PHYSICS

Daily Practice Problems

Target IIT JEE-2021
GLADIATOR

BATCH : G-1 TIME : 35 MIN. DPP-1

Q.1 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 – 2. The velocity of the particle when t = 2 is:
(A) 8 ms–1 (B) 4 ms–1 (C) 12 ms–1 (D) 0

Q.2 A particle moves in a circle of radius 18 m. Its linear speed is given by v = kt. [where k is positive
constant with value k = 2m/s2] where t is in second and v in m/s. The acceleration of the particle at t = 3s
is
(A) 2 m/s2 (B) 4 m/s2 (C) 4.2 m/s2 (D) 2.8 m/s2

Q.3 The co-ordinates of a moving particle at a time t, are given by, x = 5 sin 10 t, y = 5 cos 10 t . The
speed of the particle is :
(A) 25 (B) 50 (C) 10 (D) None

Q.4 The position of a particle moving along a straight line is given by x = 3sint where t is in sec.& x is in m.
1 3
Find the distance travelled by the particle from t = to t = sec.
4 4
FG
1 IJ   
(A) 0 m (B) 6 1 
H2 K (C) 4.2 m (D) 6
 2 
 1

Q.5 A man moves along the x-axis such that its velocity is v =1/x. If he is initially at x = 2 m, find the time when
he reaches x = 4 m
(A) 6 sec (B) 4 sec (C) 3 sec (D) he can't reach x = 4 m

Q.6 The approximate value of x where x = sin 2° cos 2°, is :

 
(A) (B) 2 (C) 1 (D)
90 45

Q.7 An object is moving in a straight line. Its position versus time graph is shown below. The detector is at the
origin (0 meters). Between t = 6 and t = 8 seconds, which statement describes the motion shown in the
graph.
Position (meters)
3

2
1

0 2 4 6 8
Time (seconds)

(A) Motion away from detector with constant velocity

(B) Motion towards detector with constant velocity
(C) Motion towards detector with decreasing velocity
(D) Motion away from detector with increasing velocity
1
Q.8 A horizontally moving rocket has velocity given by v = 10 – 2t. Its acceleration is given by
(A) 10 (B) – 2 (C) – 4 (D) + 4

Q.9 A particle starts from rest on a straight path. Its acceleration is linearly varying with time, such that
velocity of the particle at t = 2 sec and t = 4 sec is 6 ms–1 and 20 ms–1 respectively. Find acceleration of
the particle at t = 1 sec.
(A) 1.5 ms–2 (B) 2 ms–2 (C) 3 ms–2 (D) 2.5 ms–2

Q.10 A particle starts from rest on a straight path. Its acceleration is linearly varying with time, such that
velocity of the particle at t = 2 sec and t = 4 sec is 6 ms–1 and 20 ms–1 respectively. Find acceleration of
the particle at t = 1 sec.
(A) 1.5 ms–2 (B) 2 ms–2 (C) 3 ms–2 (D) 2.5 ms–2

Q.11 If y = t3/3 – t2 – 3t +1 then graph of y2 versus t will be, where y2 = d2y/dt2

(A) (B) (C) (D)

t2
Q.12 A particle moves along the curve y2 = 2x where x  & y > 0. What is the acceleration of the particle
2
at t = 2s?
(A) î (B) ˆj (C) î  ˆj (D) 2 î

Q.13 The relation between the time t and position x for a particle moving on x-axis is given by
t = px2 + qx (p & q are constant)
The relation between velocity and magnitude of acceleration -
(A) a  v3 (B) a  v2 (C) a  v4 (D) a  v

Q.14 The displacement of a particle after time t is given by x  ( k / b 2 )(1  e  bt ) , where b is a constant. What
is the acceleration of the particle ?
k bt  k bt
(A) ke bt (B)  kebt (C) 2 e (D) 2 e
b b
Q.15 The displacement of a body is given by 2s = gt2 where g is a constant. The velocity of the body at any
time t is:
(A) gt (B) gt/2 (C) gt2/2 (D) gt3/6

Q.16 The angle  through which a pulley turns with time t is specified by the function
d
 = t2 + 3t  5. Find the angular velocity = at t = 5 sec .
dt

2
 ANSWER AND SOLUTION
Q.1 (C)

  dX
Sol. X = 3t2 – 2, v = 6t = 6 × 2 = 12 m/s
dt

Q.2 (D)
Sol. at = k = 2
v2 (2  3) 2
aR = = =2
R 18
aT = 2
2 2  2 2 = 2 2 = 2.8 m/s

Q.3 (B)

Sol. Position vector = r = 5 sin 10 t î  5 cos 10 t ˆj

 dr
v= = 50 cos10 t î  50 sin 10 t ˆj
dt

| v |= (50) 2 cos2 10t  (50) 2 sin 2 10t = (50) 2{cos 2 10t  sin 2 10t} = 50

Q.4 (B)
Sol. x = 3xint
at t = 1/2, v = 0
Total distance
x = |x1| + |x2|

1 1  1 
x1 = [3sin × – 3sin × ] = 3 1 
2 4  2 
t = 1/4 to 1/2
 1 
x2 = [3sin(3/4) – 3sin(/2)] = 3  1
 2 

 1   1   1 
x = 31   + 3 1 = 6 1  
 2  2   2

Q.5 (A)
Sol.  x dx =  dt
x2 4
– =t
2 2
x2 = 2t + 4
16 – 4 = 2t  t = 6 sec

3
Q.6 (A)

Sol. 2° = rad
90
   
x = sin   cos  
 90   90 
  
sin  cos  1
90 90 90

x=
90

Q.7 (B)
Sol. t = 6 to 8 sec, x is decreasing and slope is constant

Q.8 (B)
dv
Sol. v = 10 – 2t a= =–2
dt

Q.9 (C)
Sol. Let t + 
t 2
v=  t  
2
t 2
at t = 0 v = 0   = 0 v=  t
2
at t = 2 v=6
6 = 2 + 2
+=3 .... (1)
at t=4 v = 20
20 = 8 + 4
2 +  = 5 ... (ii)
from (i) & (ii)  = 2,  = 1
 at t = 1 a = 2t + 1 = 3

Q.10 (C)
Sol. Let t + 
t 2
v=  t  
2
t 2
at t = 0 v = 0   = 0 v=  t
2
at t = 2 v=6
6 = 2 + 2
+=3 .... (1)
at t=4 v = 20
20 = 8 + 4
2 +  = 5 ... (ii)
from (i) & (ii)
 = 2,  = 1
 at t = 1 a = 2t + 1 = 3
4
Q.11 (A)
dy
Sol. = t2 – 2t – 3
dt
d2y
= 2t – 2
dt 2

Q.12 (A)
t2
Sol. x
2
2t
vx   ax = î
2
t2
y2 = 2   y = t  ay = 0
2
  
a  a x  a y  î

Q.13 (A)
dt dx 1
Sol. = 2px + q  v= 
dx dt 2px  q
3
– 2p  1 
a= a    a  v3
(2px  q)3  2px  q 

Q.14 (B)
dx k
Sol. v  2  e bt  b
dt b
dv
 ke bt
dt

Q.15 (A)
Sol. 2S = gt2

dS
2 = g 2t  v = gt
dt
Q.16 13 rad/s
Sol.  = t2 + 3t  5
d
= = 2t + 3
dt
at t = 5 sec
 = 13 rad/s