacement from t = 0 s to t = 4 s
= -
® ®
( ) r r
2 1
= + - + ( ) ( )
^ ^ ^ ^
6 4 1 2 i j i j
= + ( )
^ ^
5 2 i j m
\ Average velocity =
+ ( )
^ ^
5 2
4
i j m
s
= + ( . . )
^ ^
125 05 i j m/s
(b) Average acceleration
=
- Final velocity Initial velocity
s 4
=
+ - + ( ) ( )
^ ^ ^ ^
2 10 4 6
4
i j i j
=
- + 2 4
4
i j
^ ^
= - + ( . )
^ ^
05i j m/s
2
(c) We cannot find the average speed as
the actual path followed by the particle
is not known.
Uniform acceleration
(a) One dimensional motion
8. If at time t the vertical displacement
between A and B is 10 m
1
2
1
2
1 10
2 2
gt gt - - = ( )
or t t
2 2
1 2 - - = ( )
or t t t
2 2
2 1 2 - - + = ( )
or 2 3 t =
t =1.5 s
9. The two bodies will meet if
Displacement of
first after attaining
=
highest
point
Displacement of
second before
attaining highest
point
Motion in One Dimension | 21
120°
O
B
A(t = 0 s)
(t = 2s)
v
B
®
v
A
®
Page 3
AIEEE Corner
Subjective Questions (Level 1)
1. (a) D v t v t = +
1 1 2 2
= ´
æ
è
ç
ö
ø
÷ + ´
æ
è
ç
ö
ø
÷ 60 1 80
1
2
km
h
h
km
h
h
=100 km
(b) Average speed
= =
100km
1.5h
66.67 km/h.
2. (a) Displacement in first two seconds
= + ( )( ) ( ) 4 2
1
2
6 2
2
=20 m
\ Average velocity = =
20
2
10
m
s
m/s
(b) Displacement in first four seconds
= + ( )( ) ( ) 4 4
1
2
6 4
2
=64 m
\ Displacement in the time interval
t =2 s to t = 4 s
= - 64 20
= 44 m
\ Average velocity = =
44
2
22
m
s
m/s.
3. Let the particle takes t time to reach
ground.
\ - = + - 64 20
1
2
10
2
t t ( )
i.e., 5 20 64 0
2
t t - - =
t = 6.1 s
If the particle goes h meter above tower
before coming down
0 20 2 10
2
= + - ( ) ( ) h
Þ h = 20 m
(a) Average speed =
Total distance moved
Time taken
=
+ + ( ) 20 20 60
6.1
=16.4 m/s
(b) Average velocity =
Net displacement
Time taken
=
-60m
6.1s
= -9.8 m/s
=9.8 m/s (downwards)
4. Average velocity =
S
S
vt
t
2.5v
vt vt vT
t t T
=
+ +
+ +
0 0
0 0
2 3
\ 5 3 3
0 0
t T t T + = + 2.5
or T t = 4
0
5. (a) Average acceleration
=
-
+
Final velocity initial velocity
s ( ) 4 8
=
- 0 0
12
=0 m/s
2
(c) a
v
= = tan
max
q
4
\ 4
4
=
v
max
(Q a = 4 m/s
2
)
i.e., v
max
=16 m/s
Displacement of particle in 12 seconds
= Area under v-t graph
= ´ 12
2
v
max
=
´ 12 16
2
=96 m
Average velocity
=
Displacement m) at
Time
(
( )
96 12
12
s
s
= +8 m/s
20 | Mechanics-1
h
+20 m/s
–
60 m
v
time
12s
8s 4s
v
max
q
(b)As the particle did not return back
distance travelled in 12 s
= Displacement at 12 s
\ Average speed = 8 m/s.
