HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics

# HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics | Physics Class 11 - NEET PDF Download

``` Page 1

3.1
SOLUTIONS TO CONCEPTS
CHAPTER – 3
1. a) Distance travelled = 50 + 40 + 20 = 110 m
b) AF = AB – BF = AB – DC = 50 – 20 = 30 M
AD = m 50 40 30 DF AF
2 2 2 2
? ? ? ?
In ?AED tan ? = DE/AE = 30/40 = 3/4
? ? = tan
–1
(3/4)
His displacement from his house to the field is 50 m,
tan
–1
(3/4) north to east.
2. O ? Starting point origin.
i) Distance travelled = 20 + 20 + 20 = 60 m
ii) Displacement is only OB = 20 m in the negative direction.
Displacement ? Distance between final and initial position.
3. a) V
ave
of plane (Distance/Time) = 260/0.5 = 520 km/hr.
b) V
ave
of bus = 320/8 = 40 km/hr.
c) plane goes in straight path
velocity =
ave
V
?
= 260/0.5 = 520 km/hr.
d) Straight path distance between plane to Ranchi is equal to the displacement of bus.
? Velocity =
ave
V
?
= 260/8 = 32.5 km/hr.
4. a) Total distance covered 12416 – 12352 = 64 km in 2 hours.
Speed = 64/2 = 32 km/h
b) As he returns to his house, the displacement is zero.
Velocity = (displacement/time) = 0 (zero).
5. Initial velocity u = 0 ( ? starts from rest)
Final velocity v = 18 km/hr = 5 sec
(i.e. max velocity)
Time interval t = 2 sec.
? Acceleration = a
ave
=
2
5
t
u v
?
?
= 2.5 m/s
2
.
6. In the interval 8 sec the velocity changes from 0 to 20 m/s.
Average acceleration = 20/8 = 2.5 m/s
2
?
?
?
?
?
?
time
velocity in change
Distance travelled S = ut + 1/2 at
2
? 0 + 1/2(2.5)8
2
= 80 m.
7. In 1
st
10 sec S
1
= ut + 1/2 at
2
? 0 + (1/2 × 5 × 10
2
) = 250 ft.
At 10 sec v = u + at = 0 + 5 × 10 = 50 ft/sec.
? From 10 to 20 sec ( ?t = 20 – 10 = 10 sec) it moves with uniform
velocity 50 ft/sec,
A
E
S
N
W
? ?
40 m
40 m
50 m
20 m
30 m
B
C
D
E
A ? Initial point
(starting point)
A
X
O
Y
B
(20 m, 0)
(–20 m, 0)
Initial velocity
u = 0
20
8 4
10
Time in sec
20 10 30
S (in ft)
0
t (sec)
250
750
1000
Page 2

3.1
SOLUTIONS TO CONCEPTS
CHAPTER – 3
1. a) Distance travelled = 50 + 40 + 20 = 110 m
b) AF = AB – BF = AB – DC = 50 – 20 = 30 M
AD = m 50 40 30 DF AF
2 2 2 2
? ? ? ?
In ?AED tan ? = DE/AE = 30/40 = 3/4
? ? = tan
–1
(3/4)
His displacement from his house to the field is 50 m,
tan
–1
(3/4) north to east.
2. O ? Starting point origin.
i) Distance travelled = 20 + 20 + 20 = 60 m
ii) Displacement is only OB = 20 m in the negative direction.
Displacement ? Distance between final and initial position.
3. a) V
ave
of plane (Distance/Time) = 260/0.5 = 520 km/hr.
b) V
ave
of bus = 320/8 = 40 km/hr.
c) plane goes in straight path
velocity =
ave
V
?
= 260/0.5 = 520 km/hr.
d) Straight path distance between plane to Ranchi is equal to the displacement of bus.
? Velocity =
ave
V
?
= 260/8 = 32.5 km/hr.
