Title 03. Motion in a Staright line Easy Ans.pdf ✅

1. (a) r xi yj zk ˆ ˆ ˆ = + +   2 2 2 r = x + y + z r 6 8 10 10 2 m 2 2 2 = + + = 2. (a) r i j ˆ 10 ˆ = 20 +   20 10 22.5 2 2 r = + = m 3. (c) From figure, OA = 0 i+ 30 j , AB = 20 i+ 0 j BC cos i o = −30 2 45 j o − 30 2 sin 45 = −30 i− 30 j  Net displacement, OC = OA + AB + BC = −10 i+ 0 j  OC  = 10 m. 4. (a) An aeroplane flies 400 m north and 300 m south so the net displacement is 100 m towards north. Then it flies 1200 m upward so 2 2 r = (100 ) + (1200 ) = 1204 m ~− 1200 m The option should be 1204 m, because this value mislead one into thinking that net displacement is in upward direction only. 5. (b) Total time of motion is 2 min 20 sec = 140 sec. As time period of circular motion is 40 sec so in 140 sec. athlete will complete 3.5 revolution i.e., He will be at diametrically opposite point i.e., Displacement = 2R. 6. (c) Horizontal distance covered by the wheel in half revolution = R. So the displacement of the point which was initially in contact with ground = AA' = 2 2 (R) + (2R) = 4 2 R  + 4 2 =  + (As R = 1m) 7. (d) As the total distance is divided into two equal parts therefore distance averaged speed 1 2 1 2 2 v v v v + = 8. (d) B A B A v v   tan tan =   = tan 60 tan 30 3 1 3 1 / 3 = = 9. (b) Distance average speed 1 2 2 1 2 v v v v + = 20 30 2 20 30 +   = 24 km / hr 5 120 = = 10. (b) Distance average speed 1 2 2 1 2 v v v v + = 2.5 4 2 2.5 4 +   = km / hr 13 40 65 200 = = 11. (c) Distance average speed 1 2 2 1 2 v v v v + = 30 50 2 30 50 +   = 37.5 km / hr 2 75 = = 12. (d) Average speed Total time Total distance = 1 2 t t x + = 1 2 / 3 2 / 3 v x v x x + = 36 k m / hr 3 60 2 3 20 1 1 =  +  = 13. (a) Time average speed = = + = + 2 80 40 2 1 2 v v 60km / hr . 14. (b) Distance travelled by train in first 1 hour is 60 km and distance in next 1/2 hour is 20 km. So Average speed 3 / 2 60 20 Total time Total distance + = = = 53.33 km / hour 15. (d) 16. (c) Total distance to be covered for crossing the bridge = length of train + length of bridge = 150m + 850m = 1000 m 80 sec 18 5 45 1000 Velocity Distance Time =  = = 17. (c) Displacement of the particle will be zero because it comes back to its starting point m s sec m 3 / 10 30 Total time Total distance Average speed = = = 18. (d) Velocity of particle Total time Total diplacemen t = 4 m / s 5 2 10 5 Diameter of circle =  = = 30 2 30 m A 20 m B 45° C O 2R A' Final A R Initial
19. (d) A man walks from his home to market with a speed of 5 km / h .Distance = 2.5 km and time hr v d 2 1 5 2.5 = = = . and he returns back with speed of 7.5 km / h in rest of time of 10 minutes. Distance 1.25 km 60 10 = 7.5  = So, Average speed Total time Total distance = k m hr hr k m / 8 45 (40 / 60) (2.5 1.25) = + = . 20. (b) 1 | distance| | displaceme nt| | Average speed | | Average velocity | =  because displacement will either be equal or less than distance. It can never be greater than distance. 21. (b) 22. (c) From given figure, it is clear that the net displacement is zero. So average velocity will be zero. 23. (c) Since displacement is always less than or equal to distance, but never greater than distance. Hence numerical ratio of displacement to the distance covered is always equal to or less than one. 24. (d) Length of train = 100 m Velocity of train k m hr 12.5 m /s 18 5 = 45 / = 45  = Length of bridge = 1 km = 1000 m Total length covered by train = 1100 m Time taken by train to cross the bridge 12.5 1100 = = 88 sec 25. (b) Time average velocity = 3 1 2 3 v + v + v 4m /s 3 3 4 5 = + + = 26. (a) When the body is projected vertically upward then at the highest point its velocity is zero but acceleration is not equal to zero ( 9.8 / ) 2 g = m s . 