Page 1 Introductory Exercise 3.1 1. Suppose a particle is moving with constant velocity v (along the axis of x). Displacement of particle in time t vt 1 1 = Displacement of particle in time t vt 2 2 = \ Displacement of the particle in the time interval Dt t t ( ) = - 2 1 = - vt vt 2 1 = - v t t ( ) 2 1 \ Average velocity in the time interval Dt v t t t t ¢ = - - ( ) ( ) 2 1 2 1 = v Now, as the particle is moving with constant velocity (i.e., with constant speed in a given direction) its velocity and speed at any instant will obviously be v. Ans. True. 2. As the stone would be free to acceleration under earth’s gravity it acceleration will be g. 3. A second hand takes 1 min i.e., 60s to complete one rotation (i.e., rotation by an angle of 2p rad). \ Angular speed of second hand = 2 60 p rad s = p 30 rad s -1 Linear speed of its tip = radius ´ angular speed = ´ - 2.0 cm rad s p 30 1 = - p 15 1 cms As the tip would be moving with constant speed. Average speed = - p 15 1 cms In 15 s the second hand would rotate through 90° i.e., the displacement of its tip will be r 2. \ Modulus of average velocity of the tip of second hand in 15 s. = r 2 15 = - 2 2 15 1 cms 4. (a) Yes. By changing direction of motion, there will be change in velocity and so acceleration. (b) (i) No. In curved path there will always be acceleration. (As explained in the previous answer no. 3) (ii) Yes. In projectile motion the path of the particle is a curved one while acceleration of the particle remains constant. (iii) Yes. In curved path the acceleration will always be there. Even if the path is circular with constant speed the direction of the acceleration of the particle would every time be changing. 3 Motion in One Dimension Page 2 Introductory Exercise 3.1 1. Suppose a particle is moving with constant velocity v (along the axis of x). Displacement of particle in time t vt 1 1 = Displacement of particle in time t vt 2 2 = \ Displacement of the particle in the time interval Dt t t ( ) = - 2 1 = - vt vt 2 1 = - v t t ( ) 2 1 \ Average velocity in the time interval Dt v t t t t ¢ = - - ( ) ( ) 2 1 2 1 = v Now, as the particle is moving with constant velocity (i.e., with constant speed in a given direction) its velocity and speed at any instant will obviously be v. Ans. True. 2. As the stone would be free to acceleration under earth’s gravity it acceleration will be g. 3. A second hand takes 1 min i.e., 60s to complete one rotation (i.e., rotation by an angle of 2p rad). \ Angular speed of second hand = 2 60 p rad s = p 30 rad s -1 Linear speed of its tip = radius ´ angular speed = ´ - 2.0 cm rad s p 30 1 = - p 15 1 cms As the tip would be moving with constant speed. Average speed = - p 15 1 cms In 15 s the second hand would rotate through 90° i.e., the displacement of its tip will be r 2. \ Modulus of average velocity of the tip of second hand in 15 s. = r 2 15 = - 2 2 15 1 cms 4. (a) Yes. By changing direction of motion, there will be change in velocity and so acceleration. (b) (i) No. In curved path there will always be acceleration. (As explained in the previous answer no. 3) (ii) Yes. In projectile motion the path of the particle is a curved one while acceleration of the particle remains constant. (iii) Yes. In curved path the acceleration will always be there. Even if the path is circular with constant speed the direction of the acceleration of the particle would every time be changing. 