4. a) Total distance covered 12416 – 12352 = 64 km in 2 hours.
Speed = 64/2 = 32 km/h
b) As he returns to his house, the displacement is zero.
Velocity = (displacement/time) = 0 (zero).
5. Initial velocity u = 0 ( ? starts from rest)
Final velocity v = 18 km/hr = 5 sec
(i.e. max velocity)
Time interval t = 2 sec.
? Acceleration = a
ave
=
2
5
t
u v
?
?
= 2.5 m/s
2
.
6. In the interval 8 sec the velocity changes from 0 to 20 m/s.
Average acceleration = 20/8 = 2.5 m/s
2
?
?
?
?
?
?
time
velocity in change
Distance travelled S = ut + 1/2 at
2
? 0 + 1/2(2.5)8
2
= 80 m.
7. In 1
st
10 sec S
1
= ut + 1/2 at
2
? 0 + (1/2 × 5 × 10
2
) = 250 ft.
At 10 sec v = u + at = 0 + 5 × 10 = 50 ft/sec.
? From 10 to 20 sec ( ?t = 20 – 10 = 10 sec) it moves with uniform
velocity 50 ft/sec,
A
E
S
N
W
? ?
40 m
40 m
50 m
20 m
30 m
B
C
D
E
A ? Initial point
(starting point)
A
X
O
Y
B
(20 m, 0)
(–20 m, 0)
Initial velocity
u = 0
20
8 4
10
Time in sec
20 10 30
S (in ft)
0
t (sec)
250
750
1000
Chapter-3
3.2
Distance S
2
= 50 × 10 = 500 ft
Between 20 sec to 30 sec acceleration is constant i.e. –5 ft/s
2
. At 20 sec velocity is 50 ft/sec.
t = 30 – 20 = 10 s
S
3
= ut + 1/2 at
2
= 50 × 10 + (1/2)(–5)(10)
2
= 250 m
Total distance travelled is 30 sec = S
1
+ S
2
+ S
3
= 250 + 500 + 250 = 1000 ft.
8. a) Initial velocity u = 2 m/s.
final velocity v = 8 m/s
time = 10 sec,
acceleration =
10
2 8
ta
u v ?
?
?
= 0.6 m/s
2
b) v
2
– u
2
= 2aS
? Distance S =
a 2
u v
2 2
?
=
6 . 0 2
2 8
2 2
?
?
= 50 m.
c) Displacement is same as distance travelled.
Displacement = 50 m.
9. a) Displacement in 0 to 10 sec is 1000 m.
time = 10 sec.
V
ave
= s/t = 100/10 = 10 m/s.
b) At 2 sec it is moving with uniform velocity 50/2.5 = 20 m/s.
at 2 sec. V
inst
= 20 m/s.
At 5 sec it is at rest.
V
inst
= zero.
At 8 sec it is moving with uniform velocity 20 m/s
V
inst
= 20 m/s
At 12 sec velocity is negative as it move towards initial position. V
inst
= – 20 m/s.
10. Distance in first 40 sec is, ? OAB + ?BCD
=
2
1
× 5 × 20 +
2
1
× 5 × 20 = 100 m.
Average velocity is 0 as the displacement is zero.
11. Consider the point B, at t = 12 sec
At t = 0 ; s = 20 m
and t = 12 sec s = 20 m
So for time interval 0 to 12 sec
Change in displacement is zero.
So, average velocity = displacement/ time = 0
? The time is 12 sec.
12. At position B instantaneous velocity has direction along BC . For
average velocity between A and B.
V
ave
= displacement / time = ) t / AB ( t = time
10
t
5
2
4
6
8
t
10
(slope of the graph at t = 2 sec)
2.5
50
100
0
t
5 7.5 15
40
t (sec)
20
5 m/s
O
A
B
C
D
20
B
10
10 12 20
4 B
2
C
4 2
x
6
y
Page 3

3.1
SOLUTIONS TO CONCEPTS
CHAPTER – 3
1. a) Distance travelled = 50 + 40 + 20 = 110 m
b) AF = AB – BF = AB – DC = 50 – 20 = 30 M
AD = m 50 40 30 DF AF
2 2 2 2
? ? ? ?