6. (a) Radius ( ) R of circle =
21
22
m
\ Circumference of circle = 2pR
= ´ ´ 2
22
7
21
22
m
=6 m
Speed ( ) v of particle = 1 m/s
\ Distance moved by particle in 2 s = 2 m
Thus, angle through which the particle
moved
= ´ = = °
2
6
2
2
3
120 p
p
Magnitude of Average velocity
=
=
Magnitude of displacement
Time s ( ) 2
= =
° AB R
2
2 60
2
sin
=R
3
2
=
21
22
3
2
=
21 3
44
m/s
(b) Magnitude of average acceleration
=
-
½
½
½
½
½
½
½
½
® ®
v v
B A
2s
=
°
2
120
2
2
vsin
[Q | | | | v v
A B
v
® ®
= = = 1 m/s)
= ° vsin 60
=
3
2
m/s
2
7. Position vector at t = 0 s
r i j
1
1 2
®
= + ( )
^ ^
m
Position vector at t = 4 s
r i j
2
6 4
®
= + ( )
^ ^
m
(a) Displacement from t = 0 s to t = 4 s
= -
® ®
( ) r r
2 1
= + - + ( ) ( )
^ ^ ^ ^
6 4 1 2 i j i j
= + ( )
^ ^
5 2 i j m
\ Average velocity =
+ ( )
^ ^
5 2
4
i j m
s
= + ( . . )
^ ^
125 05 i j m/s
(b) Average acceleration
=
- Final velocity Initial velocity
s 4
=
+ - + ( ) ( )
^ ^ ^ ^
2 10 4 6
4
i j i j
=
- + 2 4
4
i j
^ ^
= - + ( . )
^ ^
05i j m/s
2
(c) We cannot find the average speed as
the actual path followed by the particle
is not known.
Uniform acceleration
(a) One dimensional motion
8. If at time t the vertical displacement
between A and B is 10 m
1
2
1
2
1 10
2 2
gt gt - - = ( )
or t t
2 2
1 2 - - = ( )
or t t t
2 2
2 1 2 - - + = ( )
or 2 3 t =
t =1.5 s
9. The two bodies will meet if
Displacement of
first after attaining
=
highest
point
Displacement of
second before
attaining highest
point
Motion in One Dimension | 21
120°
O
B
A(t = 0 s)
(t = 2s)
v
B
®
v
A
®
v t gt v t t g t t
0
2
0 0 0
2
1
2
1
2
- = - - - ( ) ( )
or 0
1
2
2
0 0 0
2
0
= - - - v t g t t t ( )
or gt t gt v t
0 0
2
0 0
1
2
= +
or t
gt v
g
=
+
1
2
0 0
= +
t v
g
0 0
2
10. 5
1
2
2
= gt
Þ t =1s
For A
H t gt = + 0
1
2
2
.
or H gt =
1
2
2
…(i)
For B
H g t - = - 25
1
2
1
2
( ) …(ii)
or
1
2
1
2
1 25
2 2
gt gt - - = ( )
[Substituting value of H from Eq. (i)]
1
2
1 25
2 2
g t t [ ( ) ] - - =
t t t
2 2
2 1 5 - - + = ( )
2 1 5 t - =
Þ t = 3 s
Substituting t = 3 s in Eq. (i)
H = ´ ´ =
1
2
10 3 45
2
m
11. s at =
1
2
0
2
Forward motion
v at =
0
Backward mo tion
- = + - s at t a t ( ) ( )
0
2
1
2
\ - = -
1
2
1
2
0
2
0
2
at at t at ( )
or - = - t t t t
0
2
0
2
2
or t t t t
2
0 0
2
2 0 - - =
t
t t t
=
± - - - 2 2 41
2
0 0
2
0
2
( ) ( )
=
± 2 2 2
2
0 0
t t
= + t t
0 0
2 (- ive sign being
absurd)
= + ( ) 1 2
0
t
=2141
0
. t
From the begining of the motion the point
mass will return to the initial position
after time 3 141
0
. t .
12. 5 2 1015
2 2
= + - u ( )
Þ u
2
325 =
(a) For H
0 2 10
2 2
= + - u H ( )
i.e., 20 325 H =
or H =1625 . m
(b) For t
0 325 10 = + - ( )t
\ t =
325
10
=18 . s
13.
