PHYSICS

Iff SUPPLEMENT
Class : XI (P)
Target IIT JEE 2014

CALCULUS
EXERCISE
Q.1 A particle is moving along x-axis such that its position 'x' varies with time (t). Find the velocity
(v) and acceleration (a) of particle if its position w.r.t time is given by :
1 1
(i) x = t2 (ii) x= (iii) x (iv) x  t3 2
t t
(v) x = t 5/2 (vi) x  2t 2 (vii) x = 5000 (viii) x = t2 + t + 5
2
(ix) x = 4t 3 + 3 (x) x = 3t +
t

Q.2 Momentum of a body moving in a straight line is p = ( t2 + 2t + 1) kg m/s. Find the force acting on a body
at t = 2sec
(A) 6 N (B) 8 N (C) 4 N (D) 2 N

Q.3 A particle moves along a straight line such that at time t its position from a fixed point O on the line is
x = 3t2 ñ 2. The velocity of the particle when t = 2 is:
(A) 8 msñ1 (B) 4 msñ1 (C) 12 msñ1 (D) 0

Q.4 A body moves with velocity v = n x m/s where x is its position. The net force acting on body is zero at:
(A) 0 m (B) x = e2 m (C) x = e m (D) x = 1 m

Application of chain rule in physics

dy
Q.5 Find
dt
(i) y = sin (t + 2) (ii) y = sin (t + ) where  and  are constant.
d d
(iii) y = cos 2, where  (iv) y = sin (2 + 3) where 
dt dt
dx
(v) y = 2x2 + 3x + 4 where = Vx (vi) y = (2t + 4)3
dt
(vii) y = sin2 t (viii) y = cos2 t

dv
Q.6 Given that a  v then find 'a' as a function of 'x' if
dx
(i) v = kx + c, where 'k' and 'c' are constant
(ii) v = k x , where 'k' is constant
(iii) v = A sin kx, where 'A' and 'k' are contant
(iv) v  1 x2
1
(v) v
1 x2
Calculus [1]
ITS
Product rule and Division rule
dy
Q.7 Find
dx
x
(i) y = x sin x (ii) y = ex cos x (iii) y= (iv) y = x sin x
1 x

d
Q.8 Given that    find '' if
d
(i)  = 22 +  + 1 (ii)  = 4 sin 2 (iii)  = 2 + cos 

Q.9 The velocity of a particle moving on the x-axis is given by v = x2 + x where v is in m/s and x is in m. Find
its acceleration in m/s2 when passing through the point x = 2m
(A) 0 (B) 5 (C) 11 (D) 30

Derivative of vectors :

  
Q.10 The position of a particle moving in an xy plane is given by r  2 t 3  5 t ài  6  7 t 4 àj . Here r is 
 
in meters and t is in seconds. At t = 2s calculate (a) r , (b) v , and (c) a

Q.11 Coordinates of a moving particle are given by x = ct2 and y = bt2. The speed of the particle is given by
(A) 2t (c + b) (B) 2 t c2  b 2 (C) t c2  b 2 (D) 2 t c2  b 2

Q.12 A particle moves in the xy plane and at time t is at the point whose coordinates are (t2, t3  2t). Then at
what instant of time will its velocity and acceleration vectors be perpendicular to each other?
(A) 1/3sec (B) 2/3 sec (C) 3/2 sec (D) never


If r  2 tài  3t àj then find
2
Q.13
 
 , where  d r   dv
(i) v v (ii) a , where a
dt dt


Q.14 If r  sin 2 tài  cos 2 tàj , then find
 
(i) v (ii) a

Q.15 A particle is moving according to the position time(x-t) graph as shown. Find velocity of particle at t = 1
sec., 3 sec., 5 sec.
x(m)

6
0 t(sec)
2 4

ñ10

Calculus [2]
The
Maxima minima :
Q.16 In the interval 0  t  1 sec, charge flowing through a conductor is given by q = t2 ñ 6t + 5. In the given
interval, the maximum charge flows in the conductor at time :
(A) 3 sec (B) 1 sec (C) 0 sec (D) none of these

Q.17 The position of a particle moving along the yaxis is given as y=3t2t3 where y is in metres and t is in sec.
The time when the particle attains maximum positive y position will be
(A) 1.5 sec (B) 4 sec (C) 2 sec (D) 3 sec

Q.18 Evaluate the following indefinite integrals.

 1 
(i)  dx (ii)  xdx (iii)   x 2 dx (iv) x
5/ 2
dx

3
 3 sin x  2dx 5 x
5/3
x
2

3
(v) x 2 dx (vi) dx (vii) (viii) dx

 x 3 2 dx
(ix)  (x
2
 2 x  1) dx (x)  2
  x  dx
 (xi)  4x
 

 5 
(xii)   2  x 2  dx

Q.19 Evaluate the following definite integrals.