In ?AED tan ? = DE/AE = 30/40 = 3/4
? ? = tan
–1
(3/4)
His displacement from his house to the field is 50 m,
tan
–1
(3/4) north to east.
2. O ? Starting point origin.
i) Distance travelled = 20 + 20 + 20 = 60 m
ii) Displacement is only OB = 20 m in the negative direction.
Displacement ? Distance between final and initial position.
3. a) V
ave
of plane (Distance/Time) = 260/0.5 = 520 km/hr.
b) V
ave
of bus = 320/8 = 40 km/hr.
c) plane goes in straight path
velocity =
ave
V
?
= 260/0.5 = 520 km/hr.
d) Straight path distance between plane to Ranchi is equal to the displacement of bus.
? Velocity =
ave
V
?
= 260/8 = 32.5 km/hr.
4. a) Total distance covered 12416 – 12352 = 64 km in 2 hours.
Speed = 64/2 = 32 km/h
b) As he returns to his house, the displacement is zero.
Velocity = (displacement/time) = 0 (zero).
5. Initial velocity u = 0 ( ? starts from rest)
Final velocity v = 18 km/hr = 5 sec
(i.e. max velocity)
Time interval t = 2 sec.
? Acceleration = a
ave
=
2
5
t
u v
?
?
= 2.5 m/s
2
.
6. In the interval 8 sec the velocity changes from 0 to 20 m/s.
Average acceleration = 20/8 = 2.5 m/s
2
?
?
?
?
?
?
time
velocity in change
Distance travelled S = ut + 1/2 at
2
? 0 + 1/2(2.5)8
2
= 80 m.
7. In 1
st
10 sec S
1
= ut + 1/2 at
2
? 0 + (1/2 × 5 × 10
2
) = 250 ft.
At 10 sec v = u + at = 0 + 5 × 10 = 50 ft/sec.
? From 10 to 20 sec ( ?t = 20 – 10 = 10 sec) it moves with uniform
velocity 50 ft/sec,
A
E
S
N
W
? ?
40 m
40 m
50 m
20 m
30 m
B
C
D
E
A ? Initial point
(starting point)
A
X
O
Y
B
(20 m, 0)
(–20 m, 0)
Initial velocity
u = 0
20
8 4
10
Time in sec
20 10 30
S (in ft)
0
t (sec)
250
750
1000
Chapter-3
3.2
Distance S
2
= 50 × 10 = 500 ft
Between 20 sec to 30 sec acceleration is constant i.e. –5 ft/s
2
. At 20 sec velocity is 50 ft/sec.
t = 30 – 20 = 10 s
S
3
= ut + 1/2 at
2
= 50 × 10 + (1/2)(–5)(10)
2
= 250 m
Total distance travelled is 30 sec = S
1
+ S
2
+ S
3
= 250 + 500 + 250 = 1000 ft.
8. a) Initial velocity u = 2 m/s.
final velocity v = 8 m/s
time = 10 sec,
acceleration =
10
2 8
ta
u v ?
?
?
= 0.6 m/s
2
b) v
2
– u
2
= 2aS
? Distance S =
a 2
u v
2 2
?
=
6 . 0 2
2 8
2 2
?
?
= 50 m.
c) Displacement is same as distance travelled.
Displacement = 50 m.
9. a) Displacement in 0 to 10 sec is 1000 m.
time = 10 sec.
V
ave
= s/t = 100/10 = 10 m/s.
b) At 2 sec it is moving with uniform velocity 50/2.5 = 20 m/s.
at 2 sec. V
inst
= 20 m/s.
At 5 sec it is at rest.
V
inst
= zero.
