Page 1 3.1 SOLUTIONS TO CONCEPTS CHAPTER – 3 1. a) Distance travelled = 50 + 40 + 20 = 110 m b) AF = AB – BF = AB – DC = 50 – 20 = 30 M His displacement is AD AD = m 50 40 30 DF AF 2 2 2 2 ? ? ? ? In ?AED tan ? = DE/AE = 30/40 = 3/4 ? ? = tan –1 (3/4) His displacement from his house to the field is 50 m, tan –1 (3/4) north to east. 2. O ? Starting point origin. i) Distance travelled = 20 + 20 + 20 = 60 m ii) Displacement is only OB = 20 m in the negative direction. Displacement ? Distance between final and initial position. 3. a) V ave of plane (Distance/Time) = 260/0.5 = 520 km/hr. b) V ave of bus = 320/8 = 40 km/hr. c) plane goes in straight path velocity = ave V ? = 260/0.5 = 520 km/hr. d) Straight path distance between plane to Ranchi is equal to the displacement of bus. ? Velocity = ave V ? = 260/8 = 32.5 km/hr. 4. a) Total distance covered 12416 – 12352 = 64 km in 2 hours. Speed = 64/2 = 32 km/h b) As he returns to his house, the displacement is zero. Velocity = (displacement/time) = 0 (zero). 5. Initial velocity u = 0 ( ? starts from rest) Final velocity v = 18 km/hr = 5 sec (i.e. max velocity) Time interval t = 2 sec. ? Acceleration = a ave = 2 5 t u v ? ? = 2.5 m/s 2 . 6. In the interval 8 sec the velocity changes from 0 to 20 m/s. Average acceleration = 20/8 = 2.5 m/s 2 ? ? ? ? ? ? time velocity in change Distance travelled S = ut + 1/2 at 2 ? 0 + 1/2(2.5)8 2 = 80 m. 7. In 1 st 10 sec S 1 = ut + 1/2 at 2 ? 0 + (1/2 × 5 × 10 2 ) = 250 ft. At 10 sec v = u + at = 0 + 5 × 10 = 50 ft/sec. ? From 10 to 20 sec ( ?t = 20 – 10 = 10 sec) it moves with uniform velocity 50 ft/sec, A E S N W ? ? 40 m 40 m 50 m 20 m 30 m B C D E A ? Initial point (starting point) A X O Y B (20 m, 0) (–20 m, 0) Initial velocity u = 0 20 8 4 10 Time in sec 20 10 30 S (in ft) 0 t (sec) 250 750 1000 Page 2 3.1 SOLUTIONS TO CONCEPTS CHAPTER – 3 1. a) Distance travelled = 50 + 40 + 20 = 110 m b) AF = AB – BF = AB – DC = 50 – 20 = 30 M His displacement is AD AD = m 50 40 30 DF AF 2 2 2 2 ? ? ? ? In ?AED tan ? = DE/AE = 30/40 = 3/4 ? ? = tan –1 (3/4) His displacement from his house to the field is 50 m, tan –1 (3/4) north to east. 2. O ? Starting point origin. i) Distance travelled = 20 + 20 + 20 = 60 m ii) Displacement is only OB = 20 m in the negative direction. Displacement ? Distance between final and initial position. 3. a) V ave of plane (Distance/Time) = 260/0.5 = 520 km/hr. b) V ave of bus = 320/8 = 40 km/hr. c) plane goes in straight path velocity = ave V ? = 260/0.5 = 520 km/hr. d) Straight path distance between plane to Ranchi is equal to the displacement of bus. ? Velocity = ave V ? = 260/8 = 32.5 km/hr. 4. a) Total distance covered 12416 – 12352 = 64 km in 2 hours. Speed = 64/2 = 32 km/h b) As he returns to his house, the displacement is zero. Velocity = (displacement/time) = 0 (zero). 5. Initial velocity u = 0 ( ? starts from rest) Final velocity v = 18 km/hr = 5 sec (i.e. max velocity) Time interval t = 2 sec. ? Acceleration = a ave = 2 5 t u v ? ? = 2.5 m/s 2 . 6. In the interval 8 sec the velocity changes from 0 to 20 m/s. Average acceleration = 20/8 = 2.5 m/s 2 ? ? ? ? ? ? time velocity in change Distance travelled S = ut + 1/2 at 2 ? 0 + 1/2(2.5)8 2 = 80 m. 7. In 1 st 10 sec S 1 = ut + 1/2 at 2 ? 