(a) 15 2 60
2 2
= + ´ u a …(i)
and 15 6 = + ´ u a …(ii)
Substituting the value of 6a from Eq. (ii)
in Eq. (i)
225 20 15
2
= + - u u ( )
22 | Mechanics-1
T
O
W
E
R
H
A
5m
25 m
u = 0
B
Time interval
t second for A
(t–1) second for B
A B
1 s
H
15m
v = 0
5 m/s
u
x 60 m
6.0 s
At rest
a u a
15 m/s
Page 4
AIEEE Corner
Subjective Questions (Level 1)
1. (a) D v t v t = +
1 1 2 2
= ´
æ
è
ç
ö
ø
÷ + ´
æ
è
ç
ö
ø
÷ 60 1 80
1
2
km
h
h
km
h
h
=100 km
(b) Average speed
= =
100km
1.5h
66.67 km/h.
2. (a) Displacement in first two seconds
= + ( )( ) ( ) 4 2
1
2
6 2
2
=20 m
\ Average velocity = =
20
2
10
m
s
m/s
(b) Displacement in first four seconds
= + ( )( ) ( ) 4 4
1
2
6 4
2
=64 m
\ Displacement in the time interval
t =2 s to t = 4 s
= - 64 20
= 44 m
\ Average velocity = =
44
2
22
m
s
m/s.
3. Let the particle takes t time to reach
ground.
\ - = + - 64 20
1
2
10
2
t t ( )
i.e., 5 20 64 0
2
t t - - =
t = 6.1 s
If the particle goes h meter above tower
before coming down
0 20 2 10
2
= + - ( ) ( ) h
Þ h = 20 m
(a) Average speed =
Total distance moved
Time taken
=
+ + ( ) 20 20 60
6.1
=16.4 m/s
(b) Average velocity =
Net displacement
Time taken
=
-60m
6.1s
= -9.8 m/s
=9.8 m/s (downwards)
4. Average velocity =
S
S
vt
t
2.5v
vt vt vT
t t T
=
+ +
+ +
0 0
0 0
2 3
\ 5 3 3
0 0
t T t T + = + 2.5
or T t = 4
0
5. (a) Average acceleration
=
-
+
Final velocity initial velocity
s ( ) 4 8
=
- 0 0
12
=0 m/s
2
(c) a
v
= = tan
max
q
4
\ 4
4
=
v
max
(Q a = 4 m/s
2
)
i.e., v
max
=16 m/s
Displacement of particle in 12 seconds
= Area under v-t graph
= ´ 12
2
v
max
=
´ 12 16
2
=96 m
Average velocity
=
Displacement m) at
Time
(
( )
96 12
12
s
s
= +8 m/s
20 | Mechanics-1
h
+20 m/s
–
60 m
v
time
12s
8s 4s
v
max
q
(b)As the particle did not return back
distance travelled in 12 s
= Displacement at 12 s
\ Average speed = 8 m/s.
6. (a) Radius ( ) R of circle =
21
22
m
\ Circumference of circle = 2pR
= ´ ´ 2
22
7
21
22
m
=6 m
Speed ( ) v of particle = 1 m/s
\ Distance moved by particle in 2 s = 2 m
Thus, angle through which the particle
moved
= ´ = = °
2
6
2
2
3
120 p
p
Magnitude of Average velocity
=
=
Magnitude of displacement
Time s ( ) 2
= =
° AB R
2
2 60
2
sin
=R
3
2
=
21
22
3
2
=
21 3
44
m/s
(b) Magnitude of average acceleration
=
-
½
½
½
½
½
½
½
½
® ®
v v
B A
2s
=
°
2
120
2
2
vsin
[Q | | | | v v
A B
v
® ®
= = = 1 m/s)
= ° vsin 60
=
3
2
m/s
2
7. Position vector at t = 0 s
r i j
1
1 2
®
= + ( )
^ ^
m
Position vector at t = 4 s
r i j
2
6 4
®
= + ( )
^ ^
m
(a) Displacement from t = 0 s to t = 4 s
= -
® ®
( ) r r
2 1
= + - + ( ) ( )
^ ^ ^ ^
6 4 1 2 i j i j
= + ( )
^ ^
5 2 i j m
\ Average velocity =
+ ( )
^ ^
5 2
4
i j m
s
= + ( . . )
^ ^
125 05 i j m/s
(b) Average acceleration
=
- Final velocity Initial velocity
s 4
=
+ - + ( ) ( )
^ ^ ^ ^
2 10 4 6
4
i j i j
=
- + 2 4
4
i j
^ ^
= - + ( . )
^ ^
05i j m/s
2
(c) We cannot find the average speed as
the actual path followed by the particle
is not known.