At 8 sec it is moving with uniform velocity 20 m/s
V
inst
= 20 m/s
At 12 sec velocity is negative as it move towards initial position. V
inst
= – 20 m/s.
10. Distance in first 40 sec is, ? OAB + ?BCD
=
2
1
× 5 × 20 +
2
1
× 5 × 20 = 100 m.
Average velocity is 0 as the displacement is zero.
11. Consider the point B, at t = 12 sec
At t = 0 ; s = 20 m
and t = 12 sec s = 20 m
So for time interval 0 to 12 sec
Change in displacement is zero.
So, average velocity = displacement/ time = 0
? The time is 12 sec.
12. At position B instantaneous velocity has direction along BC . For
average velocity between A and B.
V
ave
= displacement / time = ) t / AB ( t = time
10
t
5
2
4
6
8
t
10
(slope of the graph at t = 2 sec)
2.5
50
100
0
t
5 7.5 15
40
t (sec)
20
5 m/s
O
A
B
C
D
20
B
10
10 12 20
4 B
2
C
4 2
x
6
y
Chapter-3
3.3
We can see that AB is along BC i.e. they are in same direction.
The point is B (5m, 3m).
13. u = 4 m/s, a = 1.2 m/s
2
, t = 5 sec
Distance = s =
2
at
2
1
ut ?
= 4(5) + 1/2 (1.2)5
2
= 35 m.
14. Initial velocity u = 43.2 km/hr = 12 m/s
u = 12 m/s, v = 0
a = –6 m/s
2
(deceleration)
Distance S =
) 6 ( 2
u v
2 2
?
?
= 12 m
Page 4

3.1
SOLUTIONS TO CONCEPTS
CHAPTER – 3
1. a) Distance travelled = 50 + 40 + 20 = 110 m
b) AF = AB – BF = AB – DC = 50 – 20 = 30 M
AD = m 50 40 30 DF AF
2 2 2 2
? ? ? ?
In ?AED tan ? = DE/AE = 30/40 = 3/4
? ? = tan
–1
(3/4)
His displacement from his house to the field is 50 m,
tan
–1
(3/4) north to east.
2. O ? Starting point origin.
i) Distance travelled = 20 + 20 + 20 = 60 m
ii) Displacement is only OB = 20 m in the negative direction.
Displacement ? Distance between final and initial position.
3. a) V
ave
of plane (Distance/Time) = 260/0.5 = 520 km/hr.
b) V
ave
of bus = 320/8 = 40 km/hr.
c) plane goes in straight path
velocity =
ave
V
?
= 260/0.5 = 520 km/hr.
d) Straight path distance between plane to Ranchi is equal to the displacement of bus.
? Velocity =
ave
V
?
= 260/8 = 32.5 km/hr.
4. a) Total distance covered 12416 – 12352 = 64 km in 2 hours.
Speed = 64/2 = 32 km/h
b) As he returns to his house, the displacement is zero.
Velocity = (displacement/time) = 0 (zero).
5. Initial velocity u = 0 ( ? starts from rest)
Final velocity v = 18 km/hr = 5 sec
(i.e. max velocity)
Time interval t = 2 sec.
? Acceleration = a
ave
=
2
5
t
u v
?
?
= 2.5 m/s
2
.
6. In the interval 8 sec the velocity changes from 0 to 20 m/s.
Average acceleration = 20/8 = 2.5 m/s
2
?
?
?
?
?
?
time
velocity in change
Distance travelled S = ut + 1/2 at
2
? 0 + 1/2(2.5)8
2
= 80 m.
7. In 1
st
10 sec S
1
= ut + 1/2 at
2
? 0 + (1/2 × 5 × 10
2
) = 250 ft.
At 10 sec v = u + at = 0 + 5 × 10 = 50 ft/sec.
? From 10 to 20 sec ( ?t = 20 – 10 = 10 sec) it moves with uniform
velocity 50 ft/sec,
A
E
S
N
W
? ?