3 Motion in One Dimension 5. (a) Time speed = Circumference Speed = ´ 2 4 1 p cm cm/s =8p =25.13 s (b)As particle is moving with constant speed of 1 cm/s, its average speed in any time interval will be 1 cm/s. | | / Average velocity = r T 2 4 = æ è ç ö ø ÷ 4 2 2 r r p speed = 2 2 p speed = 2 2 p cms -1 = 0.9 cms -1 | | / Average acceleration = v T 2 4 (where v = speed) = 4 2 2 v r v p = 2 2 2 p v r = 2 2 1 4 2 p ( ) = - 0.23 cm s 2 6. Distance = Speed ´ time D v t 1 1 1 = D v t 2 2 2 = Average speed = + + = + + D D t t v t v t t t 1 2 1 2 1 1 2 2 1 2 = ´ + ´ + ( ) ( ) 4 2 6 3 2 3 = - 5.2ms 1 Introductory Exercise 3.2 1. Acceleration (due to gravity). 2. s u at a t = + - 1 2 is physically correct as it gives the displacement of the particle in t th second (or any time unit). s t = Displacement in t seconds - displacement in ( ) t - 1 seconds = + é ë ê ù û ú - - + - é ë ê ù û ú ut at u t a t 1 2 1 1 2 1 2 2 ( ) ( ) Therefore, the given equation is dimensionally incorrect. 3. Yes. When a particle executing simple harmonic motion returns from maximum amplitude position to its mean position the value of its acceleration decreases while speed increases. 4. v t = 3 4 / (given) ds dt t = 3 4 / …(i) \ s t dt = ò 3 4 / = + + + t c 3 4 1 3 4 1 or s t c = + 4 7 7 4 / i.e., s t µ 74 / Differentiating Eq. (i) w.r.t. time t, d s dt t 2 2 3 4 1 3 4 = - Þ a t µ -1 4 / 5. Displacement (s) of the particle s= ´ + - ( ) ( ) 40 6 1 2 106 2 = - 240 180 =60 m (in the upward direction) Distance covered ( ) D by the particle Time to attain maximum height Motion in One Dimension | 15 Displacement Velocity st = u + at – 1 2 a Acceleration Page 3 Introductory Exercise 3.1 1. Suppose a particle is moving with constant velocity v (along the axis of x). Displacement of particle in time t vt 1 1 = Displacement of particle in time t vt 2 2 = \ Displacement of the particle in the time interval Dt t t ( ) = - 2 1 = - vt vt 2 1 = - v t t ( ) 2 1 \ Average velocity in the time interval Dt v t t t t ¢ = - - ( ) ( ) 2 1 2 1 = v Now, as the particle is moving with constant velocity (i.e., with constant speed in a given direction) its velocity and speed at any instant will obviously be v. Ans. True. 2. As the stone would be free to acceleration under earth’s gravity it acceleration will be g. 3. A second hand takes 1 min i.e., 60s to complete one rotation (i.e., rotation by an angle of 2p rad). \ Angular speed of second hand = 2 60 p rad s = p 30 rad s -1 Linear speed of its tip = radius ´ angular speed = ´ - 2.0 cm rad s p 30 1 = - p 15 1 cms As the tip would be moving with constant speed. Average speed = - p 15 1 cms In 15 s the second hand would rotate through 90° i.e., the displacement of its tip will be r 2. \ Modulus of average velocity of the tip of second hand in 15 s. = r 2 15 = - 2 2 15 1 cms 4. (a) Yes. By changing direction of motion, there will be change in velocity and so acceleration. (b) (i) No. In curved path there will always be acceleration. (As explained in the previous answer no. 3) (ii) Yes. In projectile motion the path of the particle is a curved one while acceleration of the particle remains constant. (iii) Yes. In curved path the acceleration will always be there. Even if the path is circular with constant speed the direction of the acceleration of the particle would every time be changing. 