1. PART - I
1.
S
A B
2u

u

2a
a
Suppose at point B (displacement S)  particle overtakes particle 
For  particle S = (2u) t +
2
1
a t2 ............. (1)
For  particle
S = u t +
2
1
(2a) t2 ....................(2)
 2ut +
2
1
a t2 = ut +
2
1
(2a) t2
ut =
2
1
a t2 t =
a
u
2
Putting this value in equation (1) we get
S = 2u ×
a
u
2
+
2
1
× a ×
2
a
u
2






=
a
u
4 2
+
a
u
2 2
=
a
u
6 2
2. Let a be the retardation produced by resistive force, ta and td be the time of ascent and time of descent
respectively.
If the particle rises upto a height h
then h =
2
1
(g + a) ta
2 and h =
2
1
(g – a) td
2

d
a
t
t
= a
g
a
g


=
2
10
2
10


=
3
2
Ans.
3
2
3. It is angle between extended line AP representing direction of average velocity and BC, representing
direction of instantaneous velocity , at P.
4
3
2
1
1 2 3 4
x
y
(in mt)
(in mt)
A
B
C
P
(4,1)
45º

tan  = º
53
3
4


2. 180º – (53º + 45º) = 82º
4. Aconstant acceleration is always towards –ve x direction so time to reach from Ato B is same as time taken
from B to A so total time tA  B  A
= 4 sec.
5. t2 = x2 – 1 x2 = 1 + t2
dt
dx
x = t
 1
dt
x
d
x
dt
dx
2
2
2








 2
2
dt
x
d
x
=
2
2
x
t
1
dt
dx
1 













 = 2
2
2
2
x
1
x
t
x


 a = 3
2
2
x
1
dt
x
d

6. u = 48 m/sec a = – 10 m/s2 so, by v = u + at 0 = 48 – 10 t so, t = 4.8 s
this means that the particle comes to rest at t = 4.8 s and turns back covering some distance backwards
for rest of the motion.
for the forward journey distance travelled in last 0.8 second before stopping and returning will be
(s4.8 – s4) where, s4.8 and s4 are distances travelled in 4.8 seconds and 4 seconds respectively.
s4.8 = 48 × 4.8 +
2
1
× –10 × 4.82 = 48 × 2.4
s4 = 48 × 4 +
2
1
× – 10 × 42 = 16 × 7
(s4.8 – s4) = (48 × 2.4 ) – (16 × 7)
Distance travelled 0.2 s during backward journey = s0.2 =
2
1
× 10 × 0.22 = 0.2 m
So, total distance travelled = (48 × 2.4 ) – (16 × 7) + 0.2 =
5
17
m. Ans.
7. Taking upward direction as positive
initial velocity can be obtained by II equ. of motion i.e. s = ut + 1/2 at2
considering motion from C to A
14 = u × 0.8 –
2
1
× 10 × 0.82
so, u =
2
43
m/s
(ii) Let velocity magnitude at point A = v
so, v = u – gt
v =
2
43
– 10 × 0.8 =
2
27
m/s
(iii) Hence time taken from A to B i.e. till same level = g
v
2
= 2.7 s
Hence the time instant at which the particle comes to same level
= 0.8 + 2.7 = 3.5 s Ans.
8. Distance travelled by each particle in last second of motion i.e. downwards is equal to the distance
travelled by it in first second of its motion i.e. upwards.
So, x1 = 10 – ½ × 10 × 12 = 5m
x2 = 20 – ½ × 10 × 12 = 15m
x3 = 30 – ½ × 10 × 12 = 25m
so, x1 : x2 : x3 = 1 : 3 : 5

3. 9. t1
=
1
g
u
2
and t2
=
2
g
u
2
where, u is initial velocity of each ball and t1
, t2
are total times of flight.
from here, g1
t1
= g2
t2
= 2u
10. When particle is thrown vertically upwards, its acceleration is constant downwards.
so, velocity initially decreases linearly to become zero at highest point and then again speed increases
linearly. So, graph is (C).
11. |Average velocity| =
time
|
nt
displaceme
|
=
1
2
time
AB
 = 2 m/s
12. (i) For uniformly accelerated/deaccelerated motion
v2
= u2
± 2gh
i.e. v-h graph will be a parabola (because equation is quadratic).
(ii) Initially velocity is downwards (-ve) and then after collision it reverses its direction with lesser
magnitude. i.e. velocity is upwards (+ve). Graph (A) satisfies both these conditions.
Therefore, correct answer is (A)
Note that time t = 0 corresponds to the point on the graph where h = d
Next time collision takes place at 3.
13. y = u (t – 2) + a(t – 2)2
Velocity of particle at time t
dt
dy
= u + 2a (t – 2)
Velocity at t = 0
dt
dy
= u – 4a
acceleration of particle
2
2
dt
y
d
= 2a  yt = 2
= 0
So correct answer is (C) and (D).
14. x = 2t – 3  x = (2t – 3)2
Velocity v =
dt
dx
= 2(2t – 3) × 2 = 4 (2t – 3)
if v = 0  t = 3/2
acceleration  a =
dt
dv
= 4 × 2 = 8
so correct answer is (A) and (B) .

