Page 1 Graphy 28. OA : slope is + ive and increasing. \ velocity is + ive and acceleration is + ive. AB : slope is + ive and constant \ velocity is +ive and acceleration is zero. BC : sope is + ive and decreases. \ velocity is + ive and increasing. CD : slope is â€“ive and increasing \ velocity is - ive and acceleration is - ive. 29. In M 1 and M 3 0 90 : ° £ < ° q \ slope is + ive i.e., acceleration is + ive. In M 2 and M 4 : 90 180 ° < £ ° q \ slope is - ive i.e., acceleration is - ive. (a) M 1 : Magnitude of velocity is increasing. M 2 : Magnitude of velocity is decreasing. M 3 : Magnitude of velocity is increasing. M 4 : Magnitude of velocity is decreasing. Ans : M 1 and M 3 . (b) P M ® 1 Q M ® 2 R M ® 3 S M ® 4 30. Case I v mt v = - 0 (st-line) \ ds mt v dt ò ò = - ( ) 0 i.e., s m t v t s = × - + 2 0 0 2 (Parabola) or s s v t at = - + 0 0 2 1 2 (Q m a = ) Further, a dv dt = = tan q As for 0 90 ° £ < ° q tan q is + ive, a is + ive. Case II v k = (constant) (st line parallel to time-axis) Þ ds kdt ò ò = s kt s = + 0 Further, a dv dt = =0 Motion in One Dimension | 27 O t s A B D C q v M 1 t q v M 2 t O q M 4 q v M 3 t O q v 0 m = tan q v s 0 t s Þ a Þ a = tan q t s 0 s t v Þ t a Þ t Page 2 Graphy 28. OA : slope is + ive and increasing. \ velocity is + ive and acceleration is + ive. AB : slope is + ive and constant \ velocity is +ive and acceleration is zero. BC : sope is + ive and decreases. \ velocity is + ive and increasing. CD : slope is â€“ive and increasing \ velocity is - ive and acceleration is - ive. 29. In M 1 and M 3 0 90 : ° £ < ° q \ slope is + ive i.e., acceleration is + ive. In M 2 and M 4 : 90 180 ° < £ ° q \ slope is - ive i.e., acceleration is - ive. (a) M 1 : Magnitude of velocity is increasing. M 2 : Magnitude of velocity is decreasing. M 3 : Magnitude of velocity is increasing. M 4 : Magnitude of velocity is decreasing. Ans : M 1 and M 3 . (b) P M ® 1 Q M ® 2 R M ® 3 S M ® 4 30. Case I v mt v = - 0 (st-line) \ ds mt v dt ò ò = - ( ) 0 i.e., s m t v t s = × - + 2 0 0 2 (Parabola) or s s v t at = - + 0 0 2 1 2 (Q m a = ) Further, a dv dt = = tan q As for 0 90 ° £ < ° q tan q is + ive, a is + ive. Case II v k = (constant) (st line parallel to time-axis) Þ ds kdt ò ò = s kt s = + 0 Further, a dv dt = =0 Motion in One Dimension | 27 O t s A B D C q v M 1 t q v M 2 t O q M 4 q v M 3 t O q v 0 m = tan q v s 0 t s Þ a Þ a = tan q t s 0 s t v Þ t a Þ t Case III. v v mt = + 0 \ ds v mt dt ò ò = + ( ) 0 i.e., s v t m t s = + + 0 2 0 2 or s s v t at = + - 0 0 2 1 2 (Q m a = - ) Further, a dv dt = = tan q As for 90 180 ° < £ ° q tan q is - ive, a is - ive. Time-displacement graph Time Area under the graph Initial Displacement Net displacement at 0 s 0 m - 10 m - 10 m 2 s 10 m - 10 m 0 m 4 s 30 m - 10 m + 20 m 6 s 40 m - 10 m + 30 m 8 s 30 m - 10 m + 20 m 10 s 20 m - 10 m + 10 m Time-acceleration graph time slope acceleration 0 - 2 s 5 5 m/s 2 2 - 4 s zero 0 m/s 2 4 - 6 s - 5 - 5 m/s 2 6 - 8 s - 5 - 5 m/s 2 8 - 10 s + 5 + 5 m/s 2 31. a = slope of v t - graph. S = area under v t - graph. Corresponding graphs are drawn in the answer sheet. 