3 Motion in One Dimension 5. (a) Time speed = Circumference Speed = ´ 2 4 1 p cm cm/s =8p =25.13 s (b)As particle is moving with constant speed of 1 cm/s, its average speed in any time interval will be 1 cm/s. | | / Average velocity = r T 2 4 = æ è ç ö ø ÷ 4 2 2 r r p speed = 2 2 p speed = 2 2 p cms -1 = 0.9 cms -1 | | / Average acceleration = v T 2 4 (where v = speed) = 4 2 2 v r v p = 2 2 2 p v r = 2 2 1 4 2 p ( ) = - 0.23 cm s 2 6. Distance = Speed ´ time D v t 1 1 1 = D v t 2 2 2 = Average speed = + + = + + D D t t v t v t t t 1 2 1 2 1 1 2 2 1 2 = ´ + ´ + ( ) ( ) 4 2 6 3 2 3 = - 5.2ms 1 Introductory Exercise 3.2 1. Acceleration (due to gravity). 2. s u at a t = + - 1 2 is physically correct as it gives the displacement of the particle in t th second (or any time unit). s t = Displacement in t seconds - displacement in ( ) t - 1 seconds = + é ë ê ù û ú - - + - é ë ê ù û ú ut at u t a t 1 2 1 1 2 1 2 2 ( ) ( ) Therefore, the given equation is dimensionally incorrect. 3. Yes. When a particle executing simple harmonic motion returns from maximum amplitude position to its mean position the value of its acceleration decreases while speed increases. 4. v t = 3 4 / (given) ds dt t = 3 4 / …(i) \ s t dt = ò 3 4 / = + + + t c 3 4 1 3 4 1 or s t c = + 4 7 7 4 / i.e., s t µ 74 / Differentiating Eq. (i) w.r.t. time t, d s dt t 2 2 3 4 1 3 4 = - Þ a t µ -1 4 / 5. Displacement (s) of the particle s= ´ + - ( ) ( ) 40 6 1 2 106 2 = - 240 180 =60 m (in the upward direction) Distance covered ( ) D by the particle Time to attain maximum height Motion in One Dimension | 15 Displacement Velocity st = u + at – 1 2 a Acceleration = = 40 10 4 s < 6 s It implies that particle has come back after attaining maximum height (h) given by h u g = 2 2 = ´ ( ) 40 2 10 2 = 80 m \ D = + - 80 80 60 ( ) =100 m 6. v t = - 40 10 \ dx dt t = - 40 10 or dx t dt = - ( ) 40 10 or x t dt = - ò ( ) 40 10 or x t t c = - + 40 5 2 As at t = 0 the value of x is zero. c =0 \ x t t = - 40 5 2 For x to be 60 m. 60 40 5 2 = - t t or t t 2 8 12 0 - + = \ t =2 s or 6 s 7. Average velocity = Displacement in time t t = + ut at t 1 2 2 = + u at 1 2 8. v v at 2 1 = + \ at v v = - 2 1 Average velocity = Displacement in time t t = + v t at t 1 2 1 2 = + v at 1 1 2 = + - v v v 1 2 1 2 = + v v 1 2 2 \ Ans. True. 9. 125 0 1 2 2 = × + t gt Þ t =25 s Average velocity = 125 5 m s (downwards) = 25 m/s (downwards) 10. v t t = + - 10 5 2 …(i ) \ a dv dt t = = - 5 2 At t =2 s a = - ´ 5 2 2 =1 m/s 2 From Eq. (i), dx dt t t = + - 10 5 2 \ x t t dt = + - ò ( ) 10 5 2 or x t t t c = + - + 10 5 2 3 2 3 As, at t = 0 the value of x is zero c =0 \ x t t t = + - 10 5 2 3 2 3 Thus, at t = 3 s x = ´ + - ( ) ( ) 10 3 5 2 3 3 3 2 3 = + - 30 9 22.5 = 43.5 m 11. u i ® = 2 ^ m/s a i j ® = ° + ° ( cos sin ) ^ ^ 2 60 2 60 m/s 2 = + ( ) ^ ^ 1 3 i j m/s 2 v u a ® ® ® = + t = + + 2 1 3 2 i i j ^ ^ ^ ( ) = + 4 2 3 i j ^ ^ 16 | Mechanics-1 60° 2 a = 2 m/s u = 2 m/s x y Page 4 Introductory Exercise 3.1 1. Suppose a particle is moving with constant velocity v (along the axis of x). Displacement of particle in time t vt 1 1 = Displacement of particle in time t vt 2 2 = \ Displacement of