0 + (1/2 × 5 × 10 2 ) = 250 ft. At 10 sec v = u + at = 0 + 5 × 10 = 50 ft/sec. ? From 10 to 20 sec ( ?t = 20 – 10 = 10 sec) it moves with uniform velocity 50 ft/sec, A E S N W ? ? 40 m 40 m 50 m 20 m 30 m B C D E A ? Initial point (starting point) A X O Y B (20 m, 0) (–20 m, 0) Initial velocity u = 0 20 8 4 10 Time in sec 20 10 30 S (in ft) 0 t (sec) 250 750 1000 Chapter-3 3.2 Distance S 2 = 50 × 10 = 500 ft Between 20 sec to 30 sec acceleration is constant i.e. –5 ft/s 2 . At 20 sec velocity is 50 ft/sec. t = 30 – 20 = 10 s S 3 = ut + 1/2 at 2 = 50 × 10 + (1/2)(–5)(10) 2 = 250 m Total distance travelled is 30 sec = S 1 + S 2 + S 3 = 250 + 500 + 250 = 1000 ft. 8. a) Initial velocity u = 2 m/s. final velocity v = 8 m/s time = 10 sec, acceleration = 10 2 8 ta u v ? ? ? = 0.6 m/s 2 b) v 2 – u 2 = 2aS ? Distance S = a 2 u v 2 2 ? = 6 . 0 2 2 8 2 2 ? ? = 50 m. c) Displacement is same as distance travelled. Displacement = 50 m. 9. a) Displacement in 0 to 10 sec is 1000 m. time = 10 sec. V ave = s/t = 100/10 = 10 m/s. b) At 2 sec it is moving with uniform velocity 50/2.5 = 20 m/s. at 2 sec. V inst = 20 m/s. At 5 sec it is at rest. V inst = zero. At 8 sec it is moving with uniform velocity 20 m/s V inst = 20 m/s At 12 sec velocity is negative as it move towards initial position. V inst = – 20 m/s. 10. Distance in first 40 sec is, ? OAB + ?BCD = 2 1 × 5 × 20 + 2 1 × 5 × 20 = 100 m. Average velocity is 0 as the displacement is zero. 11. Consider the point B, at t = 12 sec At t = 0 ; s = 20 m and t = 12 sec s = 20 m So for time interval 0 to 12 sec Change in displacement is zero. So, average velocity = displacement/ time = 0 ? The time is 12 sec. 12. At position B instantaneous velocity has direction along BC . For average velocity between A and B. V ave = displacement / time = ) t / AB ( t = time 10 t 5 2 4 6 8 t 10 (slope of the graph at t = 2 sec) 2.5 50 100 0 t 5 7.5 15 40 t (sec) 20 5 m/s O A B C D 20 B 10 10 12 20 4 B 2 C 4 2 x 6 y Page 3 3.1 SOLUTIONS TO CONCEPTS CHAPTER – 3 1. a) Distance travelled = 50 + 40 + 20 = 110 m b) AF = AB – BF = AB – DC = 50 – 20 = 30 M His displacement is AD AD = m 50 40 30 DF AF 2 2 2 2 ? ? ? ? In ?AED tan ? = DE/AE = 30/40 = 3/4 ? ? = tan –1 (3/4) His displacement from his house to the field is 50 m, tan –1 (3/4) north to east. 2. O ? Starting point origin. i) Distance travelled = 20 + 20 + 20 = 60 m ii) Displacement is only OB = 20 m in the negative direction. Displacement ? Distance between final and initial position. 3. a) V ave of plane (Distance/Time) = 260/0.5 = 520 km/hr. b) V ave of bus = 320/8 = 40 km/hr. c) plane goes in straight path velocity = ave V ? = 260/0.5 = 520 km/hr. d) Straight path distance between plane to Ranchi is equal to the displacement of bus. ? Velocity = ave V ? = 260/8 = 32.5 km/hr. 4. a) Total distance covered 12416 – 12352 = 64 km in 2 hours. Speed = 64/2 = 32 km/h b) As he returns to his house, the displacement is zero. Velocity = (displacement/time) = 0 (zero). 5. Initial velocity u = 0 ( ? starts from rest) Final velocity v = 18 km/hr = 5 sec (i.e. max velocity) Time interval t = 2 sec. ? Acceleration = a ave = 2 5 t u v ? ? = 2.5 m/s 2 . 