40 m
40 m
50 m
20 m
30 m
B
C
D
E
A ? Initial point
(starting point)
A
X
O
Y
B
(20 m, 0)
(–20 m, 0)
Initial velocity
u = 0
20
8 4
10
Time in sec
20 10 30
S (in ft)
0
t (sec)
250
750
1000
Chapter-3
3.2
Distance S
2
= 50 × 10 = 500 ft
Between 20 sec to 30 sec acceleration is constant i.e. –5 ft/s
2
. At 20 sec velocity is 50 ft/sec.
t = 30 – 20 = 10 s
S
3
= ut + 1/2 at
2
= 50 × 10 + (1/2)(–5)(10)
2
= 250 m
Total distance travelled is 30 sec = S
1
+ S
2
+ S
3
= 250 + 500 + 250 = 1000 ft.
8. a) Initial velocity u = 2 m/s.
final velocity v = 8 m/s
time = 10 sec,
acceleration =
10
2 8
ta
u v ?
?
?
= 0.6 m/s
2
b) v
2
– u
2
= 2aS
? Distance S =
a 2
u v
2 2
?
=
6 . 0 2
2 8
2 2
?
?
= 50 m.
c) Displacement is same as distance travelled.
Displacement = 50 m.
9. a) Displacement in 0 to 10 sec is 1000 m.
time = 10 sec.
V
ave
= s/t = 100/10 = 10 m/s.
b) At 2 sec it is moving with uniform velocity 50/2.5 = 20 m/s.
at 2 sec. V
inst
= 20 m/s.
At 5 sec it is at rest.
V
inst
= zero.
At 8 sec it is moving with uniform velocity 20 m/s
V
inst
= 20 m/s
At 12 sec velocity is negative as it move towards initial position. V
inst
= – 20 m/s.
10. Distance in first 40 sec is, ? OAB + ?BCD
=
2
1
× 5 × 20 +
2
1
× 5 × 20 = 100 m.
Average velocity is 0 as the displacement is zero.
11. Consider the point B, at t = 12 sec
At t = 0 ; s = 20 m
and t = 12 sec s = 20 m
So for time interval 0 to 12 sec
Change in displacement is zero.
So, average velocity = displacement/ time = 0
? The time is 12 sec.
12. At position B instantaneous velocity has direction along BC . For
average velocity between A and B.
V
ave
= displacement / time = ) t / AB ( t = time
10
t
5
2
4
6
8
t
10
(slope of the graph at t = 2 sec)
2.5
50
100
0
t
5 7.5 15
40
t (sec)
20
5 m/s
O
A
B
C
D
20
B
10
10 12 20
4 B
2
C
4 2
x
6
y
Chapter-3
3.3
We can see that AB is along BC i.e. they are in same direction.
The point is B (5m, 3m).
13. u = 4 m/s, a = 1.2 m/s
2
, t = 5 sec
Distance = s =
2
at
2
1
ut ?
= 4(5) + 1/2 (1.2)5
2
= 35 m.
14. Initial velocity u = 43.2 km/hr = 12 m/s
u = 12 m/s, v = 0
a = –6 m/s
2
(deceleration)
Distance S =
) 6 ( 2
u v
2 2
?
?
= 12 m
Chapter-3
3.4
15. Initial velocity u = 0
Acceleration a = 2 m/s
2
. Let final velocity be v (before applying breaks)
t = 30 sec
v = u + at ? 0 + 2 × 30 = 60 m/s
a) S
1
=
2
at
2
1
ut ? = 900 m
when breaks are applied u ? = 60 m/s
v ? = 0, t = 60 sec (1 min)
Declaration a ? = (v – u)/t = = (0 – 60)/60 = –1 m/s
2
.
S
2
=
a 2
u v
2 2
?
? ? ?
= 1800 m
Total S = S
1
+ S
2
= 1800 + 900 = 2700 m = 2.7 km.
b) The maximum speed attained by train v = 60 m/s
c) Half the maximum speed = 60/2= 30 m/s
Distance S =
a 2
u v
2 2
?
