SOLUTIONS

6. u = 6 m/ s
a = 2.5 m/ s
1.1
s = 44 m
220
1. (a) Average speed of aeroplane = = 440 km/h (a) v = u + 2 as
1
2 v = 6 + 2(2.5)(44)
250 v = 256 = 16 m/ s
(b) Average speed of bus = = 50 km/h
5 (b) v = u + at
(c) Average velocity of aeroplane =
220
= 440 km/h 16 = 6 + 2.5t
1 t=4s
2 7. u = 16 m/ s
220
(d) Average velocity of bus = = 44 km/h s = 0.4 m
5 v=0
Distance travelled 64
2. (a) Average speed = = = 32 km/h v = u + 2 as
Time taken 2
0 = 16 + 2 a(0.4)
(b) Average velocity = 0
256
3. Let total distance to be 2d. a=− = − 320 m/ s
0.8
v1 v2
v = u + at
d d 0 = 16 − 320t
1
d d t= s = 0.05 s
Total time taken = + 20
v v
s
Total distance 8. Average velocity =
Average speed = t
Total time
s
2d 2v v 60 = ⇒ s = 3000 m
= = 50
d d v +v
+ (a) v = u + at
v v
v = 0 + 2.4 × 50 = 120 m/ s
4. Let total time of journey to be 2t. 1
(b) s = ut + at
v1 v2 2
1
t t 3000 = 0 + (a)(50)
2
Total distance travelled = v t + v t a = 2.4 m/ s
v t+v t v +v
Average speed = = (c) Length of runway = 3000 m = 3 km
2t 2 1
5. Let total distance to be 3d. (d) 1500 = (2.4)t
2
v1 v2 v3
t = 25 2 s
d d d 9. u = 30 m/ s
d d d a = − 5 m/ s
Total time taken = + +
v v v (a) v = u + 2 as
3d
Average speed = 0 = 30 + 2(−5)s
d d d
+ + s = 90 m
v v v
v = u + at
3v v v
= 0 = 30 + (−5)t
v v +v v +v v
t =6s

1
 (b) If initial speed is doubled stopping distance will (c) v = u + 2 as
become 4 times and stopping time will get
(−4) = 2 + 2(−8) s
doubled.
10. Time taken by the woman to reach river s = − 0.75 m
4 km (d) v = u + 2 as
=
2.5 km / h 0 = 2 + 2(−8) s
Distance travelled by dog = ut s = 0.25 m
 4 Total distance = 0.25 + 0.25 + 0.75 = 1.25 m
= (4.5)   = 7.2 km
 2.5  s −0.75
(e) Average velocity = =
11. t=0 a=2m/s2 t=30s t=90s
t 0.75
u=0 v v=0 = − 1 m/ s = 1 m/ s, West
1.25 5
(f) Average speed = = m/ s
From t = 0 to t = 30 s, 0.75 3
v = u + at 13. u = 72 km/h = 20 m/ s
v = 0 + 2(30) v=0
v = 60 m/ s a = − 2 m/ s
v = u + 2 as
(a) v = u + 2 as
60 = 0 + 2(2) s
0 = 20 + 2(−2) s
s = 900 m
s = 100 m
From t = 30 s to t = 90 s,
(b) v = u + at
v = u + at 0 = 20 + (−2)t
0 = 60 + a(60) t = 10 s
a = − 1 m/ s 1
(c) s n = u + a (2 n − 1)
v = u + 2 as 2
0 = 60 + 2(−1)s 1
s = 20 + (−2)[2(1) − 1] = 19 m
2
s = 1800 m
1
(a) Total distance = 900 + 1800 = 2700 m s = 20 + (−2)[2(3) − 1] = 15 m
2
(b) Maximum speed = 60 m/ s [ v = 60 m/ s]
14. Let total time to be T.
(c) v = u + 2 as
1
30 = 0 + 2(2) s s = ut + at
2
s = 225 m 1 T  4T
v = u + 2 as s=0+ ×8  =
2  3 9
30 = 60 + 2(−1)s  T  8T
v = u + at = 0 + 8   =
s = 1350 m  3 3
So, position of train at half the maximum speed
4T
= 225 m and (1350 + 900) m Remaining distance = 100 −
9
= 225 m and 2250 m
1
12. u = 2 m/ s s = ut +
at
2
v = − 4m /s
4T 8T  2T 
t = 0.75 s 100 − =  
9 3 3
(a) v = u + at
− 4 = 2 + a(0.75) 20T
= 100
a = − 8 m/ s 9
T = 45 = 3 5 s
= 8m/ s , West
1
(b) v = u + at 15. Distance travelled in first 5 s = ut + at
0 = 2 − 8t 2
1 25 a
t = 0.25 s = 0 + (a)(5) =
2 2

2
 Distance travelled in first 10 s v = u + 2 as
1 100 a 17 = v + 2 ax
= 0 + a(10) =
2 2
2 ax = 289 − v …(ii)
Distance travelled in first 15 s
1 225 a From Eqs. (i) and (ii), we get
= 0 + a(15) = 2 v = 49 + 289
2 2
25 a v = 169 ⇒ v = 13 m/ s
s =
2  v + u   17 + 7 
(b) v =   =  = 12 m/ s
s =
100 a 25 a 75 a
− =  2   2 
2 2 2
20. Distance travelled by bike in 10 s
225 a 100 a 125 a
s = − = = 20 × 10 = 200 m
2 2 2
Lets say bike travels a further x distance before it is
s : s : s =1:3:5 caught by police,
1 x = 20t
16. s n = u + a(2 n − 1) …(i)
2 and 200 + x = 25t …(ii)
1
20 = u + a(2 × 7 − 1) Solving Eqs. (i) and (ii), we get
2
t = 40 s
2u + 13 a = 40 …(i)
x = 800 m
1
24 = u + a(2 × 9 − 1) Distance from turning = 200 + 800 = 1000 m
2
21. Lets say bus travel a distance x. In the same time boy
2u + 17 a = 48 …(ii)
will travel (40 + x )m.
Solving Eqs. (i) and (ii), we get
Boy,
u = 7 m/ s, a = 2 m/ s 1
s = ut + at
1 2
s = 7 + × 2[2(15) − 1] = 36 m
2 40 + x = 9t …(i)
17. v = u + at Bus,
1
20 = u + a(10) …(i) s = ut + at
1 2
s n = u + a(2 n − 1) 1
2 x = (1)t …(ii)
1 2
10 = u + a(19) Solving Eqs. (i) and (ii), we get
2
2u + 19 a = 20 …(ii) t − 18t + 80 = 0
Solving Eqs. (i) and (ii), we get t = 8 s and t = 10 s
u = − 180 m/ s and a = 20 m/ s 1.2
18. u = 54 km/ h = 15 m/s 1. (a) s = ut + 1 at
Distance travelled before brakes start acting 2
= 15 × 0.2 = 3 m 1
After brakes start acting, 80 = 0 + (10)t
2
v = u + 2 as t=4s
0 = 15 + 2(−6) s (b) v = u + 2 as
s = 18.75 m v = 0 + 2(10)(80)
Total distance = 3 + 18.75 = 21.75 m v = 40 m/ s
19. 1
7m/s v 17m/s (c) s = ut + at
P R Q
2
x x 1
40 = 0 + (10)t
(a) v = u + 2 as 2
v = 7 + 2 ax t =2 2 s
2 ax = v − 49 …(i) v = u + 2 as

