Open App

NEET Exam > NEET Notes > Physics Class 11 > HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics

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

Download, print and study this document offline

Please wait while the PDF view is loading

 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
His displacement is AD
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
His displacement is AD
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
His displacement is AD
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
His displacement is AD
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
His displacement is AD
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 97 videos\|378 docs\|103 tests

Physics Class 11

97 videos|378 docs|103 tests

Join Course for Free

Top Courses for NEET

View all

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.

Related Exams

NEET JEE Class 12 Class 11 Airforce X Y / Indian Navy SSR

About this Document

	6.3K Views
	4.93/5 Rating
	Dec 18, 2024 Last updated

Document Description: HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics for NEET 2024 is part of Physics Class 11 preparation. The notes and questions for HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics have been prepared according to the NEET exam syllabus. Information about HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics covers topics like and HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics Example, for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics.

Introduction of HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics in English is available as part of our Physics Class 11 for NEET & HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics in Hindi for Physics Class 11 course. Download more important topics related with notes, lectures and mock test series for NEET Exam by signing up for free. NEET: HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics | Physics Class 11 - NEET

Description

Full syllabus notes, lecture & questions for HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics | Physics Class 11 - NEET - NEET | Plus excerises question with solution to help you revise complete syllabus for Physics Class 11 | Best notes, free PDF download

Information about HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics

In this doc you can find the meaning of HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics defined & explained in the simplest way possible. Besides explaining types of HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics theory, EduRev gives you an ample number of questions to practice HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics tests, examples and also practice NEET tests

	Physics Class 11 97 videos\|378 docs\|103 tests

Physics Class 11

97 videos|378 docs|103 tests

Join Course for Free

Download as PDF

Explore Courses for NEET exam

Top Courses for NEET

Explore Courses

Signup for Free!

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.

Start learning for Free

10M+ students study on EduRev

Semester Notes

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

study material

shortcuts and tricks

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

ppt

Exam

mock tests for examination

Previous Year Questions with Solutions

Important questions

MCQs

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

Summary

pdf

Extra Questions

Free

Objective type Questions

Viva Questions

Sample Paper

past year papers

practice quizzes

video lectures

;

Additional Information about HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics for NEET Preparation

HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics Free PDF Download

The HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics is an invaluable resource that delves deep into the core of the NEET exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective. With the help of these notes, you can grasp complex subjects quickly, revise important points easily, and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner, allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material, or toppers' notes, this PDF has got you covered. Download the HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics now and kickstart your journey towards success in the NEET exam.

Importance of HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics

The importance of HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics cannot be overstated, especially for NEET aspirants. This document holds the key to success in the NEET exam. It offers a detailed understanding of the concept, providing invaluable insights into the topic. By knowing the concepts well in advance, students can plan their preparation effectively. Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.

HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics Notes

HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics. It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation. Practice papers and question papers enable you to assess your progress effectively. Additionally, the paper analysis provides valuable tips for tackling the exam strategically. Access to Toppers' notes gives you an edge in understanding complex concepts. Whether you're a beginner or aiming for advanced proficiency, HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics Notes on EduRev are your ultimate resource for success.

HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics NEET Questions

The "HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics NEET Questions" guide is a valuable resource for all aspiring students preparing for the NEET exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

Study HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics on the App

Students of NEET can study HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics, 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 HC Verma Solutions: Chapter 3 - Rest & Motion Kinematics is prepared as per the latest NEET syllabus.

Education Revolution