=
2 2
0 30
2 2
?
?
= 225 m from starting point
When it accelerates the distance travelled is 900 m. Then again declarates and attain 30
m/s.
? u = 60 m/s, v = 30 m/s, a = –1 m/s
2
Distance =
a 2
u v
2 2
?
=
) 1 ( 2
60 30
2 2
?
?
= 1350 m
Position is 900 + 1350 = 2250 = 2.25 km from starting point.
16. u = 16 m/s (initial), v = 0, s = 0.4 m.
Deceleration a =
s 2
u v
2 2
?
= –320 m/s
2
.
Time = t =
320
16 0
a
u v
?
?
?
?
= 0.05 sec.
17. u = 350 m/s, s = 5 cm = 0.05 m, v = 0
Deceleration = a =
s 2
u v
2 2
?
=
05 . 0 2
) 350 ( 0
2
?
?
= –12.2 × 10
5
m/s
2
.
Deceleration is 12.2 × 10
5
m/s
2
.
18. u = 0, v = 18 km/hr = 5 m/s, t = 5 sec
a =
5
0 5
t
u v ?
?
?
= 1 m/s
2
.
s =
2
at
2
1
ut ? = 12.5 m
a) Average velocity V
ave
= (12.5)/5 = 2.5 m/s.
b) Distance travelled is 12.5 m.
19. In reaction time the body moves with the speed 54 km/hr = 15 m/sec (constant speed)
Distance travelled in this time is S
1
= 15 × 0.2 = 3 m.
When brakes are applied,
u = 15 m/s, v = 0, a = –6 m/s
2
(deceleration)
Page 5

3.1
SOLUTIONS TO CONCEPTS
CHAPTER – 3
1. a) Distance travelled = 50 + 40 + 20 = 110 m
b) AF = AB – BF = AB – DC = 50 – 20 = 30 M
AD = m 50 40 30 DF AF
2 2 2 2
? ? ? ?
In ?AED tan ? = DE/AE = 30/40 = 3/4
? ? = tan
–1
(3/4)
His displacement from his house to the field is 50 m,
tan
–1
(3/4) north to east.
2. O ? Starting point origin.
i) Distance travelled = 20 + 20 + 20 = 60 m
ii) Displacement is only OB = 20 m in the negative direction.
Displacement ? Distance between final and initial position.
3. a) V
ave
of plane (Distance/Time) = 260/0.5 = 520 km/hr.
b) V
ave
of bus = 320/8 = 40 km/hr.
c) plane goes in straight path
velocity =
ave
V
?
= 260/0.5 = 520 km/hr.
d) Straight path distance between plane to Ranchi is equal to the displacement of bus.
? Velocity =
ave
V
?
= 260/8 = 32.5 km/hr.
4. a) Total distance covered 12416 – 12352 = 64 km in 2 hours.
Speed = 64/2 = 32 km/h
b) As he returns to his house, the displacement is zero.
Velocity = (displacement/time) = 0 (zero).
5. Initial velocity u = 0 ( ? starts from rest)
Final velocity v = 18 km/hr = 5 sec
(i.e. max velocity)
Time interval t = 2 sec.
? Acceleration = a
ave
=
2
5
t
u v
?
?
= 2.5 m/s
2
.
6. In the interval 8 sec the velocity changes from 0 to 20 m/s.
Average acceleration = 20/8 = 2.5 m/s
2
?
?
?
?
?
?
time
velocity in change
Distance travelled S = ut + 1/2 at
2
? 0 + 1/2(2.5)8
2
= 80 m.
7. In 1
st
10 sec S
1
= ut + 1/2 at
2
? 0 + (1/2 × 5 × 10
2
) = 250 ft.
At 10 sec v = u + at = 0 + 5 × 10 = 50 ft/sec.