3
 v = 0 + 2(10)(40) Average speed =
145
m/ s
v = 20 2 m/s 7
105
(d) v = u + 2 as Average velocity = m / s = 15 m/s
7
v = 0 + 2(10)(45) 1
(e) s = ut + at
v = 30 m/ s 2
(e) s = 80 m 1
105 = 50t + (−10)t
1 1 2
s = ut + at = (10)(2) = 20 m
2 2 t = 3 s and 7 s
Distance travelled in last 2 s = s − s = 60 m 4. (a) v = u + 2 as
1
2. (a) s = ut + at 0 = 20 + 2(−10)(H)
2
H = 20 m from tower or 180 m from ground.
1
160 = 20t + (10)t ⇒ t = 4 s 1
2 (b) s = ut + at
2
(b) v = u + 2 as 1
−160 = 20T + (−10) T
v = 20 + 2(10)(160) 2
v = 60 m/ s T = − 4 s and T = 8 s
1 200
(c) s = ut + at (c) Average speed = = 25 m/ s
2 8
1 5. At t = 8 s
60 = 20t + (10)t
2
t =2s 10 m/s
(d) v = u + 2 as
45 = 20 + 2(10) s t = 8s

s = 81.25 m
Height from ground = 160 − 81.25 = 78.75 m
40 m
1 1
(e) s n = u + a(2 n − 1) = 20 + (10)(2(4) − 1) = 55 m
2 2
3. (a) v = u + 2 as
For balloon,
0 = 50 + 2(−10)H
v = u + at
H = 125 m
v = 0 + 1.25 × 8
(b) v = u + at
v = 10 m/ s
0 = 50 + (−10)t 1
t =5s s = ut + at
2
(c) v = u + 2 as 1
 125  s = 0 + (1.25)(8)
v = 50 + 2(−10)   2
 2  s = 40 m
v = 3750 (a) For stone,
1
v = 3750 m/ s or 25 2 m/s s = ut +
at
2
(d) t=5s 1
v=0 −40 = 10T + (−10)T
2
20 m
T =4s
t=7s (b) For balloon,
1
125 m s = ut + at
2
1
s = × 1.25 × 12
2
= 90 m

4
 u  Solving Eqs. (i), (ii) and (iii), we get
(c) Distance = 40 + 2   = 40 + 10 = 50 m
 2g  t =t t

1 t = tt
6. H= gT …(i)
2 10. First stone,
1 1
H − 5.8 = g (T − 0.2) …(ii) h= gT …(i)
2 2
Solving Eqs. (i) and (ii), we get Second stone,
H = 45 m 1
h − 20 = g (T − 1) …(ii)
1 2
7. s = ut + at Solving Eqs. (i) and (ii), we get
2
h = 31.25 m
1
s = ut + (20)(60) = 36 km 11. First stone,
2
1
v = u + at s=0+ gt …(i)
2
v = 0 + 20(60) = 1200 m/ s
Second stone,
v = u + 2 as 1
s = 30(t − 2) + g (t − 2) …(ii)
0 = (1200) + 2 (−10)H 2
Solving Eqs. (i) and (ii), we get
H = 72 km
t=4s
(a) Maximum height reacted = 36 + 72 = 108 km
1
(b) v = u + at 12. For first particle, x = gt …(i)
0 = 1200 + (−10)t 2
t=0
t = 120 s u=0
x
Total time = 60 + 120 = 180 s = 3 min
8. For free fall, t=t
v = u + 2 as 100 m
100–x
v = 0 + 2(10)(500)
50 m/s
v = 100 m/ s
v = u + at
100 = 0 + 10t
For second particle,
t = 10 s 1
100 − x = 50t − gt …(ii)
After parachut opens, v = u + 2 as 2
8 = 100 + 2(−2)s Solving Eqs. (i) and (ii), we get
100 = 50t
s = 2484 m
t =2s
v = u + at
x = 20 m
8 = 100 + (−2)t 13. Lift,
t = 46 s 1
s = ut + at
(a) Total time = 10 + 46 = 56 s 2
1
(b) Height = 500 + 2484 = 2984 m 200 = (4)t
2
1 t = 10 s
9. s = ut + at
2 Object,
1
− H = ut + (− g )t …(i) At t = 5s,
2 v = u + at = 0 + 4(5) = 20 m/ s
1 1
+ H = ut + (+ g )t …(ii) s = 0 + (4)(5) = 50 m
2 2
1 1
+ H = gt …(iii) s = ut + at
2 2

5
 1 6. In first 5 s, distance travelled is 40 m,
150 = 20t + (10)t
2 1
s = ut + at
t + 4t − 30 = 0 2
−4 + 16 + 120 1
40 = u(5) + a(5) …(i)
t= 2
2
In first 10 s, distance travelled is (40 + 65 = 105 m).
t = − 2 + 34
1
Time taken by object to reach ground 105 = u(10) + a(10) …(ii)
2
= (− 2 + 34 ) + 5 Solving Eqs. (i) and (ii), we get
= (3 + 34 ) s u = 5.5 m/ s and a = 1 m/ s
7. l
1 Train u

75 2 Train w
1. Time taken = =5s
15
3 Train v
Displacement = 25 m
25 Pole
Average velocity = = 5 m/ s
5
Between (1) and (2),
2. Let distance between A and B = d v = u + 2 as
d d
Total time taken = + 2 al
20 30 w =u + …(i)
2
2d
Average speed = = 24 km/h Between (2) and (3),
d d
+ v = u + 2 as
20 30
l
s /3 s /3 s /3 u = w + 2a   …(ii)
3. Total time = + +  2
v 2v 3v
s 18 v Solving Eqs. (i) and (ii), we get
Average speed = =
s / 3 s / 3 s / 3 11 u +v
+ + w=
v 2v 3v 2
4. v = u + 2 as 8. Let bus travels a distance x.
In the same time, cyclist will travel (96 + x ).
(2u) = u + 2 a(10) …(i)
Cyclist,
v = (2u) + 2 a(10) …(ii) 1
s = ut + at
Solving Eqs. (i) and (ii), we get 2
v= 7u 96 + x = 20t …(i)
Bus,
1
5. s = ut + at 1
s = ut + at
2 2
1 1
first car s = 4t + (1)t …(1) x = (2)t …(ii)
2 2
1 Solving Eqs. (i) and (ii), we get
second car s = 2t + (2)t …(2)
2 t = 8 s and t = 12 s
From Eqs. (i) and (ii), we get 9. v = u + 2 as
t
4t + = 2t + t 0 = v + 2 as …(i)
2
0 = (nv) + 2 as ′ …(ii)
t=4s
1 Solving Eqs. (i) and (ii), we get
s = 4(4) + (1)(4) = 24 m s′ = n s
2