Uniform acceleration
(a) One dimensional motion
8. If at time t the vertical displacement
between A and B is 10 m
1
2
1
2
1 10
2 2
gt gt - - = ( )
or t t
2 2
1 2 - - = ( )
or t t t
2 2
2 1 2 - - + = ( )
or 2 3 t =
t =1.5 s
9. The two bodies will meet if
Displacement of
first after attaining
=
highest
point
Displacement of
second before
attaining highest
point
Motion in One Dimension | 21
120°
O
B
A(t = 0 s)
(t = 2s)
v
B
®
v
A
®
v t gt v t t g t t
0
2
0 0 0
2
1
2
1
2
- = - - - ( ) ( )
or 0
1
2
2
0 0 0
2
0
= - - - v t g t t t ( )
or gt t gt v t
0 0
2
0 0
1
2
= +
or t
gt v
g
=
+
1
2
0 0
= +
t v
g
0 0
2
10. 5
1
2
2
= gt
Þ t =1s
For A
H t gt = + 0
1
2
2
.
or H gt =
1
2
2
…(i)
For B
H g t - = - 25
1
2
1
2
( ) …(ii)
or
1
2
1
2
1 25
2 2
gt gt - - = ( )
[Substituting value of H from Eq. (i)]
1
2
1 25
2 2
g t t [ ( ) ] - - =
t t t
2 2
2 1 5 - - + = ( )
2 1 5 t - =
Þ t = 3 s
Substituting t = 3 s in Eq. (i)
H = ´ ´ =
1
2
10 3 45
2
m
11. s at =
1
2
0
2
Forward motion
v at =
0
Backward mo tion
- = + - s at t a t ( ) ( )
0
2
1
2
\ - = -
1
2
1
2
0
2
0
2
at at t at ( )
or - = - t t t t
0
2
0
2
2
or t t t t
2
0 0
2
2 0 - - =
t
t t t
=
± - - - 2 2 41
2
0 0
2
0
2
( ) ( )
=
± 2 2 2
2
0 0
t t
= + t t
0 0
2 (- ive sign being
absurd)
= + ( ) 1 2
0
t
=2141
0
. t
From the begining of the motion the point
mass will return to the initial position
after time 3 141
0
. t .
12. 5 2 1015
2 2
= + - u ( )
Þ u
2
325 =
(a) For H
0 2 10
2 2
= + - u H ( )
i.e., 20 325 H =
or H =1625 . m
(b) For t
0 325 10 = + - ( )t
\ t =
325
10
=18 . s
13.
(a) 15 2 60
2 2
= + ´ u a …(i)
and 15 6 = + ´ u a …(ii)
Substituting the value of 6a from Eq. (ii)
in Eq. (i)
225 20 15
2
= + - u u ( )
22 | Mechanics-1
T
O
W
E
R
H
A
5m
25 m
u = 0
B
Time interval
t second for A
(t–1) second for B
A B
1 s
H
15m
v = 0
5 m/s
u
x 60 m
6.0 s
At rest
a u a
15 m/s
i.e., u u
2
20 75 0 - + =
( )( ) u u - - = 15 5 0
\ u = 5 m/s
(15 m/s being not possible)
(b) Using Eq. (ii)
a =
5
3
m / s
2
(c) u ax
2 2
0 2 = +
i.e., x
u
a
= =
´
2 2
2
5
2
5
3
( )
=75 . m
(d) s at =
1
2
2
= ´ ´
1
2
5
3
2
t
=
5
6
2
t …(iii)
t( ) s 0 1 3 6 9 12
v (m/s) 0 5/6 7.5 30 67.5 120
Differentiating Eq. (iii) w.r.t. time t
v t =
5
3
t( ) s 0 3 6 9 12
v (m/s) 0 5 10 15 20
14.
