IIT JEE 2014 Physics Assignment Kinematics Solution

Class Assignment (Batch 2014) Solution Topics: Kinematics

PHYSICS 1.

A person travels along the straight road for half the distance with velocity v1 and the remaining half distance with velocity v2. Then average velocity is given by : (B) v22 / v12

(A) v1v2

(C)

v1  v2  / 2

(D) 2v1v2 / v1  v2 

Sol. (D)

vav 

2.

 2v v  total distance d   1 2  total time  d d   v1  v2      2v1 2v2 

The velocity acquired by a body moving with uniform acceleration is 30 ms–1 in 2 seconds and 60 m s–1 in 4 seconds. The initial velocity is : (A) zero (B) 2 ms– 1 (C) 4 m s–2 (D) 10 ms–2

Sol. (A) v = u + at 30 = u + a × 2 60 = u + a ×4 3.

... (i) ... (ii)

by solving, u = 0. When the speed of a car is v, the minimum distance over which it can be stopped is s. If the speed becomes nu, what will be the minimum distance over which it can be stopped during same time? (A) s/n

(B) ns

(C) s/n2

(D) n2s

Sol. (D)

v 2  u 2  2as S 4.

2  0  u  2as

u2 2a

 s  u2

A stone is dropped from a certain height which can reach the ground in 5 sec. It is stopped after three seconds of its fall and then is again released. The total time taken by the stone to reach the ground will be (A) 6 s (B) 6.5 s (C) 7 s (D) 7.5 s

Sol. (C)

1 h  ut  gt 2 2

1  h5  0  5   10  25  125. 2

Distance travelled in 3 sec.

1 h3  10  (3)2  45m 2 Remaining height = 125 – 45 = 80 m Time to cover this height 

2h 2  80   4 sec g 10

Total time = 3 + 4 = 7 sec.

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Class Assignment (Batch 2014) Solution Topics: Kinematics

PHYSICS 5.

A body falls freely from rest. It covers as much distance in the last second of its motion as covered in the first three seconds. The body has fallen for a time of : (A) 3 s

(B) 5 s

(C) 7s

(D) 9 s

Sol. (B) Let it takes t second to strike ground distance covered in first 3 sec.

1 h  0  g (3)2  45 2 Distance travelled in last second

1 h '  0  g (2t  1) 2 1 45  10(2t  1) 2 t = 5 sec. 6.

A body is projected with a velocity u. If passes through a certain point above the ground after t 1 sec. The time after which the body passes through the same point during the return journey is : u  (A)  – t12   g 

u2  (C) 3 – t1   g 

u  (B) 2  – t1   g 

 u2  (D) 3 2 – t1   g 

B

Sol. (B) Total time from A to B to C

T

2u g

t1 D E

 2u   2t1  g 

Time from D – B – E  T  2t1   7.

u A C

A stone is thrown vertically upward. On its way up it passes point A with speed of v, and point B 3m higher than A, with speed

v . 2

The maximum height reached by stone above point B is: (A) 1 m

(B) 2 m

(C) 3 m

(D) 5 m

Sol. (A)

v 2  u 2  2 gh

3 2  60  v 4 8.

 (v / 2) 2  v 2  2  10  3  v 2  80 .



v  h 2 2g

2

 1m

A particle starts moving in a straight line with a constant acceleration. At a time t1 seconds after the beginning of motion, the acceleration changes sign, remaining the same in magnitude. Determine the time from the beginning of motion, till it returns to the starting point :





(A) t1 2  2 sec www.vidyamandir.com



(B) t1 1 



2 sec

(C) t1 2 sec

(D) 2 2 t1 sec

2

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Class Assignment (Batch 2014) Solution Topics: Kinematics

PHYSICS Sol. (A)

a

B

a

A

v

C Let acceleration changes sign at B & particle comes to rest at C.

t AB  t BC  t1 , AB  BC

 AC  AB  BC  2 AB

1 2 2  AC  2  at1  at1 2

Let it takes t time from C to A.