6. In the interval 8 sec the velocity changes from 0 to 20 m/s. Average acceleration = 20/8 = 2.5 m/s 2 ? ? ? ? ? ? time velocity in change Distance travelled S = ut + 1/2 at 2 ? 0 + 1/2(2.5)8 2 = 80 m. 7. In 1 st 10 sec S 1 = ut + 1/2 at 2 ? 0 + (1/2 × 5 × 10 2 ) = 250 ft. At 10 sec v = u + at = 0 + 5 × 10 = 50 ft/sec. ? From 10 to 20 sec ( ?t = 20 – 10 = 10 sec) it moves with uniform velocity 50 ft/sec, A E S N W ? ? 40 m 40 m 50 m 20 m 30 m B C D E A ? Initial point (starting point) A X O Y B (20 m, 0) (–20 m, 0) Initial velocity u = 0 20 8 4 10 Time in sec 20 10 30 S (in ft) 0 t (sec) 250 750 1000 Chapter-3 3.2 Distance S 2 = 50 × 10 = 500 ft Between 20 sec to 30 sec acceleration is constant i.e. –5 ft/s 2 . At 20 sec velocity is 50 ft/sec. t = 30 – 20 = 10 s S 3 = ut + 1/2 at 2 = 50 × 10 + (1/2)(–5)(10) 2 = 250 m Total distance travelled is 30 sec = S 1 + S 2 + S 3 = 250 + 500 + 250 = 1000 ft. 8. a) Initial velocity u = 2 m/s. final velocity v = 8 m/s time = 10 sec, acceleration = 10 2 8 ta u v ? ? ? = 0.6 m/s 2 b) v 2 – u 2 = 2aS ? Distance S = a 2 u v 2 2 ? = 6 . 0 2 2 8 2 2 ? ? = 50 m. c) Displacement is same as distance travelled. Displacement = 50 m. 9. a) Displacement in 0 to 10 sec is 1000 m. time = 10 sec. V ave = s/t = 100/10 = 10 m/s. b) At 2 sec it is moving with uniform velocity 50/2.5 = 20 m/s. at 2 sec. V inst = 20 m/s. At 5 sec it is at rest. V inst = zero. At 8 sec it is moving with uniform velocity 20 m/s V inst = 20 m/s At 12 sec velocity is negative as it move towards initial position. V inst = – 20 m/s. 10. Distance in first 40 sec is, ? OAB + ?BCD = 2 1 × 5 × 20 + 2 1 × 5 × 20 = 100 m. Average velocity is 0 as the displacement is zero. 11. Consider the point B, at t = 12 sec At t = 0 ; s = 20 m and t = 12 sec s = 20 m So for time interval 0 to 12 sec Change in displacement is zero. So, average velocity = displacement/ time = 0 ? The time is 12 sec. 12. At position B instantaneous velocity has direction along BC . For average velocity between A and B. V ave = displacement / time = ) t / AB ( t = time 10 t 5 2 4 6 8 t 10 (slope of the graph at t = 2 sec) 2.5 50 100 0 t 5 7.5 15 40 t (sec) 20 5 m/s O A B C D 20 B 10 10 12 20 4 B 2 C 4 2 x 6 y Chapter-3 3.3 We can see that AB is along BC i.e. they are in same direction. The point is B (5m, 3m). 13. u = 4 m/s, a = 1.2 m/s 2 , t = 5 sec Distance = s = 2 at 2 1 ut ? = 4(5) + 1/2 (1.2)5 2 = 35 m. 14. Initial velocity u = 43.2 km/hr = 12 m/s u = 12 m/s, v = 0 a = –6 m/s 2 (deceleration) Distance S = ) 6 ( 2 u v 2 2 ? ? = 12 m Page 4 3.1 SOLUTIONS TO CONCEPTS CHAPTER – 3 1. a) Distance travelled = 50 + 40 + 20 = 110 m b) AF = AB – BF = AB – DC = 50 – 20 = 30 M His displacement is AD AD = m 50 40 30 DF AF 2 2 2 2 ? ? ? ? In ?AED tan ? = DE/AE = 30/40 = 3/4 ? ? = tan –1 (3/4) His displacement from his house to the field is 50 m, tan –1 (3/4) north to east. 2. O ? Starting point origin. i) Distance travelled = 20 + 20 + 20 = 60 m ii) Displacement is only OB = 20 m in the negative direction. Displacement ? Distance between final and initial position. 