the particle in the time interval Dt t t ( ) = - 2 1 = - vt vt 2 1 = - v t t ( ) 2 1 \ Average velocity in the time interval Dt v t t t t ¢ = - - ( ) ( ) 2 1 2 1 = v Now, as the particle is moving with constant velocity (i.e., with constant speed in a given direction) its velocity and speed at any instant will obviously be v. Ans. True. 2. As the stone would be free to acceleration under earth’s gravity it acceleration will be g. 3. A second hand takes 1 min i.e., 60s to complete one rotation (i.e., rotation by an angle of 2p rad). \ Angular speed of second hand = 2 60 p rad s = p 30 rad s -1 Linear speed of its tip = radius ´ angular speed = ´ - 2.0 cm rad s p 30 1 = - p 15 1 cms As the tip would be moving with constant speed. Average speed = - p 15 1 cms In 15 s the second hand would rotate through 90° i.e., the displacement of its tip will be r 2. \ Modulus of average velocity of the tip of second hand in 15 s. = r 2 15 = - 2 2 15 1 cms 4. (a) Yes. By changing direction of motion, there will be change in velocity and so acceleration. (b) (i) No. In curved path there will always be acceleration. (As explained in the previous answer no. 3) (ii) Yes. In projectile motion the path of the particle is a curved one while acceleration of the particle remains constant. (iii) Yes. In curved path the acceleration will always be there. Even if the path is circular with constant speed the direction of the acceleration of the particle would every time be changing. 3 Motion in One Dimension 5. (a) Time speed = Circumference Speed = ´ 2 4 1 p cm cm/s =8p =25.13 s (b)As particle is moving with constant speed of 1 cm/s, its average speed in any time interval will be 1 cm/s. | | / Average velocity = r T 2 4 = æ è ç ö ø ÷ 4 2 2 r r p speed = 2 2 p speed = 2 2 p cms -1 = 0.9 cms -1 | | / Average acceleration = v T 2 4 (where v = speed) = 4 2 2 v r v p = 2 2 2 p v r = 2 2 1 4 2 p ( ) = - 0.23 cm s 2 6. Distance = Speed ´ time D v t 1 1 1 = D v t 2 2 2 = Average speed = + + = + + D D t t v t v t t t 1 2 1 2 1 1 2 2 1 2 = ´ + ´ + ( ) ( ) 4 2 6 3 2 3 = - 5.2ms 1 Introductory Exercise 3.2 1. Acceleration (due to gravity). 2. s u at a t = + - 1 2 is physically correct as it gives the displacement of the particle in t th second (or any time unit). s t = Displacement in t seconds - displacement in ( ) t - 1 seconds = + é ë ê ù û ú - - + - é ë ê ù û ú ut at u t a t 1 2 1 1 2 1 2 2 ( ) ( ) Therefore, the given equation is dimensionally incorrect. 3. Yes. When a particle executing simple harmonic motion returns from maximum amplitude position to its mean position the value of its acceleration decreases while speed increases. 4. v t = 3 4 / (given) ds dt t = 3 4 / …(i) \ s t dt = ò 3 4 / = + + + t c 3 4 1 3 4 1 or s t c = + 4 7 7 4 / i.e., s t µ 74 / Differentiating Eq. (i) w.r.t. time t, d s dt t 2 2 3 4 1 3 4 = - Þ a t µ -1 4 / 5. Displacement (s) of the particle s= ´ + - ( ) ( ) 40 6 1 2 106 2 = - 240 180 =60 m (in the upward direction) Distance covered ( ) D by the particle Time to attain maximum height Motion in One Dimension | 15 Displacement Velocity st = u + at – 1 2 a Acceleration = = 40 10 4 s < 6 s It implies that particle has come back after attaining maximum height (h) given by h u g = 2 2 = ´ ( ) 40 2 10 2 = 80 m \ D = + - 80 80 60 ( ) =100 m 6. v t = - 40 10 \ dx dt t = - 40 10 or dx t dt = - ( ) 40 10 or x t dt = - ò ( ) 40 10 or x t t c = - + 40 5 2 As at t = 0 the value of x is zero. c =0 \ x t t = - 40 5 2 For x to be 60 m. 60 40 5 2 = - t t or t t 2 8 12 0 - + = \ t =2 s or 6 s 7. Average velocity = Displacement in time t t = + ut at t 1 2 2 = + u at 1 2 8. v v at 2 1 = + \ at v v = - 2 1 Average velocity = Displacement in time t t = + v t at t 1 2 1 2 = + v at 1 1 2 = + - v v v 1 2 1 2 = + v v 1 2 2 \ Ans. True. 9. 