1 CA  0  at 2 2 1 2 2  at1  at 2



Total time  2t1  2t1 9.

 t  2 t1



A body freely falling from rest has a velocity v after it falls through distance h. The distance it has to fall down further for its velocity to become double is : (A) h (B) 2h (C) 3h (D) 4h

Sol. (C)

v 2  2 gh



 2v  2  2 g  h  h ' 

 h '  3h

10. Two identical balls are shot upward one after another at an interval of 2s along the same vertical line with same initial velocity of 40 m/s. The height at which the balls collide is : (A) 50 m Sol. (B)

(B) 75 m

(C) 100 m

(D) 125 m

Let height be h

1 1 2 h  ut  gt 2  u  t  2   g  t  2  2 2 *11.A car is moving with uniform acceleration along a straight line between two stops X and Y. Its speed at X and Y are 2 m/s and 14 m/s. Then : (A) its speed at mid point of XY is 10 m/s (B) its speed at a point A such that XA : AY = 1 : 3 is 5 m/s (C) the time to go from X to the mid point of XY is double of that to go from mid point to Y (D) the distance travelled in first half of the total time is half of the distance travelled in the second half of the time Sol. (AC) 2 m/s x

14 m/s y

Acceleration is a

(14) 2  (2) 2  2a ( xy ) www.vidyamandir.com

96  xy     a  3

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Class Assignment (Batch 2014) Solution Topics: Kinematics

PHYSICS

Speed at midpoing

 xy  v 2  (2) 2  2a    2

 v = 10 m/s

Time from x to mid-point

8  t1  a

10 = 2 + at1 Time from mid point to y-

14  10  at2 t2  4 / a 12. A particle is thrown upwards from ground. It experiences a constant resistance force which can produce a retardation of 2 m/s2. The ratio of time of ascent to the time of descent is (g = 10 m/s2) (A) 1 : 1

(B)

2 3

(C)

2 3

(D)

3 2

Sol. (B)

t

2h a

tas  tds

ade ( g  a) 8  2     aas ( g  a) 12  3 

13. A balloon starts rising from the ground with an acceleration of 1.25 m/s2. After 8s, a stone is released from the balloon. The stone will (A) cover a distance of 40 m (B) have a displacement of 50 m (C) reach the ground in 4 s Sol. (C)

(D) begin to move down after being released

Height of stone as it released and speed

1 1 h  at 2   1.25  (8)2  40 m 2 2 v  u  at  0  1.25  4  10 m/s

1 s  ut  at 2 2 1 40  10t  10t 2 2 t = 4 sec

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Class Assignment (Batch 2014) Solution Topics: Kinematics

PHYSICS

a (m/s2) 14. The acceleration-time graph of a particle moving along a straight line is as shown in the figure. At what time the particle acquires its initial velocity (A) 12 sec (B) 5 sec (C) 8 sec

10

(D) 16 sec

t (sec) 4 Sol. (C)Area of a–t graph gives changes in velocity time at which Area will be zero, is 8 sec. *15.The figure shows the velocity (v) of a particle plotted against time (t). (A) The particle changes its direction of motion at some point (B) The acceleration of the particle remains constant (C) The displacement of the particle is zero (D) The initial and final speeds of the particle are the same Sol. (ABCD) 16. The figure shows velocity-time graph of a particle moving along a straight line. Identify the correct statement. (A) The particle starts from the origin (B) The particle crosses it initial position at t = 2s (C) The average speed of the particle in the time interval, 0 < t  2s is zero (D) All of the above Sol. (B) Area of v-t graph = Displacement 17. Find the average velocity of the particle between t = 2s and t = 7s. v(m/s) 4 13 8 m/s m/s (A) (B) 5 5 2 (C)

9 m/s 5

(D)

11 m /s 5

7 2 4

6

t (s)

–2

Sol. (C)

vav 

total displacment area under v-t curve  total time total time

18. A particle moves along a straight line such that its position at any time t is given by x  t 3 – 3t 2  2  m The displacement when the acceleration becomes zero is : (A) 0 m (B) 2m Sol. (D) 3

2

x  t  3t  2

 a

d 2x  (6t  6) dt 2

(C) – 3 m

(D) – 2 m

 a  0 at t  1

at t = 1, x = 0. at t = 0, x = 2. Displacement = – 2

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Class Assignment (Batch 2014) Solution Topics: Kinematics

PHYSICS 19. If the velocity of a particle is given by v = (180 – 16x) 1/2 m/s then its acceleration will be : (A) zero (B) 8m/s2 (C) – 8m/s2 Sol. (B)

(D) – 4m/s2

 dv  a  v.    dx  20. An object starts from rest at t = 0 & accelerates at a rate given by a = 6t then its velocity & displacement after 2 seconds will be (A) 10 m/sec, 8 m (B) 12 m/sec, 4 m (C) 4 m/sec, 6 m (D) 12 m/sec, 8 m Sol. (D)

a  6t

dx  3t 2  dt

dv  6t  dt x

v



t

 dv   6t dt 0

0

 v  3t 2

t

2   dx   3t dt 0

0

x = t3 at t = 2 sec, v = 12 m/s, x = 8 m.

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