Jour ney A to P
v xt
max
= + 0
1
…(i)
and v xs
max
2
1
2 = …(ii)
Þ t
v
x
1
=
max
Jour ney P to B
0
2
= + - v y t
max
( ) …(iii)
and v ys
max
2
2
2 = …(iv)
Þ t
v
y
2
=
max
\
v
x
v
y
t t
max max
+ = + =
1 2
4
or v
x y
max
1 1
4 +
é
ë
ê
ù
û
ú
= …(v)
From Eq. (ii) and Eq. (iv)
s s
v
x
v
y
1 2
2 2
2 2
+ = +
max max
or 4
2
1 1
2
= +
é
ë
ê
ù
û
ú
v
x y
max
…(vi)
Dividing Eq. (vi) by Eq. (v)
v
max
= 2
Substituting the value of v
max
in Eq. (v)
1 1
2
x y
+ = (Proved)
15. Let acceleration of the particle be a using
v u at = +
0 6 = + u a
\ a
u
= -
6
(a) At t = 10 s, s = - 2 m
- = ´ + ´ 2 10
1
2
10
2
u a
or - = - + 2 6 10 50 ( ) a a
or - = - 10 2 a
or a = 0.2 m/s
2
Motion in One Dimension | 23
20
15
10
5
3 6 9 12
v (m/s)
t (s)
Starts
a = + x
t
1
t
2
a = –y
Stops
4 km
B
s
1
s
2
4 min
B A
120
90
60
30
15
3 6 9 12
t (s)
s (m)
2m
4 0
t = 10 s t = 6 s
t = 0 s
v = 0
+ive
x-axis
Page 5
AIEEE Corner
Subjective Questions (Level 1)
1. (a) D v t v t = +
1 1 2 2
= ´
æ
è
ç
ö
ø
÷ + ´
æ
è
ç
ö
ø
÷ 60 1 80
1
2
km
h
h
km
h
h
=100 km
(b) Average speed
= =
100km
1.5h
66.67 km/h.
2. (a) Displacement in first two seconds
= + ( )( ) ( ) 4 2
1
2
6 2
2
=20 m
\ Average velocity = =
20
2
10
m
s
m/s
(b) Displacement in first four seconds
= + ( )( ) ( ) 4 4
1
2
6 4
2
=64 m
\ Displacement in the time interval
t =2 s to t = 4 s
= - 64 20
= 44 m
\ Average velocity = =
44
2
22
m
s
m/s.
3. Let the particle takes t time to reach
ground.
\ - = + - 64 20
1
2
10
2
t t ( )
i.e., 5 20 64 0
2
t t - - =
t = 6.1 s
If the particle goes h meter above tower
before coming down
0 20 2 10
2
= + - ( ) ( ) h
Þ h = 20 m
(a) Average speed =
Total distance moved
Time taken
=
+ + ( ) 20 20 60
6.1
=16.4 m/s
(b) Average velocity =
Net displacement
Time taken
=
-60m
6.1s
= -9.8 m/s
=9.8 m/s (downwards)
4. Average velocity =
S
S
vt
t
2.5v
vt vt vT
t t T
=
+ +
+ +
0 0
0 0
2 3
\ 5 3 3
0 0
t T t T + = + 2.5
or T t = 4
0
5. (a) Average acceleration
=
-
+
Final velocity initial velocity
s ( ) 4 8
=
- 0 0
12
=0 m/s
2
(c) a
v
= = tan
max
q
4
\ 4
4
=
v
max
(Q a = 4 m/s
2
)
i.e., v
max
=16 m/s
Displacement of particle in 12 seconds
= Area under v-t graph
= ´ 12
2
v
max
=
´ 12 16
2
=96 m
Average velocity
=
Displacement m) at
Time
(
( )
96 12
12
s
s
= +8 m/s
20 | Mechanics-1
h
+20 m/s
–
60 m
v
time
12s
8s 4s
v
max
q
(b)As the particle did not return back
distance travelled in 12 s
= Displacement at 12 s
\ Average speed = 8 m/s.
