Hints & Solutions

1. Rectilinear Motion
1.1 (B) Displacement vector is 10î  10ˆj  10k̂

 Magnitude = 10 2  10 2  10 2 = 10 3 Ans.

1 1
1.2 (C) x1  a (10) 2  50a  x 2  (10a ) (10)  a (10) 2  150a
2 2
1
x 3  (20a ) (10)  a (10) 2  250a  x1 : x 2 : x 3 : : 1 : 3 : 5 Ans
2
( u  v)
1.3 (B) s  t
2
(v  vB )
3 T  0.5
2
vT + vB = 12 m/s
Also vB = vT + (9.8) (0.5) ...............(2)
vB  vT = 4.9 m/s

1.4 (D) Initial distance between trains is 300m. Displacement of 1st train is calculated by area under V-t.
1
curve of train 1  x 10 x 40  200 m
2
1
Displacement of train 2  x 8 x ( 20)   80 m .
2
Which means it moves towards left.
 Distance between the two is 20 m.

T 3T
1.5 (D) At t  and t  , the stone is at same height,
2 2
Hence average velocity in this time interval is zero.
Change in velocity in same time interval is same for a particle moving with constant acceleration.
T 3
Let H be maximum height attained by stone, then distance travelled from t = 0 to t  is H and
4 4
T 3T H
from t  to t  distance travelled is .
4 4 2
T T 3T H
From t  to t = T sec distance travelled is H and from t  to t  distance vravelled is
2 2 4 4
dv
1.6 (C) The retardation is given by   av 2
dt
v t
dv 1 1
integrating between proper limits   v 2  a dt
  or  at 
u 0 v 4
dt 1 u dt
  at   dx 
dx u 1  aut
 s t
u dt 1
integrating between proper limits  dx    S ln (1  aut )
0 0
1  aut a
1.7 (B) Let a be the retardation produced by resistive force, ta and td be the time of ascent and descent
respectively. It the particle rises upto a height h
1 2 1 2
then h  (g  a ) t a and h  (g  a ) t d
2 2
ta ga 10  2 2 2
    Ans.
td ga 10  2 3 3
1.8 (A) The linear relationship betwen V and x is
V = - mx + C where m and C are positive constants.
 Acceleration
dV
av   m (  mx  C)
dx
 a  m 2 x  mC
Hence the graph relating a to x is.
1.9 (A) xA = xB
1 2
10.5 + 10t = at a = tan45° = 1
2
20  400  84
t2 - 20t - 21 = 0 t
2
t = 21 sec.
AC BC u1 AC 2 5
1.10 (D) u  u or  
1 2 u 2 BC 4
1.11 (D) A will be ahead of B when xA > xB
1
40 ( t  10)  (0) t  ( 2) t 2
2
as A is 10 sec. late than B.
 t 2  40 t  400  0
 ( t  20) 2  0
Which is not possible. So A will never be ahead at B.
1.12 (B) From given graphs : ax is +ve & ay is -ve as vx is increasing in +ve direction and vy in -ve direction.
1.13 (A) Distance travelled from time ‘t-1’ sec to ‘t’ sec is
a
S  u  (2 t  1) .................................. (1)
2
from given condition S = t .......................(2)
a a
(1) & (2)  t  u  ( 2 t  1)  u   t (1  a ) .
2 2
Since u and a are arbitrary constants, and they must be constant for every time.
 coefficient of t must be equal to zero.
1
 1  a  0  a  1 for a  1, u  unit
2
1
Initial speed is unit. Ans.
2