3. a) V ave of plane (Distance/Time) = 260/0.5 = 520 km/hr. b) V ave of bus = 320/8 = 40 km/hr. c) plane goes in straight path velocity = ave V ? = 260/0.5 = 520 km/hr. d) Straight path distance between plane to Ranchi is equal to the displacement of bus. ? Velocity = ave V ? = 260/8 = 32.5 km/hr. 4. a) Total distance covered 12416 – 12352 = 64 km in 2 hours. Speed = 64/2 = 32 km/h b) As he returns to his house, the displacement is zero. Velocity = (displacement/time) = 0 (zero). 5. Initial velocity u = 0 ( ? starts from rest) Final velocity v = 18 km/hr = 5 sec (i.e. max velocity) Time interval t = 2 sec. ? Acceleration = a ave = 2 5 t u v ? ? = 2.5 m/s 2 . 6. In the interval 8 sec the velocity changes from 0 to 20 m/s. Average acceleration = 20/8 = 2.5 m/s 2 ? ? ? ? ? ? time velocity in change Distance travelled S = ut + 1/2 at 2 ? 0 + 1/2(2.5)8 2 = 80 m. 7. In 1 st 10 sec S 1 = ut + 1/2 at 2 ? 0 + (1/2 × 5 × 10 2 ) = 250 ft. At 10 sec v = u + at = 0 + 5 × 10 = 50 ft/sec. ? From 10 to 20 sec ( ?t = 20 – 10 = 10 sec) it moves with uniform velocity 50 ft/sec, A E S N W ? ? 40 m 40 m 50 m 20 m 30 m B C D E A ? Initial point (starting point) A X O Y B (20 m, 0) (–20 m, 0) Initial velocity u = 0 20 8 4 10 Time in sec 20 10 30 S (in ft) 0 t (sec) 250 750 1000 Chapter-3 3.2 Distance S 2 = 50 × 10 = 500 ft Between 20 sec to 30 sec acceleration is constant i.e. –5 ft/s 2 . At 20 sec velocity is 50 ft/sec. t = 30 – 20 = 10 s S 3 = ut + 1/2 at 2 = 50 × 10 + (1/2)(–5)(10) 2 = 250 m Total distance travelled is 30 sec = S 1 + S 2 + S 3 = 250 + 500 + 250 = 1000 ft. 8. a) Initial velocity u = 2 m/s. final velocity v = 8 m/s time = 10 sec, acceleration = 10 2 8 ta u v ? ? ? = 0.6 m/s 2 b) v 2 – u 2 = 2aS ? Distance S = a 2 u v 2 2 ? = 6 . 0 2 2 8 2 2 ? ? = 50 m. c) Displacement is same as distance travelled. Displacement = 50 m. 9. a) Displacement in 0 to 10 sec is 1000 m. time = 10 sec. V ave = s/t = 100/10 = 10 m/s. b) At 2 sec it is moving with uniform velocity 50/2.5 = 20 m/s. at 2 sec. V inst = 20 m/s. At 5 sec it is at rest. V inst = zero. At 8 sec it is moving with uniform velocity 20 m/s V inst = 20 m/s At 12 sec velocity is negative as it move towards initial position. V inst = – 20 m/s. 10. Distance in first 40 sec is, ? OAB + ?BCD = 2 1 × 5 × 20 + 2 1 × 5 × 20 = 100 m. Average velocity is 0 as the displacement is zero. 11. Consider the point B, at t = 12 sec At t = 0 ; s = 20 m and t = 12 sec s = 20 m So for time interval 0 to 12 sec Change in displacement is zero. So, average velocity = displacement/ time = 0 ? The time is 12 sec. 12. At position B instantaneous velocity has direction along BC . For average velocity between A and B. V ave = displacement / time = ) t / AB ( t = time 10 t 5 2 4 6 8 t 10 (slope of the graph at t = 2 sec) 2.5 50 100 0 t 5 7.5 15 40 t (sec) 20 5 m/s O A B C D 20 B 10 10 12 20 4 B 2 C 4 2 x 6 y Chapter-3 3.3 We can see that AB is along BC i.e. they are in same direction. The point is B (5m, 3m). 13. u = 4 m/s, a = 1.2 m/s 2 , t = 5 sec Distance = s = 2 at 2 1 ut ? = 4(5) + 1/2 (1.2)5 2 = 35 m. 14. Initial velocity u = 43.2 km/hr = 12 m/s u = 12 m/s, v = 0 a = –6 m/s 2 (deceleration) Distance S = ) 6 ( 2 u v 2 2 ? ? = 12 m Chapter-3 3.4 15. Initial velocity u = 0 Acceleration a = 2 m/s 2 . Let final velocity be v (before applying breaks) t = 30 sec v = u + at ? 