6. (a) Radius ( ) R of circle =
21
22
m
\ Circumference of circle = 2pR
= ´ ´ 2
22
7
21
22
m
=6 m
Speed ( ) v of particle = 1 m/s
\ Distance moved by particle in 2 s = 2 m
Thus, angle through which the particle
moved
= ´ = = °
2
6
2
2
3
120 p
p
Magnitude of Average velocity
=
=
Magnitude of displacement
Time s ( ) 2
= =
° AB R
2
2 60
2
sin
=R
3
2
=
21
22
3
2
=
21 3
44
m/s
(b) Magnitude of average acceleration
=
-
½
½
½
½
½
½
½
½
® ®
v v
B A
2s
=
°
2
120
2
2
vsin
[Q | | | | v v
A B
v
® ®
= = = 1 m/s)
= ° vsin 60
=
3
2
m/s
2
7. Position vector at t = 0 s
r i j
1
1 2
®
= + ( )
^ ^
m
Position vector at t = 4 s
r i j
2
6 4
®
= + ( )
^ ^
m
(a) Displacement from t = 0 s to t = 4 s
= -
® ®
( ) r r
2 1
= + - + ( ) ( )
^ ^ ^ ^
6 4 1 2 i j i j
= + ( )
^ ^
5 2 i j m
\ Average velocity =
+ ( )
^ ^
5 2
4
i j m
s
= + ( . . )
^ ^
125 05 i j m/s
(b) Average acceleration
=
- Final velocity Initial velocity
s 4
=
+ - + ( ) ( )
^ ^ ^ ^
2 10 4 6
4
i j i j
=
- + 2 4
4
i j
^ ^
= - + ( . )
^ ^
05i j m/s
2
(c) We cannot find the average speed as
the actual path followed by the particle
is not known.
Uniform acceleration
(a) One dimensional motion
8. If at time t the vertical displacement
between A and B is 10 m
1
2
1
2
1 10
2 2
gt gt - - = ( )
or t t
2 2
1 2 - - = ( )
or t t t
2 2
2 1 2 - - + = ( )
or 2 3 t =
t =1.5 s
9. The two bodies will meet if
Displacement of
first after attaining
=
highest
point
Displacement of
second before
attaining highest
point
Motion in One Dimension | 21
120°
O
B
A(t = 0 s)
(t = 2s)
v
B
®
v
A
®
v t gt v t t g t t
0
2
0 0 0
2
1
2
1
2
- = - - - ( ) ( )
or 0
1
2
2
0 0 0
2
0
= - - - v t g t t t ( )
or gt t gt v t
0 0
2
0 0
1
2
= +
or t
gt v
g
=
+
1
2
0 0
= +
t v
g
0 0
2
10. 5
1
2
2
= gt
Þ t =1s
For A
H t gt = + 0
1
2
2
.
or H gt =
1
2
2
…(i)
For B
H g t - = - 25
1
2
1
2
( ) …(ii)
or
1
2
1
2
1 25
2 2
gt gt - - = ( )
[Substituting value of H from Eq. (i)]
1
2
1 25
2 2
g t t [ ( ) ] - - =
t t t
2 2
2 1 5 - - + = ( )
2 1 5 t - =
Þ t = 3 s
Substituting t = 3 s in Eq. (i)
H = ´ ´ =
1
2
10 3 45
2
m
11. s at =
1
2
0
2
Forward motion
v at =
0
Backward mo tion
- = + - s at t a t ( ) ( )
0
2
1
2
\ - = -
1
2
1
2
0
2
0
2
at at t at ( )
or - = - t t t t
0
2
0
2
2
or t t t t
2
0 0
2
2 0 - - =
t
t t t
=
± - - - 2 2 41
2
0 0
2
0
2
( ) ( )
=
± 2 2 2
2
0 0
t t
= + t t
0 0
2 (- ive sign being
absurd)
= + ( ) 1 2
0
t
=2141
0
. t
From the begining of the motion the point
mass will return to the initial position
after time 3 141
0
. t .