6
10. u=0 a =x v a=–3x 16. Car A,
v=0 s = 40t …(i)
t1 t2 Car B,
1
v = u + at s = (4)t …(ii)
2
v = 0 + xt …(i)
Solving Eqs. (i) and (ii), we get
v = u + at
t = 20 s
0 = v − 3 xt …(ii)
17. (a) A body having zero velocity can have acceleration.
Solving Eqs. (i) and (ii), we get
1
t = 3t and t + t = 8 18. s n = u + a(2 n − 1)
2
⇒ t = 2 s and t = 6 s
1
11. Let bus travels a distance y and in the same time 12 = u + a [2(2) − 1] …(i)
2
passenger travels y + x
1
1 20 = u + a(2(4) − 1) …(ii)
y = at and x + y = ut 2
2
1 Solving Eqs. (i) and (ii), we get
x + at = ut u = 6 m/s and a = 4 m/s
2
1 1
2u
t − t+
2x
=0 s = ut + at = 6(4) + (4)(4) = 56 m
a a 2 2
1
For real roots D > 0 s = 6(9) + × 4 × 9 = 216 m
2
 2y   2x 
  −4 >0 s − s = 216 − 56 = 160 m
 a  a
19. At t = t ; v = u + at
u > 2 ax v=0+a t
u > 2 ax At t = 2t , v = u + at
12. v = u + 2 as v=a t −a t =0
At t = 3t , v = u + at
 5
0 = 70 ×  + 2(−5)s …(i) v=0+a t
 18 
:
 5 :
0 = 60 ×  + 2(−5)s …(ii)
 18  :
Since, s + s < 80 m At t = 9t s, v = a t
Collision will be averted. 20. T
13. x = 9t …(i) D
100 + x = 10t …(ii) D
Solving Eqs. (i) and (ii), we get 3
t = 100 s 4m/s
14. v = u + 2 as D /3 D
Time taken for first one third of distance = =
0 = (200) + 2 a(4) …(i) 4 12
0 = u + 2 a(9) D
…(ii) Remaining time = T −
12
Solving Eqs. (i) and (ii), we get
2D 1  D  1  D 
u = 300 cm/ s = 2  T −   + 6  T −  
3 2  
12  2  12  
15. u = 126 km/h = 35 m/ s
D
x = 35t …(i) ⇒ = 4.5 m/ s
T
65 + x = 100t …(ii)
1
Solving Eqs. (i) and (ii), we get 21. s = 0 + a(p − 1)
t =1s 2
1
65 + x = 100 m s = 0 + ap
2

7
 sp =0+
1
a[2(p − p + 1) − 1] s = 2s
−p+ 2 s = 15 s − s − 2 s
1 = 12 s = 2 fs t
= a(2 p − 2 p + 1) = s + s
2 1
1 s= ft
22. s = u(2) + a(2) = 200 cm …(i) 72
2 27. u=0 v v=0
1
s = u(6) + a(6) = 420 cm …(ii)
2 t1 t2
Solving Eqs. (i) and (ii), we get
v = 0 + 2t …(i)
u = 115 cm/ s and a = − 15 cm/ s 0 = v − 4t …(ii)
v = u + at Solving Eqs. (i) and (ii), we get
v = 115 + (−15)(7) t = 2t and t + t = 3
v = 10 cm/ s ⇒ t = 2s and t = 1s
23. v = u + at ⇒ v = 4 m/ s
v = 0 + an …(i) 28. v = u + at
1 0 = 48 + (−10)t
s n = un + a(n)
2 t = 4.8 s
1 t = 4 s to t = 4.8 s
s n − = u(n − 2) + a (n − 2)
2 v = u + 2 as
Displacement in last 2 s
1 0 = 8 + 2(−10)s
= s n − s n − = a[n − (n − 2) ] 16
2 s = m
1v 2 v(n − 1) 5
= (2 n − 2)(2) =
2n n t = 4.8 s to t = 5 s
1
1
24. s = ut + at = 0 + at
1 s = 0 + (−10)(0.2)
2 2 2
v = u + at = at, 1
s =− m
1 5
s = ut + at
2 16 1 17
Total distance in the 5th second = + = m
1 1 5 5 5
− at = (at)t + (− a) t
2 2 1 1
t − 2tt − t = 0 29. 2ut + at = ut + (2 a)t
2 2
2t + 4t + 4t 1
t = = t + 2t ut = at
2 2
Total time = t + (t + 2t) = (2 + 2 )t 2u
t=
1 a
25. x = (1.5)t …(i)
2  2u  1  2u  6u
s = 2u   + a   =
1  a 2  a a
100 + x = (2)t …(ii)
2 30. Lets take time are t,8t & t.
Solving Eqs. (i) and (ii), we get 1  1 
Total distance = at + 60(8t) + 60t − at 
t = 20 s 2  2 
540t
26. s =
1
ft ⇒ t =
2s Average speed = = 54 km/h
2 f 10t
1
v = u + 2 as ⇒ v = 2 fs 31. s n = 0 + a(2 n − 1)
2
s = ( 2 fs )t 1
s n = 0 + an
v = u + 2 as 2
f sn 2n − 1
0 = ( 2 fs ) − 2   s =
 2 sn n

8
 1  2000 
32. 15 = u(3) + a(3) …(i)
2   = 0 + 2 × 10 × h
 2 
1
0 = 4(4) + a(4) …(ii)
2 h = 25 m from tower = 75 m from ground
Solving Eqs. (i) and (ii), we get 38. v = u + 2 as
u = 20 m/ s  39.2 
v = 0 + 2(9.8)  
33. 3 = 0 + 2 gh …(i)  2 
v = 4 + 2 gh …(ii) v = 19.6 m/ s
Solving Eqs. (i) and (ii), we get 39. v = u + at
v = 5 km/h v = 0 + 10(30) = 300 m/s
34. v = u + 2 as v = u + at
0 = 300 + (−10)t
(3 v) = v + 2 gh
t = 30 s
4v Total time = 30 + 30 = 60 s
h=
g 40. For first 10 s,
1 v = u + at
35. s = ut + at
2 v = 0 + 10(10) = 100 m/ s
1 1
H = 0 + × 10 × 5 = 125 m s = ut + at
2 2
1 1
Distance travelled in first 3 s = 0 + (10)(3) = 45 m s = 0 + (10)(10) = 500 m
2 2
Remaining distance = 125 − 45 = 80 m After parachute opens
1 v = u + 2 as
s = ut + at
2
1 v = 100 + 2(−2.5) (2495 − 500)
80 = 0 + (10)t
2 v = 100 − 9975 = 25
t=4s v = 5 m/ s
Total time = 3 + 4 = 7 s 1
41. s = ut + at
1 2
36. s = ut + at
2 1
h = ut − gt
t=0 2
u=0 2u 2h
t − t+ =0
g g
H t=T–1 Lets take roots of the quadratic equation to be t and
25 m t .
2u
t=T t +t = = 12 ⇒ u = 60 m/ s
g
1 At t = 2 s,
H=0+ gT …(i) 1
2 s = ut + at
1 2
H − 25 = 0 + g (T − 1) …(ii) 1
2 h′ = 60(2) + (−10)(2)
Solving Eqs. (i) and (ii), we get 2
T = 3 s and H = 45 m = 100 m