4. 15. (A) The upper part of graph shows that acceleration is positive and becomes zero but just after this
instant, the displacement cannot become negative suddenly. So, (A) is wrong.
(B) Displacement is not positive when velocity is negative. So, (B) is wrong.
(C) Velocity cannot increase if acceleration is negative. So, (C) is wrong.
(D) If velocity is negative then displacement will also be negative. So, (D) is correct.
16.* s  t2
 s = ct2
where c = constant
(i) v =
dt
ds
= 2 ct
 v  t
(ii) a =
dt
dv
= 2c
so, a = constant.
PART - II
1.
x = t3
– 3t2
– 10
 =
dt
dx
= 3t2
– 6t
v = 0 ; gives
t = 0 & t = 2 sec.
Velocity will become zero at t = 2 sec., so particle will change direction after t = 2 sec.
At t = 0
x(0 sec)
= – 10
At t = 2 sec.
x(2 sec)
= 23
– 3(2)2
– 10
= 8 – 12 – 10 = – 14
At t = 4 sec.
x(4 sec)
= 43
– 3(4)2
– 10
= 64 – 48 – 10 = 6
Distance travelled = x1
+ x2
= |– 14 – (–10)| + |6 – (–14)|
= 4 + 20 = 24
Ans. Distance Travelled = 24 units.
2. L =
a
2
u
v 2
2

................ (1)
2
L
=
a
2
u
´
v 2
2

................(2)
dividing (1) / (2)
L
u
point
v
v´
L2
2 = 2
2
2
2
u
´
v
u
v


2v´2
– 2u2
= v2
– u2
v´ =
2
v
u 2
2


5. 3. 75 km/s 120 km/s
rocket
(i) aw
=
6
75
120 
=
6
45
km/sec2
= 7.5 km/sec.
(ii)
10
6
75
120
t 
distance travelled in 10 sec = Area under curve
d = 75 × 6 + (120 – 75) × 6 ×
2
1
+ 120 × 4
= 450 + 45 × 3 + 480
= 450 + 135 + 480 = 1065 K
4. here, vmax
= v is the maximum velocity which can
be achieved for the given path
from Ist
part, tan 1
= 10 =
1
t
v
 t1
=
10
v
from IInd
part, tan 2
= 5 =
2
t
v
 t2
=
5
v
now, area under the graph is equal to total displacement
so,   1000
t
t
v
2
1
2
1 

1000
5
v
10
v
v
2
1








so, vmax
= v =
3
2
100
m/s = 81.6 m/s (approx)
(i) The maximum speed is 70 m/s which is lesser than maximum possible speed v,
hence the train will move with uniform speed for some time on the path.
The motion of train will be as shown
Let Ist
part of path has length s1
then, by v2
= u2
+ 2as, we get
702
= 02
+ 2 × 10 × s1
, so s1
= 245 m
Similarly by IIIrd
equation of motion
02
= 702
– 2 × 5 × s3
, so s3
= 490 m
Hence, s2
= 1000 – (490 + 245) = 265 m
for part 1 of the path, time taken = t1
from v = u + at, we get
70 = 0 + 10 t1
so, t1
= 7 seconds
for part 2 of the path, time taken = t2
=
70
s2
=
70
265
=
14
53
seconds
for 3rd part of the path, 0 = 70 – 5 × t3
so, t3
= 14 seconds.
Total time taken = t1
+ t2
+ t3
= 7 +
14
53
+ 14 =
14
347
seconds

6. (ii) Here, the maximum speed given i.e. 85 m/s is more than the maximum possible speed
so, for minimum time, the maximum speed attained will be vmax
= 81.6 m/s.
graph of motion will be
minimum time taken = tmin
= t1
+ t2
=
5
v
10
v max
max 
where, vmax
= v =
3
2
100
m/s = 81.6 m/s
so, tmin
=
3
2
100






10
3
=
3
2
30 s
5. Velocity of car on highway = v
Velocity of car on field =

v
Let CD = x and AD = b
T = tAC
+ tCB
=
v
x
b 
+
)
/
v
(
x2
2




dx
dT
=0  –
v
1
+
v









 2
2
x
2
x
2

= 0  x =
1
2



6. (a) V = x

dt
dx
= x
  
x
0
x
dx
= 
t
0
dt
 x =
4
t2
2


dt
dx
= V =
2
t
2

Also a =
dt
dv
=
2
2

.
(b) Let t0
be the time taken to cover the first s metre
 s =
4
t
2
0
2

 t0
=

s
2
 < v > =


0
0
t
0
t
0
dt
dt
v
 < v > =
0
t
0
2
t
dt
2
t
0


= 0
2
t
2
1
.
2

=

 s
2
.
4
2
=
2
s

.
Aliter. : v = at , v2
= 2
t
= t
2
2

= x
2
2
2

 a =
2
2

and < v > =
a
/
S
2
S
t
S
 =
S
2
a
S
= S
2
2
S
2
S 2
2



.