12. 5 2 1015
2 2
= + - u ( )
Þ u
2
325 =
(a) For H
0 2 10
2 2
= + - u H ( )
i.e., 20 325 H =
or H =1625 . m
(b) For t
0 325 10 = + - ( )t
\ t =
325
10
=18 . s
13.
(a) 15 2 60
2 2
= + ´ u a …(i)
and 15 6 = + ´ u a …(ii)
Substituting the value of 6a from Eq. (ii)
in Eq. (i)
225 20 15
2
= + - u u ( )
22 | Mechanics-1
T
O
W
E
R
H
A
5m
25 m
u = 0
B
Time interval
t second for A
(t–1) second for B
A B
1 s
H
15m
v = 0
5 m/s
u
x 60 m
6.0 s
At rest
a u a
15 m/s
i.e., u u
2
20 75 0 - + =
( )( ) u u - - = 15 5 0
\ u = 5 m/s
(15 m/s being not possible)
(b) Using Eq. (ii)
a =
5
3
m / s
2
(c) u ax
2 2
0 2 = +
i.e., x
u
a
= =
´
2 2
2
5
2
5
3
( )
=75 . m
(d) s at =
1
2
2
= ´ ´
1
2
5
3
2
t
=
5
6
2
t …(iii)
t( ) s 0 1 3 6 9 12
v (m/s) 0 5/6 7.5 30 67.5 120
Differentiating Eq. (iii) w.r.t. time t
v t =
5
3
t( ) s 0 3 6 9 12
v (m/s) 0 5 10 15 20
14.
Jour ney A to P
v xt
max
= + 0
1
…(i)
and v xs
max
2
1
2 = …(ii)
Þ t
v
x
1
=
max
Jour ney P to B
0
2
= + - v y t
max
( ) …(iii)
and v ys
max
2
2
2 = …(iv)
Þ t
v
y
2
=
max
\
v
x
v
y
t t
max max
+ = + =
1 2
4
or v
x y
max
1 1
4 +
é
ë
ê
ù
û
ú
= …(v)
From Eq. (ii) and Eq. (iv)
s s
v
x
v
y
1 2
2 2
2 2
+ = +
max max
or 4
2
1 1
2
= +
é
ë
ê
ù
û
ú
v
x y
max
…(vi)
Dividing Eq. (vi) by Eq. (v)
v
max
= 2
Substituting the value of v
max
in Eq. (v)
1 1
2
x y
+ = (Proved)
15. Let acceleration of the particle be a using
v u at = +
0 6 = + u a
\ a
u
= -
6
(a) At t = 10 s, s = - 2 m
- = ´ + ´ 2 10
1
2
10
2
u a
or - = - + 2 6 10 50 ( ) a a
or - = - 10 2 a
or a = 0.2 m/s
2
Motion in One Dimension | 23
20
15
10
5
3 6 9 12
v (m/s)
t (s)
Starts
a = + x
t
1
t
2
a = –y
Stops
4 km
B
s
1
s
2
4 min
B A
120
90
60
30
15
3 6 9 12
t (s)
s (m)
2m
4 0
t = 10 s t = 6 s
t = 0 s
v = 0
+ive
x-axis
(b) v t u a ( ) at s = = + 10 10
= - + 6 10 a a
= 4a
=08 . m/s
(c) Two or three dimensional motion
16. a
F ®
®
= =
m
10
2
N north
kg
=5 m/s
2
, north
=5j
^
m/s
2
u
®
=10 m/s, east
=10i
^
m/s
using v u a
® ® ®
= + t
v i j
®
= + ´ 10 5 2
^ ^
( )
= + 10 10 i j
^ ^
| | v
®
=10 2 m/s
v
®
=10 2, north-east
using s u a
® ® ®
= + t t
1
2
2
= ´ + ( ) ( )
^ ^
10 2
1
2
5 2
2
i j
= + ( )
^ ^
20 10 i j m
| | s
®
= + 20 10
2 2
=10 5 m
cotq = =
20
10
2
q =
-
cot
1
2
s
®
=10 5 m at cot ( )
-1
2 from east to north.