v = u + 2 as u
37. 42. = 20 ⇒ u = 20 m/ s
2g
v = 0 + 2 × 10 × 100
2u
v = 2000 m/ s T= =4s
g
v = u + 2 as

9
43. s = ut +
1 1
at − h = 4(4) + (−9.8) (4) t =1s
2 2 x = 25 m
h = 62.4 m Height from ground = 15 m
1 1
44. s t = 0 + × 10(2t − 1) 49. s = ut + at
2 2
1 1
st+ = 0 + × 10[2(t + 1) − 1] −H = u(9) + (−10)(9) …(i)
2 2
1 1 1
100 = × 10(2t − 1) + × 10 (2t + 1) + H = u(4) + (+ 10)(4) …(ii)
2 2 2
t =5s 1
+ H = (+ 10)t …(iii)
v = u + at 2
v = 0 + 10(5) = 50 m/ s Solving Eqs. (i), (ii) and (iii), we get
1
45. H = gT …(i) t =6s
2 1
1 50. s = ut + at
H − 40 = g (T − 2) …(ii) 2
2 1
− h = 4.9(3) + (−9.8) (3)
Solving Eqs. (i) and (ii), we get 2
H = 45 m h = 29.4 m
1
46. Distance covered in first 3 s = 0 + (10)(3) = 45 m 51. Time taken by stone to hit water =
2h
=2s
2 g
1
H = gT …(i) Distance 19.6
2 Velocity of sound = = = 392 m/ s
Time 2.05 − 2
1
H − 45 = g (T − 1) …(ii) 1
2 52. s = ut + at
Solving Eqs. (i) and (ii), we get 2
H = 125 m 1
−h = vt + (− g )t
47. v = u + 2 as 2
2v 2h
30 = u + 2(−10)(80) t − t− =0
g g
u = 50 m/ s
2v 4v 8h
v=0 6s + +
g g g
t=
2
30 m/s

v v 2h v  2 gh 
30 m/s = + + = 1+
g g g g  v 
80m

u 53. Since, he is throwing two balls every second, time

interval between two throwings
1
= s = 0.5 s
1 2
48. x = 20t + × 10 × t
2 v = u + at
1 0 = u + (−10)(0.5)
40 − x = 20t − × 10t
2 u = 5 m/ s
B
u 5
20 m/s
H = = = 1.25 m
x 2 g 20
1
40 m B A 54. s = 0 + (10)(1) = 5 m
2
20 m/s 1
s = 0 + × 10 (2(2) − 1) = 15 m
2
A

10
 1 2u
s =0+ × 10(2(3) − 1) = 25 m 60. = 12 t=6s
2 g
s :s : s :: 1:3:5 u = 60 m/ s
t=3s
1 t=9s
55. 0.4 = 0 + × 9.8 t 1
2 s = ut + at
2
0.8
⇒ t = H
9.8 1
H = 60(3) + (−10)(3)
1 2
0.9 = 0 + × 9.8 t
2 H = 135 m t=0 t=12s
1.8
⇒ t =
9.8
1.8 0.8 61. Total time of flight = t + t
t −t = −
9.8 9.8 2u
=t +t
9 4 3 2 1 g
= − = − = s
49 49 7 7 7 g (t + t )
u=
2
1 2
56. 1 = 0 + gt ⇒ t = 1
2 g s = ut + at
2
1 4  g (t + t ) 1
2=0+ gt ⇒ t = h=  t − gt
2 g  2  2
1 6 1
3=0+ gt ⇒ t = h = gt t
2 g 2
Time taken for successive 1m distance 1
62. s = ut + at
= t : (t − t ): (t − t ) 2
57. v = u + 2 as 1
14 = u(0.8) + (−10)(0.8)
2
H
(9.8) = u − 2(9.8)   …(i) 17.2 43
 2 u= =
0.8 2
0 = u − 2(9.8)(H) …(ii) 2u 2  43  43
T= =  = s
Solving Eqs. (i) and (ii), we get g 10  2  10
H = 9.8 m
t +t =T
58. For stone, 43
1 0.8 + t =
s = ut + at 10
2
43 4 35 7
1 t = − = = s
⇒ − 76 = u(6) + (−10)6 10 5 10 2
2
52 2h 2(9)
⇒ u = m/ s upwards 63. Time taken by a drop to reach ground = =
3 g 10
52
So, initial velocity of balloon = m/ s upwards 1 2h
3 Time interval between successive dropping =
3 g
59. At t = 5s, it will be at its highest point t=5s
u 1  1 2h 
=5 Distance of 2nd drop = 0 + g  =1m
g 2 3 g 
t=4s t=6s
u = 49 m/ s
1  2 2h 
Distance of 3rd drop = 0 + g  = 4m
2 3 g 

11
64. v = u + 2 as 0 = 10 + 12(−4)s
(2 v) = 0 + 2 gx …(i) s = 12.5 s
v = u − 2 g (H − x ) …(ii) Total distance = 6 + 12.5 = 18.5 m
A 2. v = u + 2 as
u=0
 3u 
v   = u + 2 a(14) …(i)
x  4

H A B  3u 
0 =   + 2 ax …(ii)
2v  4
H–x
u Solving Eqs. (i) and (ii), we get
B x = 18 cm
1  v + u
s = ut + at 3. s= t
2  2 
1  25 + 10 
x = 0 + gt …(iii) sA =  t
2  2 
H − x = ut − gt
1  v + 15 
…(iv) sB =  t
2  2 
Solving Eqs. (i), (ii), (iii) and (iv), we get sA = sB
H−x 2
= ⇒ v = 20 m/ s
H 3
4. a=–4 m/s2
65. g A = 9g B t=0
u u=8 m/s
H=
2g When particle comes to rest momentarily
u v = u + at
2= …(i)
2g A ⇒ 0 = 8 – 4t
u ⇒ t =2s
h= …(ii) v = u + 2 as
2g B
Solving Eqs. (i) and (ii), we get ⇒ 0 = 8 + 2(−4)s
h = 18 m ⇒ s =8m
1 Now, from t = 2 to t = 5,
66. H = 0 + gT …(i) 1
2 s = ut + at
1 2
H − 25 = 0 + g (T − 1) …(ii) 1
2 s = 0 + (−4)(3) = − 18 m
Solving Eqs. (i) and (ii), we get 2
H = 45 m Total distance = 8 + 18 = 26 m
u 4
67. = 6 ⇒ u = 60 m/ s 5. Time taken to reach max. height = = 4.5 s
g g
1 t=4.5s
s = 60 + (−10)[2(1) − 1] = 55 m
2
1
s = 60 + (−10)[2(7) − 1] = − 5 m t=4s t=5s
2
Ratio = 11 : 1

u
1. u = 36 km/h = 10 m/ s
Before brakes, s = 10 × 0.6 = 6 m
After brakes, v = u + 2 as From t = 4.5s to t = 5s

