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

1. (c) Kinematics is that branch of mechanics that deals with the study of motion without going into the causes of motion. 2. (b) In one dimensional motion, the path length and magnitude of displacement are equal only if the object does not change its directions during the course of motions. In all other cases, the path length is greater than the magnitude of displacement. 3. (b) Displacement = final positions – Initial positions x = x2 − x1 Then , (i) x = 7m− (−3m) =10m (ii) x = −3m− 7m = −10m (iii) x = 3m− (−7m) =10m 4. (b) Displacement in first eight steps = 5 -3 = 2 m Time taken for first eight steps = 8s Time taken by the drunkard to cover first 6 m of journey 6 24s 2 8 =  = If the drunkard takes 5 steps more, he will fall into the pit. So the times taken by the drunkard to cover last five step = 5 s Total time taken = 24s + 5s = 29s 5. (d) Distances  Displacement 6. (a) Displacement 2 2 = AC = (AB) + (BC) 2 2 = (10m) + (20m) = 100+ 400 = 22.5m 7. (c) When a particle returns to its starting point its displacement is zero. 8. (b) The positions- times graph of a particle moving with negative velocity is a shown in the figure. 9. (a) Both speed and velocity are constant in the case of a particle moving with uniform velocity. A particle moving with uniform velocity has zero accelerations. 10. (c) Average velocity Time int erval Displacement = , A particle moving in a given directions with non-zero velocity cannot have zero speed. In general, average speed is not equal to magnitude of average velocity. However it can be so if the motion is along a straight line without changes in directions. 11. (d) For O to P and come back to Q Average speed t OP PQ Time int erval Path length  + = = 1 20ms 24s (360 120)m − = + = Average velocity Time int erval Displacement = 1 10ms 24s 240m − = = From O to P and come back to O Average speed 1 20ms 36s 360m 360m − = + = Average velocity zero 36s 0m = = 12. (d) The area under the velocity – time graph represent the displacement over a given times interval. 13. (c) Average velocity 1 2ms 40 2 40 t 2R Time taken Displacement − =  = = = 14. (d) All the four graphs are impossible. (i) A particle cannot have two different positions at the same time. (ii) A particle cannot have velocity in opposite direction at the same time. (iii) Speed is always non-negative. (iv) Total path length of a particle can never decrease with time. Note : The arrows on the graphs are meaningless. 15. (c) Speedometer of the car measures the instantaneous speed of the car. 16. (d) Given 2 v = 2t(3− t)orv= 6t −2t 6 4t dt dv = − At maximum velocity, 0 dt dy = s 2 3 6 − 4t = 0 ort = 17. (b) A particle moving with uniform velocity has zero accelerations.
18. (c) Distance = Area under speed – time graph over a given time interval.  Distance traversed by the particle between t = 0s to t = 10 s will be = Area of OAB 10 12 2 1 =   60 m 19. (c) The slope of the tangent at any point on the displacement – times graph given instantaneous velocity at that instant. In the given graph. The slope is negative at point E. 20. (b) Speedometer measures the speed of the car in km 1 h − 21. (d) The displacement – times graph is a straight line inclined to time axis upto times 0 t indicates a uniform velocity . After times 0 t the displacement- times graph is a straight line parallel to time axis indicates particle at rest. 22. (b) 6 3v dv 6 3v or dt dt dv − = − = Integrating both sides, we get in (6 3v) C 3 1 t = − − + Where C is a constant of integrations At t = 0 , v = 0  ln 6 3 1 C =       −  = − 6 6 3v ln 3 1 t       − − = 6 6 3v 3t ln ( ) 3t 3t v or v 2 1 e 2 1 e 1 − − = − = − 23. (c) For zero accelerations, the positions- time graph is a straight line. 24. (a) The slope of the tangent drawn on velocity – times graph at any instant of times is equal to the instantaneous accelerations. 