125 0 1 2 2 = × + t gt Þ t =25 s Average velocity = 125 5 m s (downwards) = 25 m/s (downwards) 10. v t t = + - 10 5 2 …(i ) \ a dv dt t = = - 5 2 At t =2 s a = - ´ 5 2 2 =1 m/s 2 From Eq. (i), dx dt t t = + - 10 5 2 \ x t t dt = + - ò ( ) 10 5 2 or x t t t c = + - + 10 5 2 3 2 3 As, at t = 0 the value of x is zero c =0 \ x t t t = + - 10 5 2 3 2 3 Thus, at t = 3 s x = ´ + - ( ) ( ) 10 3 5 2 3 3 3 2 3 = + - 30 9 22.5 = 43.5 m 11. u i ® = 2 ^ m/s a i j ® = ° + ° ( cos sin ) ^ ^ 2 60 2 60 m/s 2 = + ( ) ^ ^ 1 3 i j m/s 2 v u a ® ® ® = + t = + + 2 1 3 2 i i j ^ ^ ^ ( ) = + 4 2 3 i j ^ ^ 16 | Mechanics-1 60° 2 a = 2 m/s u = 2 m/s x y | | v ® = + 4 12 2 =2 7 m/s s u a ® ® ® = + t t 1 2 2 = + + ( ) ( ) ^ ^ ^ 2 2 1 2 1 3 2 2 i i j = + + 4 2 2 3 i i j ^ ^ ^ = + 6 2 3 i j ^ ^ \ | | s ® = + 36 12 = 4 3 m 12. Part I v i j ® = + ( ) ^ ^ 2 2t m/s …(i) dv j ¾® = dt 2 ^ a j ® =2 ^ m/s 2 From Eq. (i), d dt t s i j ® = + ( ) ^ ^ 2 2 \ s i j ® = + ò ( ) ^ ^ 2 2t dt s i j ® = + + 2 2 t t c ^ ^ Taking initial displacement to be zero. s ® (at t = 1 s) = + ( ) ^ ^ 2 i j m Part II Yes. As explained below. v i j ® = + 2 2 ^ ^ t implies that initial velocity of the particle is 2 i ^ m/s 2 and the acceleration is 2 $ j m/s 2 \ s ® (at t = 1 s) = ´ + ( ) ( ) ^ ^ 2 1 1 2 2 1 2 i j = + ( ) ^ ^ 2 i j m 13. x t = 2 and y t = 2 \ y x = æ è ç ö ø ÷ 2 2 or, x y 2 4 = (The above is the equation to trajectory) x t = 2 \ dx dt = 2 i.e., v i x ® = 2 ^ y t = 2 \ dy dt t = 2 i.e., v j y t ® + 2 ^ Thus, v v v ® ® ® = + x y = + ( ) ^ ^ 2 2 i j t m/s a dv j ® ¾® = = dt 2 ^ m/s 2 Introductory Exercise 3.3 1. At t t = 1 v = tan q As q < ° 90 , v t 1 is + ive. At t t = 2 v t 2 = f tan As f > ° 90 , v t 2 is - ive. Corresponding v-t graph will be Acceleration at t t = 1 : a t 1 = tan a As a < ° 90 , a t 1 is + ive constant. Acceleration at t t = 2 a t 2 = tan b As b < ° 90 , a t 2 is + ive constant. 2. Let the particle strike ground at time t velocity of particle when it touches ground Motion in One Dimension | 17 s f q 0 t 1 t 2 t v t 1 t 2 b t a b Page 5 Introductory Exercise 3.1 1. Suppose a particle is moving with constant velocity v (along the axis of x). Displacement of particle in time t vt 1 1 = Displacement of particle in time t vt 2 2 = \ Displacement of the particle in the time interval Dt t t ( ) = - 2 1 = - vt vt 2 1 = - v t t ( ) 2 1 \ Average velocity in the time interval Dt v t t t t ¢ = - - ( ) ( ) 2 1 2 1 = v Now, as the particle is moving with constant velocity (i.e., with constant speed in a given direction) its velocity and speed at any instant will obviously be v. Ans. True. 2. As the stone would be free to acceleration under earth’s gravity it acceleration will be g. 3. A second hand takes 1 min i.e., 60s to complete one rotation (i.e., rotation by an angle of 2p rad). \ Angular speed of second hand = 2 