7. 7.
B
A
C
54 km/hr
54 km/hr
36 km/hr
Time taken by C to overtake A is =
h
/
km
)
54
36
(
km
1

=
90
1
hr =
3
2
× 60 sec. = 40 sec.
Let the minimum acceleration of B is a to overtake A before C then
S = ut +
2
1
at2
1 km = (54 – 36) ×
90
1
+
2
90
90
1


× a 
2
90
90
1


× a = 1 –
90
18
=
90
72
a =
90
72
× 90 × 90 × 2 km/h2

60
60
60
60
1000
2
90
70






m/sec2
= 1 m/sec2
.
8. At t = 0, raft (a float of timber) and motor boat are at point A. The velocity of raft is equal to velocity of
stream.
At t =  = 60 min, the motor boat is at point P and raft is at point B.
 The time taken by raft to reach from A to B = the time taken by motor boat to reach at P from A.
At t =  + t0
, both meet at point C,
So, the time taken by raft to reach at C from B is equal to the time taken by the motor boat to reach at
C from P in upstream motion. This time is equal to t0
.
Let
vA = actual velocity of motor boat,
vB = actual velocity of stream = velocity of raft
 During down stream,
vc = relative velocity of motor with respect to stream.
 v0 = vA – vB
 vA
= v0
+ vB
  =
A
v
AP
=
B
0 v
v
AP

But AB = distance travelled by raft in time  = vB
During upstream,
v0 = vA + vB
 vA = v0 – vB
 PC = distance travelled by motor boat in upstream in time t0 = (v0 – vB) t0
BC = distance travelled by raft in time t0 = vBt0
According to fig.
 AP – PC = AC = 
or (v0
+ vB
)  – (v0
– vB
) t0
= 
or v0
t +vB
t – v0
t0
+ vB
t0
=  ..........(i)
Also,  AB + BC =  or vB
t + vB
t0
= 
or vB
=
0
t



..........(ii)
From equation (i) and (ii) we get
v0
t +vB
t – v0
t0
+ vB
t0
= 
or 










 0
0
0
0
0
0 t
t
t
v
–
t
v or v0
2
+  – v0
t0
2
+ t0
=  ( + t0
)
or v0
2
– v0
t0
2
=  ( + t0
) –  – t0
 or  = t0
From (i) we have
v0
 +vB
 – v0
t0
+ vB
t0
= 
putting  = t0 we get, vB =

2

.

8. 9. (a) From graph, obviously engine stopped at its highest velocity i.e., 190 ft/s. Ans.
(b) The engine burned upto the instant it reached to its maximum velocity. Hence it burned for 2s. Ans.
(c) The rocket reached its highest point for the time upto which the velocity is positive. Hence, from graph,
rocket reached its highest point in 8 s.
ymax

dt
dy
= 0  velocity in y direction = vy
= 0 m/s.
(d) When the parachute opened up, the velocity of rocket starts increasing. Hence, at t = 10.85 (from
graph), parachute was opened up. At that moment the velocity of the rocket falling down was 90 ft/s.
(e) The rocket startsfalling when its velocity becomesnegative. From the graph hence timetaken by rocket to
fall before the parachute opened will be (10.8 – 8) s = 2.8 s.
(f) Rocket's acceleration was greatest when the slope of tangent in V – t graph was maximum. As t = 2 sec,
the tangent is vertical i.e, slope is infinity hence the rocket's acceleration was greatest at t = 2 s.
(g) The acceleration is constant when V – t graph is linear. Hence, the acceleration was constant between
2 and 10.8 s. Its value is given by slope = –
2
–
8
190
= – 32 ft/s2
(nearest to integer) Ans.
10. F(x) = 2
x
2
k

k and x2
both are positive hence F(x) is always negative (whether x is positive negative .)
x = 0 x = 0.5m
u = v
x = 1.0m
At t = 0
v = 0
F(x) A
B
Applying work energy theorem between points A and B :
Change in kinetic energy between A and B = work done by the force between A and B

2
1
mv2
= 


5
.
0
x
m
0
.
1
x
)
dx
.(
F =  




 
5
.
0
0
.
1 2
)
dx
(
x
2
k
=
2
k


5
.
0
0
.
1 2
x
dx
=
2
k
5
.
0
0
.
1
x
1






= 





2
k







0
.
1
1
5
.
0
1
=
2
k
 v =
m
k

Substituting the values
v = 
kg
10
m
N
10
2
2
2


=  1 m /s
Therefore, velocity of particle at x = 1.0 m is v = 1.0 m /s
Negative sign indicates that velocity is in negative x-direction.
(b) Applying work energy that theorem between any intermediate value x = x, we get
2
1
mv2
= 

x
0
.
1 2
x
2
dx
k
=
x
0
.
1
x
1
2
k






= 





1
x
1
2
k
 v2
= 





1
x
1
m
k
 v = 1
x
1
 =
x
1
x 
but v = – 





dt
dx
=
x
1
x 
 dx
x
1
x

= – dx
or  
25
.
0
1
dx
x
1
x
= – 
t
d
dx
Solving this, we get t = 1.48s