12
 1 1 1
s = ut + at = (10)(0.5) = 1.25 m s = ut + at
2 2 2
Distance travelled in the 5 second = 2 × 1.25 = 2.5 m 1
6 = 0 + a( 2 )
6. v=0 2
a = 6 m/ s
v
2h h
9. 4.25 = +
g 320
h = 80 m
5 ft 10. Between A and B,
v = u + 2 as
u
u/4
C

1 u/3
s = ut + at B
2
1
⇒ 5 = u(0.5) + (−32) (0.5) u/2
2 A
⇒ u = 18 ft/ s  u  u
⇒ v = u + 2 as   =   + 2 a (AB) …(i)
 3  2
⇒ v = 18 + 2(−32)(5) Between B and C,
⇒ v = 324 − 320  u  u
  =   + 2 a(BC) …(ii)
⇒ v = 2 ft/ s  4  3
v 2 1 Solving Eqs. (i) and (ii), we get
Height above the window = = = ft
2 g 2 × 32 16 AB 20
=
7. BC 7
u=0
1
11. AB = gt
2
h 1
AC = g (t + 2)
2
1
u=√8gh
AD = g (t + 3)
2
A u=0
u t
= 4h
2g B
u = 8 gh 2s
1 C
s = ut + at
2 1s
1
x = gt …(i) D
2 BC = CD
1 AC − AB = AD − AC
h − x = ( 8 gh )t − gt …(ii)
2 2AC = AB + AD
Solving Eqs. (i) and (ii), we get 1  1 1
2  g (t + 2)  = gt + g (t + 3)
t=
h
=
h 2  2 2
8 gh 8g
2t + 8t + 8 = t + t + 9 + 6t
2h 2(11.5 − 1.5) 2t = 1
8. t= = = 2s
g 10 1
t=
2

13
12. 10 m/s v 20 m/s

P O Q
x x 1.
Between P and O,
v = u + 2 as B
v = 10 + 2(a)(x ) …(i)
Between O and Q, A
v = u + 2 as
20 = v + 2 ax …(ii)
Solving Eqs. (i) and (ii), we get
We can open the cube as shown in figure.
v = 250 = 15.8 m/ s
Shortest path length = (2a) + a = 5a
u
13. = 10 5a
2g Time taken by ant =
u
u = 196
2. Let s be distance covered by each car. Let times taken
u = 14 m/ s by the 2 cars to complete journey be t and t and
14. velocities at finishing point be v and v .
P
9.8 m/s v − v = v ,t − t = t
v v −v 2a s − 2a s
= = = aa
2h Q
9.8 m/s t t −t 2s 2s
−
a a
h
3. v = u + 2 as
50 = 10 + 2 as …(i)
1 v = 50 + 2(− a)(− s)
For P, s = ut + at …(ii)
2
Solving Eqs. (i) and (ii), we get
1
⇒ 2h = (9.8)T + (+ 9.8) T …(i) v = 4900 = 70 m/ s
2
1 4. Lets t be time when engine is switched off.
For Q, s = ut + at
2 1
s = ut + at
1 2
⇒ − h = 9.8T + (− 9.8)T …(ii)
2 1
Height of helicopter = s = 0 + (4)t
Solving Eqs. (i) and (ii), we get 2
T =6s 1
(4)t
15. For balloon, u = 0, a = 4.9 m/s Time taken by sound = 2 = 50 − t
320
v = u + at t = 8000 − 160t
At t = 2, v = 0 + (4.9)2 = 9.8 m/s
1 t + 160t − 8000 = 0
s = ut + at
2 t + 200t − 40t − 8000 = 0
1
At t = 2; s = 0 + (4.9)(2) = 9.8 m t(t + 200) − 40(t + 200) = 0
2 t = 40 s
Initial velocity of ball will be same as that of the balloon. For helicopter, v = u + at
For ball, v = u + 2 as v = 0 + 4(40) = 160 m/ s
0 = (9.8) + 2 (−9.8)(+ h) 5. Distance travelled by policeman in 12 s
 5 
h = 4.9 m  v + u  150 × + 0
18
Maximum height of ball from ground = 9.8 + 4.9 = 14.70 m = t=  12 = 250 m
 2   2 
 

14
 Thereafter, policeman travel with constant speed. 8. t =10 s
 5 a =0
1500 = 250 + 150 ×  (T − 12 − 2) ⇒ T = 44 s A u =40 m/s
 18 
1500 1500 18
Speed of speeder = m/ s = × km/h B
44 44 5 u =0
= 122.7 km/h a = 2 m/s2
t=0
6. At t = 5 s, balloon will be at a height of 25 × 5 = 125 m
v B = uB + aB t
Lets say bullet hits balloon in further time t and balloon
travels ‘x’ distance further. 40 = 0 + 2(t)
For bullet, t = 20 s
1
1 In 20 s, distance travelled by B = (2)(20) = 400 m
s = ut + at 2
2
Distance travelled by A = 40 × 10 = 400 m
125 + x = ut − 5t
So, A will be never ahead of B.
For balloon,
9. D
1
s = ut + at D D
2 2 2
x = 25t v1 v2 v3
125 + 25t = ut − 5t Total distance = D
5t + (25 − u)t + 125 ≥ 0 Total time = T
D D /2
For real roots D ≥ 0 Time taken to cover first distance =
2 v
(25 − u) − 2500 ≥ 0
D
(25 − u) ≥ 2500 Remaining time = T −
2v
|25 − u|≥ 50 1 
D D  1  D 
u − 25 ≥ 50 = v  T −   + v  T − 
2 2  2v   2  2v  
u ≥ 75 m/s
D 2 v (v + v )
7. Lets T be time taken by each of them. Average speed = =
T 2v + v + v
Distance travelled by first particle
 1 T    T  T 1 T  5
10. u y
s =  0 + a    +   a  + (2 a)    s = aT
     2 2 2    x v=0
 2 2  2 8
y
Distance travelled by second particle = s v
s 1
= 0 + (2 a) T …(i) Displacement =
1
(Distance)
2 2 3
s 1 x + 2y
= [(2 a) T ] T + aT …(ii) x= ⇒x =y
2 2 3
T +T =T …(iii)
u 2 a (x + y )
Solving Eqs. (i), (ii) and (iii), we get = = 2
5T − 2TT − T = 0 v 2 ay
(1 + 6 )T 11. b − a = un +
1
a′ n
T = …(i)
5 2
s 1 1
= (2 a)T c − a = u(2 n) + a ′ (2 n) …(ii)
2 2 2
2 a(7 + 2 6 ) 1
s = T d − a = u(3 n) + a ′ (3 n)
25 2
5 Solving Eqs. (i) and (ii), we get
aT c − 2b + a
s 35 − 10 6
= 8 = a′ =
s 2 a(7 + 2 6 ) 16 n
T
25