17. s i j
®
= +
0
2 4 ( )
^ ^
m
a i
®
=
1
2
^
m/s
2
(t = 0 s to t = 2 s) t
1
2 = s
u 0
® ®
= m/s
a j
®
= -
2
4
^
m/s
2
(t = 2 s to t = 4 s) t
2
2 = s
(a) Velocity
v u a
® ® ®
= +
1
1
1
t
= +
®
0 i ( )
^
2 2
= 4i
^
v v a a
® ® ® ®
= + +
2 1 1 2 2
( )t
= + - 4 2 4 2 i i j
^ ^ ^
( )
or v i j
®
= -
2
8 8 ( )
^ ^
m/s
(b) Co-ordinate of particle
s s u a
® ® ®
®
= + + 1 0
1 1 1
2
1
2
t t
= + + + ( ) ( )( ) ( )
^ ^ ^
2 4 0 2
1
2
2 2
2
i j i
= + + 2 4 4 i j i
^ ^ ^
= + 6 4 i j
^ ^
s s v
® ® ®
= + + +
2 1
1 2 1 2 2
2
1
2
t a a t ( )
= + + + - 6 4 4 2
1
2
2 4 2
2
i j i i j
^ ^ ^ ^ ^
( ) ( )
= + + + - 6 4 8 4 8 i j i i j
^ ^ ^ ^ ^
= - 18 4 i j
^ ^
Co-ordinate of the particle
[ , ] 18 4 m m -
18. u i j
®
= - ( )
^ ^
2 4 m/s, s 0
® ®
=
0
m
a i j
®
= + ( )
^ ^
4 m/s
2
(a) Velocity
v u a
® ® ®
= + t
= - + + ( ) ( )
^ ^ ^ ^
2 4 4 2 i j i j
= - ( )
^ ^
10 2 i j m/s
(b) Co-ordinates of the particle
s s u a
® ® ® ®
= + +
0
1
2
2
t t
= + - + +
®
0 i j i j ( ) (
$
)
^ ^ ^
2 4
1
2
4 2
2
24 | Mechanics-1
O
North
East
q
s
®
j
^
i
^
Read More
FAQs on DC Pandey Solutions: Motion in One Dimension- 2 - 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. In this type of motion, the object can only move forward or backward, and its position is described by a single coordinate.
| 2. How is motion in one dimension different from motion in two or three dimensions? | |
Ans. Motion in one dimension only involves movement along a straight line, while motion in two or three dimensions allows movement in multiple directions. In one-dimensional motion, the object's position is described by a single coordinate, whereas in higher dimensions, multiple coordinates are required.
| 3. What are the key equations used to describe motion in one dimension? | |
Ans. The key equations used to describe motion in one dimension are the equations of motion. These include the displacement equation (Δx = x_f - x_i), the average velocity equation (v_avg = Δx/Δt), the final velocity equation (v_f = v_i + at), and the position equation (x = x_i + v_i*t + 0.5*a*t^2), among others.
| 4. How can we calculate the velocity and acceleration of an object in one-dimensional motion? | |
Ans. The velocity of an object in one-dimensional motion can be calculated by dividing the change in position (displacement) by the change in time. The acceleration can be calculated by dividing the change in velocity by the change in time. Mathematically, velocity (v) is equal to displacement (Δx) divided by time (Δt), and acceleration (a) is equal to change in velocity (Δv) divided by time (Δt).
| 5. What are some examples of motion in one dimension? | |
Ans. Some examples of motion in one dimension include a car moving along a straight road, a ball thrown vertically upwards and falling back down, an elevator moving up or down a building, and a pendulum swinging back and forth in a straight line.
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their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
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Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
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students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of DC Pandey Solutions: Motion in One Dimension- 2 is prepared as per the latest NEET syllabus.
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