60 p rad s = p 30 rad s -1 Linear speed of its tip = radius ´ angular speed = ´ - 2.0 cm rad s p 30 1 = - p 15 1 cms As the tip would be moving with constant speed. Average speed = - p 15 1 cms In 15 s the second hand would rotate through 90° i.e., the displacement of its tip will be r 2. \ Modulus of average velocity of the tip of second hand in 15 s. = r 2 15 = - 2 2 15 1 cms 4. (a) Yes. By changing direction of motion, there will be change in velocity and so acceleration. (b) (i) No. In curved path there will always be acceleration. (As explained in the previous answer no. 3) (ii) Yes. In projectile motion the path of the particle is a curved one while acceleration of the particle remains constant. (iii) Yes. In curved path the acceleration will always be there. Even if the path is circular with constant speed the direction of the acceleration of the particle would every time be changing. 3 Motion in One Dimension 5. (a) Time speed = Circumference Speed = ´ 2 4 1 p cm cm/s =8p =25.13 s (b)As particle is moving with constant speed of 1 cm/s, its average speed in any time interval will be 1 cm/s. | | / Average velocity = r T 2 4 = æ è ç ö ø ÷ 4 2 2 r r p speed = 2 2 p speed = 2 2 p cms -1 = 0.9 cms -1 | | / Average acceleration = v T 2 4 (where v = speed) = 4 2 2 v r v p = 2 2 2 p v r = 2 2 1 4 2 p ( ) = - 0.23 cm s 2 6. Distance = Speed ´ time D v t 1 1 1 = D v t 2 2 2 = Average speed = + + = + + D D t t v t v t t t 1 2 1 2 1 1 2 2 1 2 = ´ + ´ + ( ) ( ) 4 2 6 3 2 3 = - 5.2ms 1 Introductory Exercise 3.2 1. Acceleration (due to gravity). 2. s u at a t = + - 1 2 is physically correct as it gives the displacement of the particle in t th second (or any time unit). s t = Displacement in t seconds - displacement in ( ) t - 1 seconds = + é ë ê ù û ú - - + - é ë ê ù û ú ut at u t a t 1 2 1 1 2 1 2 2 ( ) ( ) Therefore, the given equation is dimensionally incorrect. 3. Yes. When a particle executing simple harmonic motion returns from maximum amplitude position to its mean position the value of its acceleration decreases while speed increases. 4. v t = 3 4 / (given) ds dt t = 3 4 / …(i) \ s t dt = ò 3 4 / = + + + t c 3 4 1 3 4 1 or s t c = + 4 7 7 4 / i.e., s t µ 74 / Differentiating Eq. (i) w.r.t. time t, d s dt t 2 2 3 4 1 3 4 = - Þ a t µ -1 4 / 5. Displacement (s) of the particle s= ´ + - ( ) ( ) 40 6 1 2 106 2 = - 240 180 =60 m (in the upward direction) Distance covered ( ) D by the particle Time to attain maximum height Motion in One Dimension | 15 Displacement Velocity st = u + at – 1 2 a Acceleration = = 40 10 4 s < 6 s It implies that particle has come back after attaining maximum height (h) given by h u g = 2 2 = ´ ( ) 40 2 10 2 = 80 m \ D = + - 80 80 60 ( ) =100 m 6. v t = - 40 10 \ dx dt t = - 40 10 or dx t dt = - ( ) 40 10 or x t dt = - ò ( ) 40 10 or x t t c = - + 40 5 2 As at t = 0 the value of x is zero. c =0 \ x t t = - 40 5 2 For x to be 60 m. 60 40 5 2 = - t t or t t 2 8 12 0 - + = \ t =2 s or 6 s 7. Average velocity = Displacement in time t t = + ut at t 1 2 2 = + u at 1 2 8. v v at 2 1 = + \ at v v = - 2 1 Average velocity = Displacement in time t t = + v t at t 1 2 1 2 = + v at 1 1 2 = + - v v v 1 2 1 2 = + v v 1 2 2 \ Ans. True. 9. 