15
12.  PQ PQ 
Y 2PQ = v  +  ⇒v = 24 m/ s
(0, v1t)  20 30 
v 3 t , v3 t 15. u + 2 gh = 2 u − 2 gh
v 2t 2 2
v2 v3 (v1t,0) u + 2 gh = 4u − 8 gh
X
v1 3u = 10 gh
v1t u 5h
=
2g 3
v
16. Maximum height =
vt vt 2g
−0 −` v t
2 = 2 1
s = ut + at
vt vt 2
−v t −0
2 2 1  3v v 
s=0+ g − 
v v − 2v 2  2g g
=
v − 2v v 1v
s=
v = v − 2v v − 2v v + 2v v 8 g
v v 5v
v =
2v v Total distance = + =
v +v 2g 8g 8g
u
13. u 17. =H
C 2g
A B v=0 2u
t +t = …(i)
D g
t 1
Between B and C, and = …(ii)
v = u + at t 3
0 = uB + (−5)(1) Solving (i) and (ii), we get
uB = 5 m/ s u
t =
Between A and B, 2g
v = u + at 3u
and t =
s = uA − 5(1) 2g
2
uA = 10 m/ s 1  u 1  u
s = ut + at = u   + (− g )  
Between A and B, 2  2g  2  2g 
v = u + 2 as 3u 3H
= =
5 = 10 + 2(−5)s 8 g 4
s = 7.5 m 18. Lets say time taken is T.
Between B and C,
v = u + 2 as
0 = 5 + 2(−5)s A B
s = 2.5 m
Total distance = 7.5 + 2.5 + 2.5 = 12.5 m
 v + u
14. s =   t = v ⋅t
 2  A B
For P to Q, PQ = 20t
1
For Q to S, QS = 30t s A = uT − gT
2
PQ = QS = 20t = 30t
1
For P to S, s B = u(T − t) − g (T − t)
2
(PQ + QS) = v (t + t ) sA = sB

16
 1 1
uT − gT = u (T − t) − g (T − t) 24. T =
2h
=
2(320)
=8s
2 2 g 10
1
ut = g (2T − t)(t) 1
2 s = ut + at
2u 2
2T − t = 1
g 320 = u(3) + (10)(3)
1  2u  2
T =  + t u(3) = 275
2g 
275
u t u= = 91.66 m/ s
T= + 3
g 2
1
1 25. s = ut + at
19. s A = u(10) − g (10) 2
2
1
 1   1  s = 0 + × 10 × 5 = 125 m
s B = (u + 8) 6 − g (6)  + u(4) + × 2 × 4  2
 2   2 
1 26.
s B = (u(10) + 64) − g (6) u=0
2 H a=g
s B − s A = 64 + 5(64) = 64 + 320 = 384 m
√2gH
20. v = u + 2 as
(20) = u − 2(10)(25) = 30 m/ s
100 m g
2h a= 2
T 12 8 2
21. = = =
T 2h 12 3 1
Inside lake, s = ut + at
8 2
1
22. 50T − gT = 45  2 gH  
2 2H  1  g   2H 
100 =   7 −  +   7 − 
T − 10T + 9 = 0  2  g  2  2  g 
T = 1 and T = 9 ⇒ H = 45 m
1
u(9 − 2) − g (9 − 2) = 45 27. A
2 u=0
7u − 245 = 45 h
290
u= m/s W B
7 I tAB = t1
1.8 m N
tAC = t2
D
u H
= 20 ⇒ u = 20 m/ s
O
23. W C
2g
v

v
D
20 m

40 m/s 1
s = ut + at
2
v = u + 2 as 1
For A to B, h = 0 + (10)(t ) …(i)
v = 40 + 2(−10)(20) 2
v = 20 3 m/ s 1
For A to C, h + 1.8 = 0 + (10)t …(ii)
2
v = u + at
20 3 = 40 − 10t Solving Eqs. (i) and (ii), we get

t = 4−2 3 5(t − t ) = 1.8 …(iii)

Total time = 8 − 4 3 t − t = 0.2 …(iv)

17
 Solving Eqs. (iii) and (iv), we get 1
s = ut + at
t = 1 s and t = 0.8 s 2
and tCD = tDC = 1 s 1
⇒ 40 = (1.25)t
So, tAD = 2 s 2
1   1  ⇒ t =8s
28. H =  gt  + 20(3) − gt 
2   2 
v = u + at
H = 60 m ⇒ v = 0 + 1.25 × 8
1
29. s n = u + a(2 n − 1) ⇒ v = 10 m/ s
2
1
1  u   3a For BOLT, s = ut + at
= u + (− a) 2  − 1 − 1 = 2
2  a   2
−40 = 10t − 5t
1
30. H = g ( 2) …(i)
2 t − 2t − 8 = 0
u = 2 g (10) = 20 g t=4s
1 Distance travelled by lift from t = 10 s to t = 12 s
−H = ( 20 g )(2) − g (2) …(ii)
2 1
s = ut + at
Solving Eqs. (i) and (ii), we get 2
0 = 2 20 g − g s = 12.5 × 2 = 25 m
2 20g = g Height of lift = 62.5 + 25 = 87.5 m
80g = g
g = 80 m/ s 1. v ⋅ a > 0 object is speeding up.
31. R v= 0 v ⋅ a < 0 object is slowing down.
2. (a,d) (a) |Instantaneous velocity| = Instantaneous speed
Q Q (d) In circular motion average velocity is zero in one
P
revolution.
P
3. For A to C,
E D

C
F
1  TP 
RP = g  B
2 2 A

1  TQ  s 3a 3v
RQ = g  v = = =
2 2 t 2a 2
g (TP − TQ ) v
PQ = RP − RQ = =H For A to D,
8
s 2a 2v
g=
8H v = = =
t 3a 3
TP − TQ
v
1
32. s = ut + at For A to F,
2 s a v
1 v = = =
⇒ s = 0 + (1.25)(10) t 5 a 5
2 v
⇒ s = 62.5 m For A to B,
v = u + at s a
v = = =v
⇒ v = 0 + 1.25 × 10 t a
⇒ v = 12.5 m/ s v