? From 10 to 20 sec ( ?t = 20 – 10 = 10 sec) it moves with uniform
velocity 50 ft/sec,
A
E
S
N
W
? ?
40 m
40 m
50 m
20 m
30 m
B
C
D
E
A ? Initial point
(starting point)
A
X
O
Y
B
(20 m, 0)
(–20 m, 0)
Initial velocity
u = 0
20
8 4
10
Time in sec
20 10 30
S (in ft)
0
t (sec)
250
750
1000
Chapter-3
3.2
Distance S
2
= 50 × 10 = 500 ft
Between 20 sec to 30 sec acceleration is constant i.e. –5 ft/s
2
. At 20 sec velocity is 50 ft/sec.
t = 30 – 20 = 10 s
S
3
= ut + 1/2 at
2
= 50 × 10 + (1/2)(–5)(10)
2
= 250 m
Total distance travelled is 30 sec = S
1
+ S
2
+ S
3
= 250 + 500 + 250 = 1000 ft.
8. a) Initial velocity u = 2 m/s.
final velocity v = 8 m/s
time = 10 sec,
acceleration =
10
2 8
ta
u v ?
?
?
= 0.6 m/s
2
b) v
2
– u
2
= 2aS
? Distance S =
a 2
u v
2 2
?
=
6 . 0 2
2 8
2 2
?
?
= 50 m.
c) Displacement is same as distance travelled.
Displacement = 50 m.
9. a) Displacement in 0 to 10 sec is 1000 m.
time = 10 sec.
V
ave
= s/t = 100/10 = 10 m/s.
b) At 2 sec it is moving with uniform velocity 50/2.5 = 20 m/s.
at 2 sec. V
inst
= 20 m/s.
At 5 sec it is at rest.
V
inst
= zero.
At 8 sec it is moving with uniform velocity 20 m/s
V
inst
= 20 m/s
At 12 sec velocity is negative as it move towards initial position. V
inst
= – 20 m/s.
10. Distance in first 40 sec is, ? OAB + ?BCD
=
2
1
× 5 × 20 +
2
1
× 5 × 20 = 100 m.
Average velocity is 0 as the displacement is zero.
11. Consider the point B, at t = 12 sec
At t = 0 ; s = 20 m
and t = 12 sec s = 20 m
So for time interval 0 to 12 sec
Change in displacement is zero.
So, average velocity = displacement/ time = 0
? The time is 12 sec.
12. At position B instantaneous velocity has direction along BC . For
average velocity between A and B.
V
ave
= displacement / time = ) t / AB ( t = time
10
t
5
2
4
6
8
t
10
(slope of the graph at t = 2 sec)
2.5
50
100
0
t
5 7.5 15
40
t (sec)
20
5 m/s
O
A
B
C
D
20
B
10
10 12 20
4 B
2
C
4 2
x
6
y
Chapter-3
3.3
We can see that AB is along BC i.e. they are in same direction.
The point is B (5m, 3m).
13. u = 4 m/s, a = 1.2 m/s
2
, t = 5 sec
Distance = s =
2
at
2
1
ut ?
= 4(5) + 1/2 (1.2)5
2
= 35 m.
14. Initial velocity u = 43.2 km/hr = 12 m/s
u = 12 m/s, v = 0
a = –6 m/s
2
(deceleration)
Distance S =
) 6 ( 2
u v
2 2
?
?
= 12 m
Chapter-3
3.4
15. Initial velocity u = 0
Acceleration a = 2 m/s
2
. Let final velocity be v (before applying breaks)
t = 30 sec
v = u + at ? 0 + 2 × 30 = 60 m/s
a) S
1
=
2
at
2
1
ut ? = 900 m
when breaks are applied u ? = 60 m/s
v ? = 0, t = 60 sec (1 min)
Declaration a ? = (v – u)/t = = (0 – 60)/60 = –1 m/s
2
.
S
2
=
a 2
u v
2 2
?
