Courses

# Chapter 3 - Motion in One Dimension (Part - 2) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

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

Created by: Ciel Knowledge

## NEET : Chapter 3 - Motion in One Dimension (Part - 2) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

``` 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
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
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
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
^
```
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

210 docs

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;