Tiêu đề 03. Motion in a Staright line Hard Ans.pdf ✅

1. (d) Let total distance travelled = x and total time taken t1 + t2 60 2 / 3 20 x / 3 x = +  Average speed 180 2 60 1 1 60 (2 / 3) 20 (1 / 3) + = + = x x x = 36km / hr 2. (b) Given 2 2 4 2 s = t + t +  velocity (v) = = 4t + 2 dt ds and acceleration (a) 2 4(1) 0 4m / s dt dv = = + = 3. (c) Given 2 3 x = at − bt  velocity 2 ( ) 2at 3bt dt dx v = = − and acceleration (a) = 2a 6bt. dt dv = − When acceleration = 0  2a − 6bt = 0  b a b a t 6 3 2 = = . 4. (c) Given 2 4 y = a + bt + ct − dt  v = 3 0 b 2ct 4dt dt dy = + + − Putting t = 0, vinitial = b So initial velocity = b Now, acceleration (a) 2 0 2c 12 d t dt dv = = + − Putting t = 0, ainitial = 2c 5. (a) Differentiating time with respect to distance = x +  dx dt 2   +  = = dt x dx v 2 1 So, acceleration (a) = dt dx dx dv dt dv = . = 2 3 2 2 . . 2 (2 ) . 2 v v v x v dx dv v      = − = − + − = 6. (a) Given that 2 x  t or 2 x = Kt (where K= constant) Velocity (v) Kt dt dx = = 2 and Acceleration (a) K dt d = = 2  It is clear that velocity is time dependent and acceleration does not depend on time. So we can say that particle is moving with uniform acceleration but variable velocity. 7. (b)  2 1     = − 2 cos 90 5 5 5 2 2 2 1 2 2 2 2  = 1 + − = + = o       = 5 2 Average acceleration 2 m/s 2 1 10 5 2 = =   = t  toward north-west (As clear from the figure). 8. (b) Given that 4t 2t 3  = −  x =  dt , x = t − t + C 4 2 , at t = 0, x = 0 C = 0 When particle is 2m away from the origin 4 2 2 = t − t  2 0 4 2 t − t − =  ( 2) ( 1) 0 2 2 t − t + =  t = 2 sec (4 2 ) 12 2 3 2 = = t − t = t − d t d d t d a   12 2 2 a = t − for t = 2 sec  12 ( 2 ) 2 2 a =  −  2 a = 22 m/s 9. (b) Let particle moves towards east and by the application of constant force it moves towards west 10 m/s 1 = +  and 2 m/s  2 = −  . Acceleration Time Change in velocity = t 2 −1 =  4 ( 2) (10) a − − = 4 −12 = 2 = −3 m/s 10. (b) 9sec 0.34 3.06 Average velocity Distance Time = = = and Time Change in velocity Acceleration = 2 0.02 / 9 0.18 = = m s . 11. (a) Let initial (t = 0) velocity of particle = u for first 5 sec of motion s 10 metre 5 = , 1 v  − 5m / s 2  =     90o 5m / s 1  = 
so by using 2 2 1 s = ut + a t 2 (5) 2 1 10 = 5u + a  2u + 5a = 4 .... (i) for first 8 sec of motion s 20 metre 8 = 2 (8) 2 1 20 = 8u + a  2u + 8a = 5 .... (ii) By solving (i) and (ii) 2 / 3 1 / 6 7 u = m s a = m s Now distance travelled by particle in total 10 sec. ( ) 2 10 10 2 1 s = u 10 + a by substituting the value of u and a we will get s10 = 28.3 m So the distance in last 2 sec = s10 – s8 = 28.3 − 20 = 8.3 m 12. (c) Since the body starts from rest. Therefore u = 0 . 2 25 (5) 2 1 2 1 a S = a = 2 100 (10) 2 1 2 1 2 a S + S = a =  2 1 2 100 S a S = − = 2 75 a 2 225 (15) 2 1 2 1 2 3 a S + S + S = a =  3 2 1 2 225 S S a S = − − = 2 125 a Thus Clearly 1 2 3 5 1 3 1 S = S = S 13. (a) (2 1) 2 1 S n = u + a n − (8) [2 5 1] 36metres 2 1 = 0 +  − = 14. (b) F = ma m a 1  if F = constant. Since the force is same and the effective mass of system becomes double m m m m a a 2 2 1 1 2 = = , 1 2 2 2 m/s 2 = = a a 15. (a) As S (2n − 1) n , 5 7 3 4 = S S 16. (a) m g u H 11.25 2 10 (15) 2 2 2 max =  = = 17. (c) Let both balls meet at point P after time t. The distance travelled by ball A ( 2 1 2 1 h ) = gt .....(i) The distance travelled by ball B 2 2 2 1 (h )= ut − gt .....(ii) By adding (i) and (ii) h + h = ut 1 2 = 400 (Given 400 . h = h1 + h2 = )  t = 400 / 50 = 8sec and h1 = 320 m, h2 = 80 m 18. (c) Maximum height of ball = 5m, So velocity of projection  u = 2gh = 210 5 =10 m /s time interval between two balls (time of ascent) = . 60 1 1sec min g u = = So no. of ball thrown per min = 60 19. (b) Let particle thrown with velocity u and its maximum height is H then g u H 2 2 = When particle is at a height H / 2 , then its speed is 10 m / s From equation v u 2gh 2 2 = − , ( ) g u u g H u g 4 2 2 10 2 2 2 2 2  = −      = −  200 2 u =  Maximum height 2 10 200 2 2  = = g u H = 10m 20. (b) Speed of stone in a vertically upward direction is 20 m/s. So for vertical downward motion we will consider u = −20 m / s v u 2gh ( 20) 2 10 200 v 65 m / s 2 2 2 = + = − +    = 21. (b) Let at point A initial velocity of body is equal to zero For path AB : v 0 2gh 2 = + ... (i) For path AC : (2v) 0 2gx 2 = +  4v 2gx 2 = ... (ii) Solving (i) and (ii) x = 4h A h1 h2 B P 400 m