25. (d) 26. (c) 27. (a) For the given velocity – displacement graph, Intercept = 0 v and slope 0 0 x V = − Thus, the equations of given lien of velocity-displacement graph is 0 0 0 x v x v v = − + Accelerations, v dx dv dt dx dx dv dt dv a = = = 0 0 x v dx dv  = −          = − − + 0 0 0 0 0 x v x v x v a (Using (i)) 0 2 0 2 0 2 0 x v x0 x v = In is a straight line with positions slope and a negative intercept. The variation of a with x is as shown in the figure. 28. (a) x = t – sin t 1 cost dt dx v = = − sin t dt dv a = =  x (t) > 0 for all values of t > 0 and v (t) can be zero for one value of t. a (t) can zero for one value of t. 29. (d) For uniform motions with zero accelerations , v – t graph is a straight line parallel to the time axis. 30. (b) Given: a = p – qx .....(i)  0 = p – qx or q p x =  Velocity is maximum at q p x =  Accelerations,       = = = = dt dx v dx dv v dt dx dx dv dt dv a  a p qx dx dv v = = − vdv = (p−qx)dx Integrating both sides of the above equations, we get 2 qx px 2 v 2 2 = − 2 2 2 v = 2px − qx or v = 2px − qx
At q P q P q q P , v v 2P q p x 2 max =         −        = = = 31. (a) As x 3  t Velocity , v 2  3t Accelerations, a  6t 32. (b) Let S1 be the distances travelled by particle in time 2 to 5 s and S2 be the distance travelled by particle in time 5 to 6s.  Total distance travelled, S= S S . 1 + 2 During the time interval 0 to 5s, the accelerations of particle is equal to the slope of line OA. i.e., 2 2.4ms 5 12 a − = = Velocity at the end of 2s will be. V = 0 + 1 2.4 2 4.8ms−  = Taking motion of particle for time interval 2s to 5s, Here, u = 4.8ms ,a 2.4ms , −1 −2 = S S ,t 5 2 3s = 1 = − = Then. S 2.4 3 25.2m 2 1 4.8 3 2 1 =  +   = Accelerations of the particle during the motion t = 5 s to t = 10s is A = Slope of line AB = 2 2.4ms 5 12 − − = − Taking motions of the particle for the time 1s (i.e. 5s to 6s), Here, u = 12m 1 s − ,a =- 2.4ms , −2 t = 1s , S = S2  ( 2.4) 1 10.8m 2 1 S 12 1 2 2 =  + −  =  S = S1 = 25.2 +10.8 = 36m 33. (a) The a – t graph show that initially the body moves with uniform velocity. Its accelerations increases for a short time and then falls to zero and their after the body moves with a constant velocity. Such a physical situations arises when a cricket ball moving with a speed is hit with a bat for a very short time interval. 34. (a) Here, u 1 10ms− = , t = 3 s , v = 16m 1 s − 2ms . 3 16 10 t v u a −2 = − = − = Now velocity at 2s, before the given instant 10 = u + 22 (v = u + at) 1 y 6ms−  = 35. (c) At the highest point velocity of the balls becomes zero, but its accelerations is equal to g. 36. (a) The given law is known as galileo’s law of odd number. This law was estabilished by galileo Galilei who was the first to make quantitative studies of free fall. 37. (c) Let 1 t be the time taken by the ball to reach the highest point. Here , 1 1 2 v = 0,u = 20ms ,a = −g = −10ms ,t = t − − As v = u +at  0 = 20 + (-10 ) 1 t or 1 t = 2s Taking vertical downward motion of the ball from highest point to ground. Here , u = 0, a = + g = 10ms , −2 s= 20 m + 25m = 45m 2 t = t As 2 at 2 1 S = ut + 2 2 (10)t 2 1 45 = 0 + 9 or t 3s 10 90 10 45 2 t 2 2 2 = = =  = Total times taken by the ball to each the ground. t t 2s 3s 5s = 1 + 2 = + = 38. (b) Free fall of object in vacuum is a case of motion with uniform acceleration. 