18
  v + u P
4. s= t 5 m/s
 2 
 v + 5
⇒ 30 =  4 Q 5m/s
 2  2h

⇒ v + 5 = 15 h
⇒ v = 10 m/ s
⇒ v = u + at
Solving Eqs. (i) and (ii), we get
⇒ 10 = 5 + a(4)
5t(1 + t)
⇒ a = 1.25 m/ s −2 =
5t(1 − t)
5. v = u + at −2 + 2t = 1 + t
v B = v T + 10(0.5) t = 3s
vB − vT = 5 h = 30 m
u + v 9. (a, b, d) As the particle goes up, it slows down.
s= t So, t > t > t .
 2 
Distance d
Average speed = =
 v + vB  Time t
3= T  (0.5)
 2  Since, distances are same, average speed is inversely
v T + v B = 12 ⇒ v B = 8.5 m/ s proportional to time.
and v T = 3.5 m/ s So, v > v > v
u Acceleration is constant.
6. H = So, a = a = a = g
2g
v = u + at ⇒ v − u = at
v = u + 2 as
⇒ ∆v = at
1  u 
v = u + 2(− g )     ⇒ ∆v ∝ t
 2  2g   So, ∆v > ∆v > ∆v
u 10. (a, c) Velocity is vector quantity. It has both magnitude
v=
2 and direction. If either magnitude or direction
changes, velocity changes.
1
7. s = ut + at Acceleration is rate of change of velocity.
2 So, when velocity changes, body has acceleration.
100 = ut − 5t 1
t 11. 140 = 15(14) − a(14)
2
5t − ut + 100 = 0 5
a = m /s
t 7
u v = u + at
t +t = 5
5 v = 15 − (14) = 5 m/ s
7
t t = 20
12. v = u + 2 as
8. Lets take time taken to reach ground to be t.
For P, 0 = 10 + 2(−5)s
1 s = 10 m
s = ut + at
2 v = u + at
⇒ 2h = 5t + 5t …(i)
0 = 10 − 5t
For Q, t =2s
1
s = ut + at From t = 2 s to t = 3 s
2 1
⇒ h = 5t − 5t …(ii) s = ut + at
2

19
 1
s = (−5)(1) = 2.5 m 10. v = u + at
2 0 = 20 + (−5)t
So distance travelled from t = 0 to t = 3 t=4s
= 10 + 2.5 = 12.5 m 20
From t = 0s to t = 4 s; s = = 40 m
1 2(5)
13. s n = u + a(2 n − 1)
2 20
1 From t = 4s to t = 8s; s = = 40 m
25 = u + a[(2(5) − 1)] … (i) 2(5)
2
Total distance = 40 + 40 = 80 m
1
33 = u + a [(2(7) − 1)] … (ii) 1
2 11. & 12. −h = u(6) − g (6)
2
Solving Eqs. (i) and (ii), we get
…(i)
u = 7 m/ s, a = 4 m/ s 1
+ h = u(2) + g (2)
14. (a,b,c) If acceleration has a component in the direction 2
of velocity, then body speeds up. If acceleration has a …(ii)
component in the direction opposite to velocity, then Solving Eqs. (i) and (ii), we get
body slows down.
h = 60 m and u = 20 m/ s

Comprehension Based Questions

1. v = u + at 1. v = u + at
0 = 40 − 10t ⇒ t = 4 s ⇒ v = 0 + 10t …(i)
2. v = u + 2 as v = u + at
⇒ 0 = v − 5(15 − t) …(ii)
0 = 40 + 2(−10)(x − 10)
Solving Eqs. (i) and (ii), we get
x = 90 m t = 5 and v = 50 m/ s
3. v = u + 2 as 1
s = ut + at
v = 40 + 2(−10)(−10) 2
1
v = − 30 2 m/ s ⇒ s = (10)(5) = 125 m
2
1
4. s = ut + at 1
s = ut + at
2 2
−10 = 40t − 5t 1
⇒ s = 50(10) + (−5)(10)
t = 4+3 2 2
1 s = 250 m
5. s = ut + at v = u + 2 as
2
1 ⇒ 25 = 50 + 2(−5) s
s= × 20 × (15 × 60) = 8100 km or 8.1 × 10 km
2 ⇒ s = 187.5 m
384000 − 8100 − 8100 Distance = 125 + 187.5 = 312.5 m
6. = 0.95
384000 125 + 250 375
Average speed = = = 25 m/ s
7. v = u + at 15 15
v = 0 + 20 (15 × 60) = 18000 m/ s 2u
2. = 10
384000 × 10 g
t= = 340 min
18000 u = 50 m/ s
Total time = 15 + 340 + 15 = 370 min 10
Time between two successive throwings = = 0.5 s
8. v = u + 2 as 20
Height of 15th ball when 20th ball is in hand
10 = 20 + 2 a(30)
1
a = − 5 m/ s = ut + at
2
9. v = u + at 1
= 50(2.5) + (−10) (2.5) = 93.75 m
v = 20 + (−5)(8) = − 20 m/ s 2

20
 Speed of 10th ball when 20th ball is in hand = u + at w −u
t =
= 50 + (−10) 5 = 0 a
Number of balls thrown per minute v = u + at
60 v = w + at
= = 120 balls/min
0.5 v−w
t =
3. A → PQ, B → R, C → S, D → S a
4. A → PQ, B → PRS, C → PR, D → PRS t = 2t
w −u v − w
=2 
a  a 
1. v = u + 2 as w − u = 2v − 2w
 9u  3w = 2v + u
  = u + 2 al (For a single plank) 2v + u
 10  w=
3
0 = u + 2 a(nl) (For n planks)
2v + u v +u
2. w= =
3 2
4 v + u + 4uv u + v
v = u + 2as 3s v = u + 2as =
t = 9 2
v = 2as 2as 0 = ( 2as )
8 v + 2u + 8uv = 9u + 9 v
1
s = ut + at + 2a′( 5s)
2
a′ = −
a v − 8uv + 7u = 0
2s 5
t =  v  v
a v = u + at   −8  +7= 0
 u  u
a
0 = 2as − t λ − 8 λ + 7 = 0 ⇒λ = 1 and λ = 7
5
2s 4. Accelerating, v = u + at
t =5 v = xt
a
v
t =
Total distance x
v =
Total time v = u + 2 as
s + 3s + 5s v = 2 xs
=
2s 3s 2s
+ +5 s =
v
a 2 as a 2x
=
9s
=
9 2
as Retarding, v = u + at
15 s 15 0=v − yt
2 a v
t =
= 3 2as y
3 v = u + 2 as
v 2 as
=5 =
3 0 =v − 2 ys
v 2 as 5
v
s =
3. A x C x B 2y
u w v t + t = 4 min
w = u + 2 ax   1 1
⇒ v  +  =4 …(i)
 Solving them, we get  x y
v = w + 2 ax 
s + s = 2 km
v +u  1 1
w= ⇒
v
2  +  =2 …(ii)
2  x y
v = u + at
From Eqs. (i) and (ii), we get
⇒ w = u + at
v =1