22. (b) 2 2 1 S = at  t  s (As a = constant ) = = =  2 / 4 1 1 2 1 2 s s s s t t s t t 2 2 4 2 1 2 = = = 23. (c) For first case of dropping . 2 1 2 h = gt For second case of downward throwing 2 1 1 gt 2 1 h = −ut + 2 2 1 = gt  ( ) 2 1 2 1 2 1 − ut = g t − t ......(i) For third case of upward throwing 2 2 2 2 2 1 2 1 h = ut + gt = gt  ( ) 2 1 2 2 2 2 ut = g t − t .......(ii) on solving these two equations : 2 2 2 2 1 2 2 1 t t t t t t − − − =  . 1 2 t = t t 24. (c) By formula (2 1) 2 1 hn = u + g n −  [2 6 1]} 2 10 [2 5 1] 2 { 2 10 u −  − = u −  −  u − 45 = 2 (u − 55)  u = 65m / s. 25. (b) Let the interval be t then from question For first drop (2 ) 5 2 1 2 g t = .....(i) For second drop 2 2 1 x = gt .....(ii) By solving (i) and (ii) 4 5 x = and hence required height h 3.75 . 4 5 = 5 − = m 26. (d) The separation between two bodies, two second after the release of second body is given by : s = ( ) 2 1 2 2 2 1 g t − t = 9.8 (3 2 ) 24.5 2 1 2 2   − = m. 27. (a) x =  2 1 t t v dt =         = =  3 0 2 3 0 2 2 2 t t dt 9 m. 28. (c)  2 1 v v dv =  2 1 t t a dt =  2 1 ( ) t t bt dt 2 1 2 2 2 1 t t bt v v          − =  2 2 2 0 0 2 2 1 bt v bt v v t = +         = +  3 0 2 0 6 1 2 dt v t bt bt S = v dt + = +   29. (d) u = at  at dt ds =  4 0 2 4 0 2         = =  t s at dt a = 8a. 30. (b) u 2 = 2gh; h/3 = 2gh t – 1/2 gt2 or gt2 – 2 2gh t + 2h/3 = 0 t = ( ) 2g 2 2gh  8gh − 8gh /3 or ( ) 2 2gh 2 2gh/3( 3 1) 2 2gh 2 2gh/3 3 1 t t 2 1 + − − − = = ( ) 3 ( 3 1) 3 3 1 + − − − = 3 2 3 2 + − 31. (d) dx/dt = - ae -t + be t as t increases - a/et decreases and be t increases. 32. (c) t = ax2 + bx or dt/dx = 2ax + b or v = dx/dt = 1/2ax + b ( ) ( ) 3 2 3 2av 2ax b 2a dt dx 2ax b 2a dt dv = − + − = + − = . 33. (d) aav = 10 i 5ˆ 5i t vf vi − = − = a = ms NW 2 1 −2 x A B C u = 0 h v 2v
34. (c) v 2 = 2gh = 2 × 10 × 50 d = 50 + ( ) 293m 2 2 3 2 10 50 2 =         − −   s 35. (a) v 2 = 2gh or v = 2gh , i.e., 2 1 2 1 h h v v =  v2 = 3 v0 36. (a) v = 0 + at1; 0 = at1- b(t – t1) or t1 = bt/a + b  vmax = abt/a + b 37. (b) 200 = (6.2 – 5.5)t or t = 285.714 s s = (6.2 × 285.714) = 1770m (faster), 1770 – 8 × 200 = 170 Thus 170 m away from the starting point along the track in the direction of run. 38. (a) x(3) = 2(3)2 – 5(3) + 6 = 93x (0) = 6 vav = ( ) ( ) 3 9 6 3 0 x 3 x 0 − = − − = 1 ms-1 dt t 3 dx = = 4t – 5 = 4(3) – 5 = 7 ms-1 39. (d) vy 2 = 2ay = 2 × 10 × 12 v = 25 + 240 = 16.2 ms-1 . 40. (c) v 2 – u 2 = 2as or s = 2 1 25 15 2 2  − = 200 m. 41. (a) vAB = 15 – 10 = 5 ms-1 xAB = 10 m; t = xAB/vAB = 2s. Road distance covered = vAs t + length of car A = 15 × 2 + 5 = 35 m. 42. (a) a = dt dy . dy dv dt dv = or   = − y 0 v v vdv kydy 0 or v0 2 – v 2 = ky2 . 43. (b)   = −  −  − t 0 v 0 dt v dv loge ( ) t v = −   −  or v =   (1 – e -t ) 44. (d) Let boat move at angle  to the normal as shown in fig. then time to cross the river = l/v cos . drift x = (2v – vsin ) l/v cos  for x to be minimum dx/d = 0 = 1(2 sec tan  - sec2 ) or sin  = 1/2 or  = 300 and  = 90 + 30 = 120 45. (d) Total distance covered = area under v – t graph. From fig. 20 × 25 = 5t1 2 + (25 – 2t1) 5t1 or 5t1 2 – 125 t1 + 500 = 0 or (t1 - 5) (t1- 20) = 0  t1 = 5s discard t1 = 20 s. 46. (a) vws = vw – vs = (15 cos 60 i ˆ + 15 sin 60 j ˆ ) – 30 i ˆ |v|= ( ) ( ) 2 2 39.5 + 7.5 = 39.7 kmh-1 tan  = 7.5/37.8 = 1/5  = tan-1 1/5 North of West.