27. (b) Let two boys meet at point C after time 't' from the starting. Then AC = v t , BC v t = 1 2 2 2 (AC) = (AB) + (BC)  2 2 1 2 2 2 v t = a +v t By solving we get 2 1 2 2 v v a t − = 28. (c) 48 . 100 2 2 40 60 1 2 1 2 kmph v v v v vav =   = + = 29. (b) As 2 2 1 S = ut + at  S a(10) 50a 2 1 2 1 = = .....(i) As v = u + at  velocity acquired by particle in 10 sec v = a10 For next 10 sec , 2 2 ( ) (10) 2 1 S = (10a)10 + a  S2 = 150 a .....(ii) From (i) and (ii) S1 = S 2 / 3 30. (c) Acceleration 2 2 2 2a dt d x = = 31. (d) Velocity along X-axis at dt dx vx = = 2 Velocity along Y-axis bt dt dy vy = = 2 Magnitude of velocity of the particle, 2 2 2 2 v = vx + vy = 2t a + b 32. (a) S v dt k t dt k t 2 9 9m 2 1 2 1 3 0 2 3 0 3 0 =   =       = = =   33. (a) 3 S = kt  kt dt d S a 6 2 2 = = i.e. a  t 34. (a) From 2 2 1 S = ut + a t 2 1 ( 1) 2 1 S = a P − and 2 2 2 1 S = a P [As u = 0 ] From (2 1) 2 = + n − a Sn u 2( 1) 1 2 2 ( 1) 2 = − + − − + P P a S th P P 2 2 1 2 2 = P − P + a It is clear that 1 2 ( 1) S 2 th S S P P = + − + 35. (d) m F a   = . If F = 0  then a = 0  . 36. (b) v 4t 2t 3 = − (given)  12 2 2 = = t − dt dv a and 4 2 0 3 0 x v dt (4t 2t) dt t t t t = = − = −   When particle is at 2m from the origin 2 4 2 t − t =  2 0 4 2 t − t − = ( 2)( 1) 0 2 2 t − t + =  t = 2 sec Acceleration at t = 2 sec given by, A v B v1 C a
12 2 2 a = t − =12 2 −2 = 2 22 m / s 37. (a)     + = +  = x x v dx dt 2 1 2  dt dx dx dv dt dv a = = . 2 3 2 2 . . 2 (2 ) .2 v v v x v dx dv a v      = − = − + − = =  Retardation 3 = 2v 38. (b) Let 1 2 3 u ,u ,u and 4 u be velocities at time 0, , ( ) 1 1 2 t = t t + t and ( ) 1 2 3 t + t + t respectively and acceleration is a then 2 and 2 , 2 3 4 3 2 3 2 1 2 1 u u v u u v u u v + = + = + = Also , ( ) 2 1 1 3 1 1 2 u = u + at u = u + a t + t and ( ) 4 1 1 2 3 u = u + a t + t + t By solving, we get ( ) ( ) 2 3 1 2 2 3 1 2 t t t t v v v v + + = − − 39. (c) Acceleration a = tan , where  is the angle of tangent drawn on the graph with the time axis. 40. (b) If acceleration is variable (depends on time) then v u f dt  = + ( ) u a t dt  = + ( ) 2 2 a t = u + 41. (a) (2 1) 2 = − n − a Sn u (2 5 1) 1 meter 2 2 = 10 −  − = 42. (b) From v u 2aS 2 2 = +  0 u 2aS 2 = +  2 2 2 20 / 2 10 (20) 2 m s S u a = −  − = − = 43. (d) v = u + at = 10 + 2  4 = 18 m /sec 44. (c) If particle starts from rest and moves with constant acceleration then in successive equal interval of time the ratio of distance covered by it will be 1 : 3 : 5 : 7 .....(2n − 1) i.e. ratio of x and y will be 1 : 3 i.e. 3 1 = y x  y = 3 x 45. (a) 2 1 2 = + n − a Sn u S th 2 5 1 7 18 25m 2 4 7 5 = +  − = + = . 46. (c) Acceleration t dt dv a = = 0.1 2 = 0.2t Which is time dependent i.e. non-uniform acceleration. 47. (b) Constant velocity means constant speed as well as same direction throughout. 48. (a) Distance travelled in 4 sec 16 2 1 24 = 4u + a  ...(i) Distance travelled in total 8 sec 64 2 1 88 = 8u + a  ...