32. Average velocity = = = - s t A A t Net area Time 1 2 Average speed = = + d t A A t 1 2 33. Average acceleration = - = - - =- v v f i time s 10 20 6 5 2 m/ v v f i - = Area under a t - graph. 34. (a) Acceleration = slope of v t - graph. 35. (b) r r s f i - = = area under v t - graph. (c) Equations are written in answer sheet. Rel a ti ve Mo ti on 35. (a) Acceleration of 1 w.r.t. 2 = - - - ( ) ( ) g g =0 m/s 2 (b) Initial velocity of 2 w.r.t. 1 = + - - ( ) ( ) 20 5 =25 m/s (c) Initial velocity of 1 w.r.t. 2 = - - + ( ) ( ) 5 20 = -25 m/s 28 | Mechanics-1 t 10 8 6 4 2 â€“ 10 m + 10 m + 20 m + 30 m s t 10 8 6 4 2 5 a q v v 0 t Þ s s 0 t Þ a t a = tan q Page 3 Graphy 28. OA : slope is + ive and increasing. \ velocity is + ive and acceleration is + ive. AB : slope is + ive and constant \ velocity is +ive and acceleration is zero. BC : sope is + ive and decreases. \ velocity is + ive and increasing. CD : slope is â€“ive and increasing \ velocity is - ive and acceleration is - ive. 29. In M 1 and M 3 0 90 : ° £ < ° q \ slope is + ive i.e., acceleration is + ive. In M 2 and M 4 : 90 180 ° < £ ° q \ slope is - ive i.e., acceleration is - ive. (a) M 1 : Magnitude of velocity is increasing. M 2 : Magnitude of velocity is decreasing. M 3 : Magnitude of velocity is increasing. M 4 : Magnitude of velocity is decreasing. Ans : M 1 and M 3 . (b) P M ® 1 Q M ® 2 R M ® 3 S M ® 4 30. Case I v mt v = - 0 (st-line) \ ds mt v dt ò ò = - ( ) 0 i.e., s m t v t s = × - + 2 0 0 2 (Parabola) or s s v t at = - + 0 0 2 1 2 (Q m a = ) Further, a dv dt = = tan q As for 0 90 ° £ < ° q tan q is + ive, a is + ive. Case II v k = (constant) (st line parallel to time-axis) Þ ds kdt ò ò = s kt s = + 0 Further, a dv dt = =0 Motion in One Dimension | 27 O t s A B D C q v M 1 t q v M 2 t O q M 4 q v M 3 t O q v 0 m = tan q v s 0 t s Þ a Þ a = tan q t s 0 s t v Þ t a Þ t Case III. v v mt = + 0 \ ds v mt dt ò ò = + ( ) 0 i.e., s v t m t s = + + 0 2 0 2 or s s v t at = + - 0 0 2 1 2 (Q m a = - ) Further, a dv dt = = tan q As for 90 180 ° < £ ° q tan q is - ive, a is - ive. Time-displacement graph Time Area under the graph Initial Displacement Net displacement at 0 s 0 m - 10 m - 10 m 2 s 10 m - 10 m 0 m 4 s 30 m - 10 m + 20 m 6 s 40 m - 10 m + 30 m 8 s 30 m - 10 m + 20 m 10 s 20 m - 10 m + 10 m Time-acceleration graph time slope acceleration 0 - 2 s 5 5 m/s 2 2 - 4 s zero 0 m/s 2 4 - 6 s - 5 - 5 m/s 2 6 - 8 s - 5 - 5 m/s 2 8 - 10 s + 5 + 5 m/s 2 31. a = slope of v t - graph. S = area under v t - graph. Corresponding graphs are drawn in the answer sheet. 32. Average velocity = = = - s t A A t Net area Time 1 2 Average speed = = + d t A A t 1 2 33. Average acceleration = - = - - =- v v f i time s 10 20 6 5 2 m/ v v f i - = Area under a t - graph. 