21
 Put v = 1 in Eq. (i), we get  u + 2 g (15) 
1 1  
+ =4  2 
x y  
15 =
5. For P, 2g
v = u + at u + 300
300 =
30 = 15 + at 4
at = 15 u = 900
For Q, u = 30 m/ s
v = u + at 1
9. 0 = vt − gt …(i)
v = 20 + (− a)t 2
v = 20 − at 1
160 = vt + gt …(ii)
= 20 − 15 = 5 2
⇒ v = 20 m/ s
6. Time taken for reaching maximum speed = t
v = u + at 10. v=0
B
20 = 0 + 1 t 10 m
t = 20 s
6m
1 O C
s = ut + at
2
1
s = (1)(20) = 200 m 6√5m
2
v0
Time taken for retardation = t ′
v = u + at A
0 = 20 + (−2)t ′ AC = 6( 5 ) − 6 = 12 m
t′ = 10 s
1 and CB = 10 − 6 = 8 m
s = ut + at
2 The maximum height attained by the ball
1
s′ = 20(10) + (−2)(10) H = AB = 12 + 8 = 20 m
2 v = u − 2 gH
s′ = 100 m 0 = v − 2 gH
Remaining distance will be covered with constant
speed of 20 m/s. v = 2 gH = 20 m/ s
1000 − 200 − 100 11.
t= = 35 s O
20
D
Total time = 20 + 35 + 10 = 65 s
C
7. On each collision speed will get halved.
 2 gh   2 gh  B
2  2 
2h  2   2 
T= + + + ....
g g g A

2h 2h  1 1 1  Stone A takes 4 s from O to A

T= + 1 + + + + ∞
8 g  2 2 2  1 1
⇒ s = ut + at ⇒ OA = 0 + (10)(4) = 80 m
  2 2
2h  1  Stone B is travelling for 3 s
T= 1 +  =6s
g  1 1
1− OB = 0 + (10)(3) = 45 m
 2 2
1
u OC = 0 + (10)(2) = 20 m
8. H = 2
2g

22
 1 From A to B,
OD = 0 + (10)(1) = 5 m
2 v = u + 2 as
So, AB = OA − OB = 80 − 45 = 35 m 20 = 40 + 2(−10)(AB)
BC = OB − OC = 45 − 20 = 25 m
AB = 60 m
CD = OC − OD = 20 − 5 = 15 m
For next 2s,
3 (AB + BC) 3(35 + 25)
= =6 1
2CD 2(15) s = ut + at
2
12. Lets say they collide after time t 1
 BC = 20 × 2 + (0)(2)
     2
 2 gh t − 2h  − 1 g t − 2h   = 55 …(i)
  g 2  g  BC = 40 m
  From C to D,
1 v = u + 2 as
g (t − 2) = h − 55 …(ii)
2
0 = 20 + 2(−20)(CD)
Solving Eqs. (i) and (ii), we get
t = 7s and h = 180 m CD = 10 m
AB + BC + CD = 60 + 40 + 10 = 110 m
13. Planet A
1 u u
s = ut + at 15. : : : 64 : 25
2 2g 2g
1 h ⇒ u :u :: 8:5
⇒ h = 0 + (2 g )t ⇒ t =
2 g 2u 2u
− =6 …(i)
v = u + 2 as g g
⇒ v = 0 + 2(2 g ) h ⇒ v = 2 gh u 8
= …(ii)
u 5
Planet B
1 Solving Eqs. (i) and (ii), we get
s = ut + at
2 u = 80 m/ s and u = 50 m / s
1 h 16.
⇒ h = 0 + (8 g )t ⇒ t =
2 4g
x y
v = u + 2 as A B
µ
⇒ v = 0 + 2(8 g )h ⇒ v = 4 gh t= 0 C V=0
D
v – v = v ⇒ 4 gh – 2 gh = u t=
3 √2
√2–1
⇒ u = 2 gh
h h Between A and C,
t –t =t ⇒ – =t
g 4g Distance travelled = x + 2 y
1 h Displacement = x
⇒ t= 1
2 g x = (x + 2 y )
3
v 2 gh
= = 4g ⇒ x=y …(i)
t 1 h
v = u + 2 as
2 g Between A and B, 0 = u + 2 a(x + y) …(ii)
⇒ u = 4 gt Between A and C, v c = u + 2 ax …(iii)
14. v=0
D Solving Eqs. (i), (ii) and (iii), we get
C u
20 m/s vc = − ⇒ v = u + at
2
B 20 m/s
Between A and C,
−u  3 2 
A 40 m/s = u + a  …(iv)
2  2 − 1

23
 Between A and D, g
2 gh ± ( 2 gh ) – 4 × (–h)
− u = u + at …(v) 2
∴ t=
Solving Eqs. (iv) and (v), we get g
2×
t = 12 s 2
17. Lets say passenger is late by T second for departure. 2 gh ± 2 gh + 2 gh 2 gh
= = (1 + 2 )
For train, u=0 g g
v = u + at 2h h h
= (1 + 2 ) = (2 + 2 ) = 3.4
⇒ v = 0 + aT ⇒ v = aT g g g
For second last boggy, Hence, correct option is (d).
1 2. Let s be distance covered by each car. Let times taken
s = ut + at
2 by the car A and car B be t and t respectively, and
1 velocities at finishing point be v and v respectively
l = (aT )(3) + a (3) …(i) v − v = v and t − t = t
2
v v −v 2a s − 2a s
For (last boggy + second last boggy), = = = aa
t t −t 2s 2s
1 −
s = ut + at a a
2
1 So, v = ( a a )t
2l = aT (5) + a(5) …(ii)
2 u
3. tAB =
Solving Eqs. (i) and (ii), we get g
u
T = 3.50 s tAD = n
g
Previous Years’ Questions tAC =
2u
g
B
1. Let t be the time taken by the packet to reach the
ground. As, the helicopter rises from rest in upward u
direction, its final velocity is A
C
v
t=0 u

h H

t=t

v = 0 + 2 gh ⇒ v = 2 gh D
From second equation of motion, u 2u u
⇒ tCD = tAD − tAC =n − = (n − 2)
1 g 8 g
s = ut + at
2 From C to D,
Here, s = –h 1
s = ut +
at
u or v = 2 gh ⇒ a = g 2
Substituting all these values in above equation, we get  (n − 2)u  1  (n − 2) u 
H =u + g 
1  g  2  g 
−h = 2 gh t + (− g )t
2 (n − 2)u  n − 2
1 H= 1 + 
⇒ gt – 2 gh t – h = 0 g  2 
2
2 gH = n(n − 2)u
This is a quadratic equation in t.

24