(ii) After solving (i) and (ii), we get u =1 m/s. 49. (c) (3 6 ) 6 6 2 = = t − t = t − dt d dt dx vx . At = 1, = 0 x t v ( 2 ) 2 2 2 = = t − t = t − dt d dt dy vy . At t = 1, vy = 0 Hence 0 2 2 v = vx + vy = 50. (a) Distance travelled in th n second (2 1) 2 = + n − a u Distance travelled in th 5 second (2 5 1) 2 8 = 0 +  − =36m 51. (d) 2 2 2 2 v = u + 2as  (9000 ) − (1000 ) = 2a4 7 2  a = 10 m /s Now a v u t − = t sec 4 7 8 10 10 9000 1000 − =  −  = 52. (b) 2 v = u + at  −2 = 10 + a 4  a = −3m /sec 53. (c) S x ux t ax t S x 6 16 48 m 2 1 2 1 2 = +  =   = Sy uy t ay t Sy 8 16 64 m 2 1 2 1 2 = +  =   = S S x S y 80m 2 2 = + = 54. (d) 2 S  u . If u becomes 3 times then S will become 9 times i.e. 9  20 = 180 m 55. (c) 2 4 y = a + bt + ct − dt 3 b 2ct 4dt dt dy v = = + − and 2 2c 12dt dt dv a = = − Hence, at t = 0, vinitial = b and ainitial = 2c. 56. (a) 2 S  u  2 2 1 2 1         = u u S S  S m S 8 4 2 1 2 2 =  = 57. (c) ( ) 2 g a h t + = 0.49 0.7 sec 11 5.4 (9.8 1.2) 2 2.7 = = = +  = As u = 0 and lift is moving upward with acceleration 58. (a) Displacement 2 5 2 x = t + t +
Velocity = = 4t +1 dt dx Acceleration 4 2 2 = = dt d x i.e. independent of time Hence acceleration 2 = 4 m /s 59. (d) Both trains will travel a distance of 1 km before to come in rest. In this case by using v u 2as 2 2 = + 0 (40) 2 1000 2  = + a  2  a = −0.8 m /s 60. (a) v = u + at  v = 0 + 5 10 = 50 m /s 61. (b) Let 'a' be the retardation of boggy then distance covered by it be S. If u is the initial velocity of boggy after detaching from train (i.e. uniform speed of train) a u v u as u as sb 2 2 0 2 2 2 2 2 = +  = −  = Time taken by boggy to stop a u v = u + at  0 = u − at  t = In this time t distance travelled by train a u s ut t 2 = = = Hence ratio 2 1 = t b s s 62. (a) (2 1) 2 (2 1) 2 = + − = n − a n a Sn u because u = 0 Hence 5 7 3 4 = S S 63. (b)   v = u + adt = u + (3t + 2t + 2)dt 2 t u t t t t t u 2 2 2 2 3 3 3 2 3 2 = + + + = + + + = 2 + 8 + 4 + 4 = 18 m /s (As t = 2 sec) 64. (d) 3 12 3 2 = = t − t + dt ds v and = = 6t −12 dt dv a For a = 0 , we have t = 2 and at 1 2, 9 − t = v = − ms 65. (d) 66. (b) 2 2 a = ax + ay 2 1 2 2 2 2 2 2                 +         = dt d y dt d x Here 0 2 2 = dt d y . Hence 2 2 2 8m / s dt d x a = = 67. (b) F = ma , If force is constant then m a 1  . So If mass is doubled then acceleration becomes half. 68. (c) (2 1) 2 = + n − a Sn u  (2 6 1) 2 1.2 = 0 +  − a  2 0.218 / 11 1.2 2 a = m s  = 69. (b) Here v = 144 km / h = 40m /s v = u + at  2 40 = 0 + 20 a  a = 2 m /s  s at 2 (20) 400 m 2 1 2 1 2 2 = =   = 70. (c) b a a bt t dt d x at bt dt dx 3 2 3 2 6 0 2 2 2 = −  = − =  = 71. (b) Stopping distance F mu 2 2 1 Retardins force Kinetic energy = = If retarding force (F) and velocity (v) are equal then stopping distance  m (mass of vehicle) As mcar  mtruck therefore car will cover less distance before coming to rest. 72. (d) u = 72 kmph = 20m /s, v = 0 By using v u 2as 2 2 = −  s u a 2 2 = 2 2 1 / 2 200 (20) = m s  = 73. (b) 2 12t 3t dt ds v = = − Velocity is zero for t = 0 and t = 4 sec 74. (a) 75. (b) Let A and B will meet after time t sec. it means the distance travelled by both will be equal. S ut t A = = 40 and 2 2 4 2 1 2 1 S at t B = =   4 20 2 1 40 2 S A = S B  t = t  t = sec 76. (b) bt dt dx x a bt , v 2 2 = + = = Instantaneous velocity v = 2  3  3 = 18 cm /sec 77. (c) If the body starts from rest and moves with constant acceleration then the ratio of distances in consecutive equal time interval S1 : S2 : S3 = 1 : 3 : 5 78. (c) b ct dt d x x at bt ct , a 2 6 2 2 2 3 = + − = = − 79. (a) Let initial (t = 0) velocity of particle = u For first 5 sec motion s 10 metre 5 =