3.1SOLUTIONS TO CONCEPTS CHAPTER – 31. a) Distance travelled = 50 + 40 + 20 = 110 m Notes | EduRev

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: 3.1SOLUTIONS TO CONCEPTS CHAPTER – 31. a) Distance travelled = 50 + 40 + 20 = 110 m Notes | EduRev

 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.
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