125 0 1 2 2 = × + t gt Þ t =25 s Average velocity = 125 5 m s (downwards) = 25 m/s (downwards) 10. v t t = + - 10 5 2 …(i ) \ a dv dt t = = - 5 2 At t =2 s a = - ´ 5 2 2 =1 m/s 2 From Eq. (i), dx dt t t = + - 10 5 2 \ x t t dt = + - ò ( ) 10 5 2 or x t t t c = + - + 10 5 2 3 2 3 As, at t = 0 the value of x is zero c =0 \ x t t t = + - 10 5 2 3 2 3 Thus, at t = 3 s x = ´ + - ( ) ( ) 10 3 5 2 3 3 3 2 3 = + - 30 9 22.5 = 43.5 m 11. u i ® = 2 ^ m/s a i j ® = ° + ° ( cos sin ) ^ ^ 2 60 2 60 m/s 2 = + ( ) ^ ^ 1 3 i j m/s 2 v u a ® ® ® = + t = + + 2 1 3 2 i i j ^ ^ ^ ( ) = + 4 2 3 i j ^ ^ 16 | Mechanics-1 60° 2 a = 2 m/s u = 2 m/s x y | | v ® = + 4 12 2 =2 7 m/s s u a ® ® ® = + t t 1 2 2 = + + ( ) ( ) ^ ^ ^ 2 2 1 2 1 3 2 2 i i j = + + 4 2 2 3 i i j ^ ^ ^ = + 6 2 3 i j ^ ^ \ | | s ® = + 36 12 = 4 3 m 12. Part I v i j ® = + ( ) ^ ^ 2 2t m/s …(i) dv j ¾® = dt 2 ^ a j ® =2 ^ m/s 2 From Eq. (i), d dt t s i j ® = + ( ) ^ ^ 2 2 \ s i j ® = + ò ( ) ^ ^ 2 2t dt s i j ® = + + 2 2 t t c ^ ^ Taking initial displacement to be zero. s ® (at t = 1 s) = + ( ) ^ ^ 2 i j m Part II Yes. As explained below. v i j ® = + 2 2 ^ ^ t implies that initial velocity of the particle is 2 i ^ m/s 2 and the acceleration is 2 $ j m/s 2 \ s ® (at t = 1 s) = ´ + ( ) ( ) ^ ^ 2 1 1 2 2 1 2 i j = + ( ) ^ ^ 2 i j m 13. x t = 2 and y t = 2 \ y x = æ è ç ö ø ÷ 2 2 or, x y 2 4 = (The above is the equation to trajectory) x t = 2 \ dx dt = 2 i.e., v i x ® = 2 ^ y t = 2 \ dy dt t = 2 i.e., v j y t ® + 2 ^ Thus, v v v ® ® ® = + x y = + ( ) ^ ^ 2 2 i j t m/s a dv j ® ¾® = = dt 2 ^ m/s 2 Introductory Exercise 3.3 1. At t t = 1 v = tan q As q < ° 90 , v t 1 is + ive. At t t = 2 v t 2 = f tan As f > ° 90 , v t 2 is - ive. Corresponding v-t graph will be Acceleration at t t = 1 : a t 1 = tan a As a < ° 90 , a t 1 is + ive constant. Acceleration at t t = 2 a t 2 = tan b As b < ° 90 , a t 2 is + ive constant. 2. Let the particle strike ground at time t velocity of particle when it touches ground Motion in One Dimension | 17 s f q 0 t 1 t 2 t v t 1 t 2 b t a b would be gt. KE of particle will be 1 2 2 2 mg t i.e., KE µ t 2 . While going up the velocity will get - ive but the KE will remain. KE will reduce to zero at time 2t when the particle reaches its initial position. KE = 1 2 2 2 mg t = 1 2 2 2 mg h g = mgh 3. Speed of ball (just before making first collision with floor) = 2gh = ´ ´ 2 10 80 = 40 m/s Time taken to reach ground = 2h g = ´ 2 80 10 = 4 s Speed of ball (just after first collision with floor) = = 40 2 20 m/s Time to attain maximum height t = - - = 20 10 2 s \ Time for the return journey to floor = 2 s. Corresponding velocity-time will be 4. h t ( ) tan ( ) - = = - 2 2 2 1 q Þ h t = - 2 2 ( ) Particle will attain its initial velocity i.e., net increase in velocity of the particle will be zero when, area under a-t graph = 0 ( ) ( ) ( ) 1 2 2 2 2 2 0 + ´ + - - = h t or 3 2 0 2 - - = ( ) t or ( ) t - = 2 3 2 or t - = ± 2 3 or t = ± 2 3 Ans . At time t = + 2 3 s (t = - 2 3 not possible). Introductory Exercise 3.4 1. Relative acceleration of A w.r.t. B a g g AB = + - + ( ) ( ) = 0 2. Velocity of A w.r.t. B v v A B = - \ Relative displacement (i.e., distance between A and B) would be s v v t a t A B AB = - + ( ) 1 2 2 18 | Mechanics-1 KE 2t time 2h g = t 8 4 t (s) Speed (m/s) 8 4 t (s) Velocity (m/s) 2 a (m/s ) 2 1 2 (t–2) t h q qRead More

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