39. (c) The sing of accelerations does not tell us whether the particle’s speed is increasing or decreasing. The sing of accelerations depends on the choice of the positive directions of the axis. For example, if the vertically upward directions is chosen to be positive directions of the axis, the accelerations due to gravity is negative. If a particle is falling under gravity, this acceleration though negative results in increase in speed. 40. (a) Time taken by body 1 A = a Accelerations of body 1 A = a Time taken by body B,t 5 2 3s 2 = − = Accelerations of body 2 B = a Distance covered by first body in 5th seconds after its start. 1 1 1 1 5 a 2 9 (2 5 1) 2 a (2t 1) 0 2 a S = u + − = +  − = Distance coveted by the second body in the 3rd second after its start, 2 2 2 2 3 a 2 5 (2 3 1) 2 a (2t 1) 0 2 a S = u + − = +  − = Since S5 = S3  a or a : a 5 : 9 2 5 a 2 9 1 = 2 1 2 = 41. (a)
42. (c) 2 at 2 1 S = vt + It is a not a kinematic equations of motions. All other are three kinematic equations of motions. 43. (a) Here, n v n v 0 t v u a = − = − = Displacement in last 2 s 2 2 n n 2 a(n 2) 2 1 an 2 1 S −S − = − − (n 1) n v = 2a(n −1) = 2 − n 2v(n −1) = 44. (b) Using, 2 at 2 1 S = ut + Here 1 2 S 0,u 50ms ,a g 10ms − − = = = − = − 2 2 ( 10)t 50t 5t 2 1 0 = 50t +  − = − t =10s 45. Here , u = 0, g = 2 10ms − , h = 1 km = 1000m As v u 2gh 2 2 − = v 2gh 2  = or 1 v 2gh 2 10 1000 100 2ms − = =   = 1 kmh 5 18 100 2 − =  1 1 360 2kmh 510kmh − − =  46. (a) Distance travelled by a body in nth second is (2n 1) 2 a Dn = u + − Here, u = 0, a = g  Distance travelled by the body in 1st seconds is g 2 1 (2 1 1) 2 g D1 = 0 +  − = Distance travelled by the body in 2nd seconds is g 2 3 (2 2 1) 2 g D2 = 0 +  − = Distances travelled by the body in 3rd seconds is g 2 5 (2 3 1) 2 g D3 = 0 +  − = and so on. Hence, the distances covered by a freely fallings body in its first, second, third .....nth second of its motions forms an arithmetic progression. 47. (c) Here, u = 0  v 2as 2 = It is a parabola of the type y 4ax 2 = Hence, option (c) represent the correct graph 48. (d) Stopping distance, 2a u d 2 2 = − Where u is the initial speed of the car. As the stopping distance is proportional to square of the initial velocity. Doubling the initial velocity increases the stopping distance by a factor of 4. 49. (a) Here , a = g – by When an object falls with constant speed v , c its accelerations becomes zero. b g g − bvc = 0or vc = 50. (c) As distance travelled in successive seconds differ by 2m each, therefore acceleration is constant = 2 2ms− For (2n 1) 2 a Dn = u + − (2 8 1) 2 2 Ds = u +  − 20 = u + 15 or u 5ms ` −1 = 51. (a) Let d, is the distance travelled by the vehicle before it stops. Here, final velocity v = 0, initial velocity = u, S = d, Using equations of motions v u 2as 2 2 = + (0) u 2ad, 2 2  = + 2a u d 2 s = 2 ds  u 52. (b) When velocity of A = velocity o B, then, relative velocity is zero.  Displacement – time graphs of A and B must have same slope (other than zero) 53. (b) In this case 1 VBG 4kmh − = 1 VBG 9kg h − =  For an observer on a stationary platform, speed of child running opposite to the directions of motions of the belt is 1 1 VCG VCB VBG 9km h 5k h − − = + = − = − - ve sing shows that the child appears to be moving from right to left with a speed of 5kmh . −1 54. (d) Distance between the parents = 50m Since the parent and child are located on the same belt, the speed of the child as observed by the stationary observer in