? ? ?
= 1800 m
Total S = S
1
+ S
2
= 1800 + 900 = 2700 m = 2.7 km.
b) The maximum speed attained by train v = 60 m/s
c) Half the maximum speed = 60/2= 30 m/s
Distance S =
a 2
u v
2 2
?
=
2 2
0 30
2 2
?
?
= 225 m from starting point
When it accelerates the distance travelled is 900 m. Then again declarates and attain 30
m/s.
? u = 60 m/s, v = 30 m/s, a = –1 m/s
2
Distance =
a 2
u v
2 2
?
=
) 1 ( 2
60 30
2 2
?
?
= 1350 m
Position is 900 + 1350 = 2250 = 2.25 km from starting point.
16. u = 16 m/s (initial), v = 0, s = 0.4 m.
Deceleration a =
s 2
u v
2 2
?
= –320 m/s
2
.
Time = t =
320
16 0
a
u v
?
?
?
?
= 0.05 sec.
17. u = 350 m/s, s = 5 cm = 0.05 m, v = 0
Deceleration = a =
s 2
u v
2 2
?
=
05 . 0 2
) 350 ( 0
2
?
?
= –12.2 × 10
5
m/s
2
.
Deceleration is 12.2 × 10
5
m/s
2
.
18. u = 0, v = 18 km/hr = 5 m/s, t = 5 sec
a =
5
0 5
t
u v ?
?
?
= 1 m/s
2
.
s =
2
at
2
1
ut ? = 12.5 m
a) Average velocity V
ave
= (12.5)/5 = 2.5 m/s.
b) Distance travelled is 12.5 m.
19. In reaction time the body moves with the speed 54 km/hr = 15 m/sec (constant speed)
Distance travelled in this time is S
1
= 15 × 0.2 = 3 m.
When brakes are applied,
u = 15 m/s, v = 0, a = –6 m/s
2
(deceleration)
Chapter-3
3.5
S
2
=
) 6 ( 2
15 0
a 2
u v
2 2 2
?
?
?
?
= 18.75 m
Total distance s = s
1
+ s
2
= 3 + 18.75 = 21.75 = 22 m.
```

## Physics Class 11

102 videos|411 docs|121 tests

## FAQs on HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics - Physics Class 11 - NEET

 1. What are the basic concepts of rest and motion in kinematics?
Ans. In kinematics, rest refers to the state of an object when it is not changing its position with respect to its surroundings. Motion, on the other hand, is the change in position of an object with respect to its surroundings. The basic concepts in kinematics related to rest and motion include displacement, velocity, acceleration, and time.
 2. How is displacement different from distance in kinematics?
Ans. Displacement refers to the change in position of an object in a particular direction, whereas distance refers to the total path covered by the object irrespective of the direction. Displacement is a vector quantity, meaning it has both magnitude and direction, while distance is a scalar quantity, having only magnitude.
 3. What is the difference between uniform motion and non-uniform motion?
Ans. Uniform motion refers to the motion of an object in which its velocity remains constant throughout, i.e., there is no change in speed or direction. Non-uniform motion, on the other hand, is the motion in which the velocity of an object changes over time, either in terms of speed, direction, or both.
 4. How can we calculate average velocity in kinematics?
Ans. Average velocity can be calculated by dividing the total displacement of an object by the total time taken. It is the ratio of change in position to the time interval in which the change occurs. Mathematically, average velocity (v_avg) is given by v_avg = Δx/Δt, where Δx represents the change in position and Δt represents the change in time.
 5. What is the equation for motion under constant acceleration in kinematics?
Ans. The equation for motion under constant acceleration in kinematics is given by the second equation of motion: v = u + at, where v represents the final velocity, u represents the initial velocity, a represents the acceleration, and t represents the time taken. This equation relates the initial and final velocities of an object with its acceleration and time taken.

## Physics Class 11

102 videos|411 docs|121 tests

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