34. (a) Acceleration = slope of v t - graph. 35. (b) r r s f i - = = area under v t - graph. (c) Equations are written in answer sheet. Rel a ti ve Mo ti on 35. (a) Acceleration of 1 w.r.t. 2 = - - - ( ) ( ) g g =0 m/s 2 (b) Initial velocity of 2 w.r.t. 1 = + - - ( ) ( ) 20 5 =25 m/s (c) Initial velocity of 1 w.r.t. 2 = - - + ( ) ( ) 5 20 = -25 m/s 28 | Mechanics-1 t 10 8 6 4 2 â€“ 10 m + 10 m + 20 m + 30 m s t 10 8 6 4 2 5 a q v v 0 t Þ s s 0 t Þ a t a = tan q \ Velocity of 1 w.r.t. 2 at time t = æ è ç ö ø ÷ 1 2 s = - + æ è ç ö ø ÷ ( ) ( ) 25 0 1 2 = -25 m/s (d) Initial relative displacement of 2 w.r.t. 1 S 0 20 ® = - m Using S S u a ® ® ® ® = + + 0 2 1 2 rel rel t t , as at time t (= time of collision of the particles) the relative displacement of 2 w.r.t. 1 will be zero (i.e., S O ® ® = ) O ® = - + - × æ è ç ö ø ÷ ( ) ( ) 20 25 1 2 Þ t =0.85 s 36. Let length of escalator L (= 15 m) walking speed of man = L 90 Speed of escalator = L 60 Time taken by man walking on a moving escalator = + L L L 90 60 =36 s Required time has been found without using the actual length of the escalator. 37. S 0 ® = Initial displacement of elevator w.r.t. ball = + 10 m. Relative velocity of elevator w.r.t. ball = - ( ) ( ) 2 18 = -16 m/s Accelerator of elevator w.r.t. ball = - - ( ) ( ) 0 10 =10 m/s 2 Using S S ® ® = + + 0 2 1 2 u t a t rel rel 0 10 16 1 2 10 2 = + + - + + ( ) ( ) ( ) t t or 5 16 10 0 2 t t - - = t =3.65 s Position of elevator when it meets ball = + ´ 5 2 ( ) 3.65 = 12.30 m level 38. (a) 1 2 60 2 ( ) 2.2 t = â€¦(i) Þ t=7.39 s (b) 1 2 60 2 ( ) 3.5 t x = + â€¦(ii) Dividing Eq. (ii) by Eq. (i), 60 60 + = x 3.5 2.2 or x 60 = 1.3 2.2 x =35.5 m (c) At the time of overtaking Speed of automobile = ( ) ( ) 3.5 7.39 =25.85 m/s Speed of truck = ( ) ( ) 2.2 7.39 = 16.25 m/s 39. Let, acceleration of lift = a (upward) \ acceleration of thrown body w.r.t. lift = - - + ( ) ( ) g a = - + ( ) g a If time of flight is t, using v u a t rel rel rel = + ( ) ( ) { ( )} - = + + - + v u a g t Þ ( ) a g t u + = 2 or a u t g = - 2 Motion in One Dimension | 29 10 m 12 m 18 m/s 5m 2 m/s Elevator T A T x 60 m 2 3.5 m /s 2 2.2 m /s time = t s time = 0 A Page 4 Graphy 28. OA : slope is + ive and increasing. \ velocity is + ive and acceleration is + ive. AB : slope is + ive and constant \ velocity is +ive and acceleration is zero. BC : sope is + ive and decreases. \ velocity is + ive and increasing. CD : slope is â€“ive and increasing \ velocity is - ive and acceleration is - ive. 29. In M 1 and M 3 0 90 : ° £ < ° q \ slope is + ive i.e., acceleration is + ive. In M 2 and M 4 : 90 180 ° < £ ° q \ slope is - ive i.e., acceleration is - ive. (a) M 1 : Magnitude of velocity is increasing. M 2 : Magnitude of velocity is decreasing. M 3 : Magnitude of velocity is increasing. M 4 : Magnitude of velocity is decreasing. Ans : M 1 and M 3 . (b) P M ® 1 Q M ® 2 R M ® 3 S M ® 4 30. Case I v mt v = - 0 (st-line) \ ds mt v dt ò ò = - ( ) 0 i.e., s m t v t s = × - + 2 0 0 2 (Parabola) or s s v t at = - + 0 0 2 1 2 (Q m a = ) Further, a dv dt = = tan q As for 0 90 ° £ < ° q tan q is + ive, a is + ive. Case II v k = (constant) (st line parallel to time-axis) Þ ds kdt ò ò = s kt s = + 0 Further, a dv dt = =0 Motion in One Dimension | 27 O t s A B D C q v M 1 t q v M 2 t O q M 4 q v M 3 t O q v 0 m = tan q v s 0 t s Þ a Þ a = tan q t s 0 s t v Þ t a Þ t Case III. v v mt = + 0 \ ds v mt dt ò ò = + ( ) 0 i.e., s v t m t s = + + 0 2 0 2 or s s v t at = + - 0 0 2 1 2 (Q m a = - ) Further, a dv dt = = tan q As for 90 180 ° < £ ° q tan q is - ive, a is - ive. Time-displacement graph Time Area under the graph Initial Displacement Net displacement at 0 s 0 m - 10 m - 10 m 2 s 10 m - 10 m 0 m 4 s 30 m - 10 m + 20 m 6 s 40 m - 10 m + 30 m 8 s 30 m - 10 m + 20 m 10 s 20 m - 10 m + 10 m Time-acceleration graph time slope acceleration 0 - 2 s 5 5 m/s 2 2 - 4 s zero 0 m/s 2 4 - 6 s - 5 - 5 m/s 2 6 - 8 s - 5 - 5 m/s 2 8 - 10 s + 5 + 5 m/s 2 31. a = slope of v t - graph. S = area under v t - graph. Corresponding graphs are drawn in the answer sheet. 32. Average velocity = = = - s t A A t Net area Time 1 2 Average speed = = + d t A A t 1 2 33. Average acceleration = - = - - =- v v f i time s 10 20 6 5 2 m/ v v f i - = Area under a t - graph. 34. (a) Acceleration = slope of v t - graph. 35. (b) r r s f i - = = area under v t - graph. (c) Equations are written in answer sheet. Rel a ti ve Mo ti on 35. (a) Acceleration of 1 w.r.t. 2 = - - - ( ) ( ) g g =0 m/s 2 (b) Initial velocity of 2 w.r.t. 1 = + - - ( ) ( ) 20 5 =25 m/s (c) Initial velocity of 1 w.r.t. 2 = - - + ( ) ( ) 5 20 = -25 m/s 28 | Mechanics-1 t 10 8 6 4 2 â€“ 10 m + 10 m + 20 m + 30 m s t 10 8 6 4 2 5 a q v v 0 t Þ s s 0 t Þ a t a = tan q \ Velocity of 1 w.r.t. 2 at time t = æ è ç ö ø ÷ 1 2 s = - + æ è ç ö ø ÷ ( ) ( ) 25 0 1 2 = -25 m/s (d) Initial relative displacement of 2 w.r.t. 1 S 0 20 ® = - m Using S S u a ® ® ® ® = + + 0 2 1 2 rel rel t t , as at time t (= time of collision of the particles) the relative displacement of 2 w.r.t. 1 will be zero (i.e., S O ® ® = ) O ® = - + - × æ è ç ö ø ÷ ( ) ( ) 20 25 1 2 Þ t =0.85 s 36. Let length of escalator L (= 15 m) walking speed of man = L 90 Speed of escalator = L 60 Time taken by man walking on a moving escalator = + L L L 90 60 =36 s Required time has been found without using the actual length of the escalator. 37. S 0 ® = Initial displacement of elevator w.r.t. ball = + 10 m. Relative velocity of elevator w.r.t. ball = - ( ) ( ) 2 18 = -16 m/s Accelerator of elevator w.r.t. ball = - - ( ) ( ) 0 10 =10 m/s 2 Using S S ® ® = + + 0 2 1 2 u t a t rel rel 0 10 16 1 2 10 2 = + + - + + ( ) ( ) ( ) t t or 5 16 10 0 2 t t - - = t =3.65 s Position of elevator when it meets ball = + ´ 5 2 ( ) 3.65 = 12.30 m level 38. (a) 1 2 60 2 ( ) 2.2 t = â€¦(i) Þ t=7.39 s (b) 1 2 60 2 ( ) 3.5 t x = + â€¦(ii) Dividing Eq. (ii) by Eq. (i), 60 60 + = x 3.5 2.2 or x 60 = 1.3 2.2 x =35.5 m (c) At the time of overtaking Speed of automobile = ( ) ( ) 3.5 7.39 =25.85 m/s Speed of truck = ( ) ( ) 2.2 7.39 = 16.25 m/s 39. Let, acceleration of lift = a (upward) \ acceleration of thrown body w.r.t. lift = - - + ( ) ( ) g a = - + ( ) g a If time of flight is t, using v u a t rel rel rel = + ( ) ( ) { ( )} - = + + - + v u a g t Þ ( ) a g t u + = 2 or a u t g = - 2 Motion in One Dimension | 29 10 m 12 m 18 m/s 5m 2 m/s Elevator T A T x 60 m 2 3.5 m /s 2 2.2 m /s time = t s time = 0 A 40. OB OA AB = + ( ) ( ) 2 2 = + ( ) ( ) 20 20 2 2 =20 2 m Speed along OB is 2 2 m/s \ Time taken to reach B = - 20 2 2 2 1 m ms =10 s 41. In D OPQ, OP = ® | | v br OP = ® | | v r PQ POQ OP OPQ sin sin Ð = Ð 2 45 4 45 sin( ) sin ° - = ° q q= æ è ç ö ø ÷- ° é ë ê ù û ú - sin 1 1 2 2 45 42. Let pilot heads point R to reach point Q sinq= RQ PR = 200 500 t t Þ q = - sin ( ) 1 0.4 ( ) ( ) ( ) PR RQ PQ 2 2 2 = + or ( ) ( ) ( ) 500 200 1000 2 2 2 t t - = or t = - 1000 500 200 2 2 ( ) ( ) = - 10 25 4 = 10 21 h Objective Questions (Level 1) Single Correct Option 1. As the packet is detached from rising balloon its acceleration will be g in the downward direction. Option (b) is correct. 2. While going up : 0 1 = - + æ è ç ö ø ÷ u mg f m T T um mg f 1 = + â€¦(i) Using v u as 2 2 2 = + 0 2 2 2 = - + æ è ç ö ø ÷ u mg f m s i.e., S mu mg f = + 2 2( ) While coming down s mg f m T = - æ è ç ö ø ÷ 1 2 2 2 or mu mg f mg f m T 2 2 2 2 1 2 ( ) + = - æ è ç ö ø ÷ 30 | Mechanics-1 f (Air resistance) S mg Stops u B A 45° O River flow q P Q 20 m B A 45° O 2Ö2 m/s OAB = OA = 20 m q East P Q R North Wind 200km/h South 1000 km Page 5 Graphy 28. OA : slope is + ive and increasing. \ velocity is + ive and acceleration is + ive. AB : slope is + ive and constant \ velocity is +ive and acceleration is zero. BC : sope is + ive and decreases. \ velocity is + ive and increasing. CD : slope is â€“ive and increasing \ velocity is - ive and acceleration is - ive. 29. In M 1 and M 3 0 90 : ° £ < ° q \ slope is + ive i.e., acceleration is + ive. In M 2 and M 4 : 90 180 ° < £ ° q \ slope is - ive i.e., acceleration is - ive. (a) M 1 : Magnitude of velocity is increasing. M 2 : Magnitude of velocity is decreasing. M 3 : Magnitude of velocity is increasing. M 4 : Magnitude of velocity is decreasing. Ans : M 1 and M 3 . (b) P M ® 1 Q M ® 2 R M ® 3 S M ® 4 30. Case I v mt v = - 0 (st-line) \ ds mt v dt ò ò = - ( ) 0 i.e., s m t v t s = × - + 2 0 0 2 (Parabola) or s s v t at = - + 0 0 2 1 2 (Q m a = ) Further, a dv dt = = tan q As for 0 90 ° £ < ° q tan q is + ive, a is + ive. Case II v k = (constant) (st line parallel to time-axis) Þ ds kdt ò ò = s kt s = + 0 Further, a dv dt = =0 Motion in One Dimension | 27 O t s A B D C q v M 1 t q v M 2 t O q M 4 q v M 3 t O q v 0 m = tan q v s 0 t s Þ a Þ a = tan q t s 0 s t v Þ t a Þ t Case III. v v mt = + 0 \ ds v mt dt ò ò = + ( ) 0 i.e., s v t m t s = + + 0 2 0 2 or s s v t at = + - 0 0 2 1 2 (Q m a = - ) Further, a dv dt = = tan q As for 90 180 ° < £ ° q tan q is - ive, a is - ive. Time-displacement graph Time Area under the graph Initial Displacement Net displacement at 0 s 0 m - 10 m - 10 m 2 s 10 m - 10 m 0 m 4 s 30 m - 10 m + 20 m 6 s 40 m - 10 m + 30 m 8 s 30 m - 10 m + 20 m 10 s 20 m - 10 m + 10 m Time-acceleration graph time slope acceleration 0 - 2 s 5 5 m/s 2 2 - 4 s zero 0 m/s 2 4 - 6 s - 5 - 5 m/s 2 6 - 8 s - 5 - 5 m/s 2 8 - 10 s + 5 + 5 m/s 2 31. a = slope of v t - graph. S = area under v t - graph. Corresponding graphs are drawn in the answer sheet. 32. Average velocity = = = - s t A A t Net area Time 1 2 Average speed = = + d t A A t 1 2 33. Average acceleration = - = - - =- v v f i time s 10 20 6 5 2 m/ v v f i - = Area under a t - graph. 34. (a) Acceleration = slope of v t - graph. 35. (b) r r s f i - = = area under v t - graph. (c) Equations are written in answer sheet. Rel a ti ve Mo ti on 35. (a) Acceleration of 1 w.r.t. 2 = - - - ( ) ( ) g g =0 m/s 2 (b) Initial velocity of 2 w.r.t. 1 = + - - ( ) ( ) 20 5 =25 m/s (c) Initial velocity of 1 w.r.t. 2 = - - + ( ) ( ) 5 20 = -25 m/s 28 | Mechanics-1 t 10 8 6 4 2 â€“ 10 m + 10 m + 20 m + 30 m s t 10 8 6 4 2 5 a q v v 0 t Þ s s 0 t Þ a t a = tan q \ Velocity of 1 w.r.t. 2 at time t = æ è ç ö ø ÷ 1 2 s = - + æ è ç ö ø ÷ ( ) ( ) 25 0 1 2 = -25 m/s (d) Initial relative displacement of 2 w.r.t. 1 S 0 20 ® = - m Using S S u a ® ® ® ® = + + 0 2 1 2 rel rel t t , as at time t (= time of collision of the particles) the relative displacement of 2 w.r.t. 1 will be zero (i.e., S O ® ® = ) O ® = - + - × æ è ç ö ø ÷ ( ) ( ) 20 25 1 2 Þ t =0.85 s 36. Let length of escalator L (= 15 m) walking speed of man = L 90 Speed of escalator = L 60 Time taken by man walking on a moving escalator = + L L L 90 60 =36 s Required time has been found without using the actual length of the escalator. 37. S 0 ® = Initial displacement of elevator w.r.t. ball = + 10 m. Relative velocity of elevator w.r.t. ball = - ( ) ( ) 2 18 = -16 m/s Accelerator of elevator w.r.t. ball = - - ( ) ( ) 0 10 =10 m/s 2 Using S S ® ® = + + 0 2 1 2 u t a t rel rel 0 10 16 1 2 10 2 = + + - + + ( ) ( ) ( ) t t or 5 16 10 0 2 t t - - = t =3.65 s Position of elevator when it meets ball = + ´ 5 2 ( ) 3.65 = 12.30 m level 38. (a) 1 2 60 2 ( ) 2.2 t = â€¦(i) Þ t=7.39 s (b) 1 2 60 2 ( ) 3.5 t x = + â€¦(ii) Dividing Eq. (ii) by Eq. (i), 60 60 + = x 3.5 2.2 or x 60 = 1.3 2.2 x =35.5 m (c) At the time of overtaking Speed of automobile = ( ) ( ) 3.5 7.39 =25.85 m/s Speed of truck = ( ) ( ) 2.2 7.39 = 16.25 m/s 39. Let, acceleration of lift = a (upward) \ acceleration of thrown body w.r.t. lift = - - + ( ) ( ) g a = - + ( ) g a If time of flight is t, using v u a t rel rel rel = + ( ) ( ) { ( )} - = + + - + v u a g t Þ ( ) a g t u + = 2 or a u t g = - 2 Motion in One Dimension | 29 10 m 12 m 18 m/s 5m 2 m/s Elevator T A T x 60 m 2 3.5 m /s 2 2.2 m /s time = t s time = 0 A 40. OB OA AB = + ( ) ( ) 2 2 = + ( ) ( ) 20 20 2 2 =20 2 m Speed along OB is 2 2 m/s \ Time taken to reach B = - 20 2 2 2 1 m ms =10 s 41. In D OPQ, OP = ® | | v br OP = ® | | v r PQ POQ OP OPQ sin sin Ð = Ð 2 45 4 45 sin( ) sin ° - = ° q q= æ è ç ö ø ÷- ° é ë ê ù û ú - sin 1 1 2 2 45 42. Let pilot heads point R to reach point Q sinq= RQ PR = 200 500 t t Þ q = - sin ( ) 1 0.4 ( ) ( ) ( ) PR RQ PQ 2 2 2 = + or ( ) ( ) ( ) 500 200 1000 2 2 2 t t - = or t = - 1000 500 200 2 2 ( ) ( ) = - 10 25 4 = 10 21 h Objective Questions (Level 1) Single Correct Option 1. As the packet is detached from rising balloon its acceleration will be g in the downward direction. Option (b) is correct. 2. While going up : 0 1 = - + æ è ç ö ø ÷ u mg f m T T um mg f 1 = + â€¦(i) Using v u as 2 2 2 = + 0 2 2 2 = - + æ è ç ö ø ÷ u mg f m s i.e., S mu mg f = + 2 2( ) While coming down s mg f m T = - æ è ç ö ø ÷ 1 2 2 2 or mu mg f mg f m T 2 2 2 2 1 2 ( ) + = - æ è ç ö ø ÷ 30 | Mechanics-1 f (Air resistance) S mg Stops u B A 45° O River flow q P Q 20 m B A 45° O 2Ö2 m/s OAB = OA = 20 m q East P Q R North Wind 200km/h South 1000 km Þ T um mg f mg f 2 = + - ( )( ) â€¦(ii) Using Eq. (i) and Eq. (ii) T T mg f mg f 2 1 = + - Þ T T 2 1 > Option (c) is correct. 3. Angular speed (w) of seconds hand = 2 60 p rad s -1 Speed of the tip of seconds hand v = ´ p 30 1 cm/s [Q v r = w and r=1cm] As in 15s the seconds hand rotates through 90°, the change in velocity of its tip in 15 s will be =v 2 = - p 2 30 1 cms Option (d) is correct. 4. Average speed = + + = - 5 5 5 30 5 60 40 1 ms Option (c) is correct. 5. Relative velocity of boat w.r.t. water = + - - - ( ) ( ) ^ ^ ^ ^ 3 4 3 4 i j i j = + 6 8 i j ^ ^ Option (b) is correct. 6. 21 18 11 42 18 42 = ´ + ´ + ( ) ( ) v Þ v = 25.29 m/s 7. x t t = - 32 8 3 3 v dx dt t = = - 32 8 2 â€¦(i) \ Particle is at rest when 32 8 0 2 - = t i.e., t =2 s Differentiating Eq. (i) w.r.t. time t a dv dt t = = - 16 \ a at time t = 2 s (when particle is at rest) = - ´ ( ) ( ) 16 2 = -32 m/s 2 Option (b) is correct. 8. For first one second 2 1 2 1 2 = ´ a Þ a = 4 m/s 2 Velocity at the end of next second v = ´ ( ) ( ) 4 2 = 8 m/s Option (b) is correct. 9. x t t = - + 3 3 \ displacement at time t (= 1 s) = - + = - 3 1 1 2 3 ( ) ( ) m and displacement at time t (= 3 s) = - + = 3 3 3 18 3 ( ) ( ) m And as such displacement in the time interval (t = 1 s to t = 3 s) = - - ( ) ( ) 18 2 m m = +20 m Option (c) is correct. 10. Acceleration a bt = \ dv dt bt = or dv btdt ò ò = or v bt C = + 1 2 2 Now, at t = 0, v v = 0 \ C v = 0 i.e., v bt v = + 1 2 2 0 or ds dt v bt = + 0 2 1 2 or ds v bt dt ò ò = + æ è ç ö ø ÷ 0 2 1 2 or s v t bt k = + + 0 3 1 6 At t =0, s =0, \ k=0 Þ s v t bt = + 0 3 1 6 Option (a) is correct. Motion in One Dimension | 31Read More

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