 2x 
2 4 4 2
1
 x dx  x dx  x dx
2 3/ 2 2
(i) (ii) (iii) (iv)  3x  1 dx
1 0 2 0

/ 2 1
1
 cos 2x  sin 2x dx
2
(v) (vi)  4  2x dx (vii)
 2  3x  dx
3
0 0
1

Integration (application in physics)

Q.20 The initial velocity of a particle is u and the acceleration is given by (kt), where k is a positive constant.
The distance travelled in time t is :
(A) s = ut2 + kt2 (B) s = ut + (kt3/6)
(C) s = ut + (kt3/2) (D) s = (ut2/2) + (kt3/6)

Q.21 Force acting on a body of mass 1 kg is related to its position x as F = (x3 ñ 3x) N. It is at rest at x = 1.
Its velocity at x = 3 can be :
(A) 4 m/s (B) 3 m/s (C) 2 m/s (D) 5 m/s

Equation of trajectory :
Q.22 A particle moves in the xy plane with velocity vx = 8t2 and vy = 2. If it passes through the point
x = 14 and y = 4 at t = 2 sec. The equation of the path is
(A) x = y2y+2 (B) x = y+2
(C) x = y +2
2
(D) x = y2+y+2

Calculus [3]
2
Area under v - t graph , i - t graph
Q.23 Figure shows a graph of velocity versus time for a particle in one dimensional motion. Which of the
following statements is correct.
v

t
t

(A) The shaded area represents distance traveled by particle in time interval t.
(B) The shaded area represents the acceleration during time interval t.
(C) The acceleration is constant during time internal t.
(D) During time interval t particle first moves away from initial position and then returns back.

Q.24 A particle is moving according to the position time(x-t) graph as shown. Find
x(m)

6
0 t(sec)
2 4

ñ10

(a) Time t = ? when particle returns to its initial position x = 0.

(b) Position ëxí of particle at t = 5 sec.

Integration of a vector.

  dv  
Q.25 a  2 t ài  3àj and a  , find v at t = 2 sec. Given that v = 0, at t = 0.
dt
B

 F B dr , where d r  dx ài  dy àj .
  
Q.26 The work done by a force in moving a particle from A to B is given by
A

It is given that F  xài  yàj ; A(1, 2) & B(ñ1, 1). Evaluate the work done.

Application of chain rule in Physics

Q.27 Due to heating a metal sphere expands such that its radius is increasing at rate 0.1 mm/sec.
Find the rate of change of volume when radius of sphere is 1m.

Calculus [4]
2
 ANSWER KEY
1 2 t 3 / 2 3t 5 / 2 3 t1 / 2 3t 1 / 2
Q.1 (i) 2t, 2 (ii)  , (iii)  , (iv) ,
t2 t3 2 4 2 4

5t 3 / 2 15t1 / 2
(v) , (vi) 2 2 t , 2 2 (vii) 0, 0 (viii) 2t + 1, 2
2 4

2 4
(ix) 12t 2, 24t (x) 3  ,
t 2 t3

Q.2 A Q.3 C Q.4 D

Q.5 (i) cos (t + 2) (ii)  cos (t + ) (iii) ñ2 sin 2 (iv) 2 cos (2 + 3)
(v) Vx (4x + 3) (vi) 6(2t + 4)2 (vii) 2 sin t cos t (viii) ñ2 cos t sin t

k2
Q.6 (i) k(kx + c) (ii) (iii) A2 k sin kx cos kx (iv) x (v) ñx(1 + x2)ñ2
2

1  sin x 
Q.7 (i) (x cos x + sin x) (ii) (ex cos x ñ ex sin x) (iii) 2 (iv)   x cos x 
(1  x ) 2 x 

Q.8 (i) (4 + 1) (22 +  + 1) (ii) (16 sin 4) (iii) (ñ sin ) (2 + cos )

Q.9 D Q.10 (a) 6ài  106àj (b) 19ài  224àj (c) 24ài  336àj ]

Q.11 D Q.12 B Q.13 (i) 2ài  6 t àj (ii) 6 àj

Q.14 (i) 2 cos 2 t ài  2 sin 2 t àj (ii)  4 sin 2 t ài  4 cos 2 t àj

Q.15 2.5 m/s, zero, ñ7.5 m/s Q.16 C Q.17 C

x2 1 2 7/2
Q.18 (i) x + c (ii) c (iii)  c (iv) x c
2 x 7

3 5/3 x3 9 8/3
(v) x c (vi) +c (vii) (2x ñ 3 cos x + c) (viii) x +c
5 3 40

Calculus [5]
If
 x3 1 x3 1  5 
(xi)  x2  x  c (x)  x  2  c (xi) n x (xii)  2 x   c 
3 4 3 4  x 

7 64 40
Q.19 (i) (ii) (iii) ln2 = 0.693 (iv)
3 5 3
n 2
(v) 1 (vi) (vii) 289.25
2

Q.20 B Q.21 A Q.22 A Q.23 A

14 
Q.24 (a) t = s (b) x = ñ2.5 m Q.25 v  4ài  6àj Q.26 ñ 3/2
3

Q.27 1.256 L 10 ñ3 m 3 /s

Calculus [6]
It