0 + 2 × 30 = 60 m/s a) S 1 = 2 at 2 1 ut ? = 900 m when breaks are applied u ? = 60 m/s v ? = 0, t = 60 sec (1 min) Declaration a ? = (v – u)/t = = (0 – 60)/60 = –1 m/s 2 . S 2 = a 2 u v 2 2 ? ? ? ? = 1800 m Total S = S 1 + S 2 = 1800 + 900 = 2700 m = 2.7 km. b) The maximum speed attained by train v = 60 m/s c) Half the maximum speed = 60/2= 30 m/s Distance S = a 2 u v 2 2 ? = 2 2 0 30 2 2 ? ? = 225 m from starting point When it accelerates the distance travelled is 900 m. Then again declarates and attain 30 m/s. ? u = 60 m/s, v = 30 m/s, a = –1 m/s 2 Distance = a 2 u v 2 2 ? = ) 1 ( 2 60 30 2 2 ? ? = 1350 m Position is 900 + 1350 = 2250 = 2.25 km from starting point. 16. u = 16 m/s (initial), v = 0, s = 0.4 m. Deceleration a = s 2 u v 2 2 ? = –320 m/s 2 . Time = t = 320 16 0 a u v ? ? ? ? = 0.05 sec. 17. u = 350 m/s, s = 5 cm = 0.05 m, v = 0 Deceleration = a = s 2 u v 2 2 ? = 05 . 0 2 ) 350 ( 0 2 ? ? = –12.2 × 10 5 m/s 2 . Deceleration is 12.2 × 10 5 m/s 2 . 18. u = 0, v = 18 km/hr = 5 m/s, t = 5 sec a = 5 0 5 t u v ? ? ? = 1 m/s 2 . s = 2 at 2 1 ut ? = 12.5 m a) Average velocity V ave = (12.5)/5 = 2.5 m/s. b) Distance travelled is 12.5 m. 19. In reaction time the body moves with the speed 54 km/hr = 15 m/sec (constant speed) Distance travelled in this time is S 1 = 15 × 0.2 = 3 m. When brakes are applied, u = 15 m/s, v = 0, a = –6 m/s 2 (deceleration) Page 5 3.1 SOLUTIONS TO CONCEPTS CHAPTER – 3 1. a) Distance travelled = 50 + 40 + 20 = 110 m b) AF = AB – BF = AB – DC = 50 – 20 = 30 M His displacement is AD AD = m 50 40 30 DF AF 2 2 2 2 ? ? ? ? In ?AED tan ? = DE/AE = 30/40 = 3/4 ? ? = tan –1 (3/4) His displacement from his house to the field is 50 m, tan –1 (3/4) north to east. 2. O ? Starting point origin. i) Distance travelled = 20 + 20 + 20 = 60 m ii) Displacement is only OB = 20 m in the negative direction. Displacement ? Distance between final and initial position. 3. a) V ave of plane (Distance/Time) = 260/0.5 = 520 km/hr. b) V ave of bus = 320/8 = 40 km/hr. c) plane goes in straight path velocity = ave V ? = 260/0.5 = 520 km/hr. d) Straight path distance between plane to Ranchi is equal to the displacement of bus. ? Velocity = ave V ? = 260/8 = 32.5 km/hr. 4. a) Total distance covered 12416 – 12352 = 64 km in 2 hours. Speed = 64/2 = 32 km/h b) As he returns to his house, the displacement is zero. Velocity = (displacement/time) = 0 (zero). 5. Initial velocity u = 0 ( ? starts from rest) Final velocity v = 18 km/hr = 5 sec (i.e. max velocity) Time interval t = 2 sec. ? Acceleration = a ave = 2 5 t u v ? ? = 2.5 m/s 2 . 6. In the interval 8 sec the velocity changes from 0 to 20 m/s. Average acceleration = 20/8 = 2.5 m/s 2 ? ? ? ? ? ? time velocity in change Distance travelled S = ut + 1/2 at 2 ? 0 + 1/2(2.5)8 2 = 80 m. 7. In 1 st 10 sec S 1 = ut + 1/2 at 2 ? 0 + (1/2 × 5 × 10 2 ) = 250 ft. At 10 sec v = u + at = 0 + 5 × 10 = 50 ft/sec. ? From 10 to 20 sec ( ?t = 20 – 10 = 10 sec) it moves with uniform velocity 50 ft/sec, A E S N W ? ? 40 m 40 m 50 m 20 m 30 m B C D E A ? Initial point (starting point) A X O Y B (20 m, 0) (–20 m, 0) Initial velocity u = 0 20 8 4 10 Time in sec 20 10 30 S (in ft) 0 t (sec) 250 750 1000 Chapter-3 3.2 Distance S 2 = 50 × 10 = 500 ft Between 20 sec to 30 sec acceleration is constant i.e. –5 ft/s 2 . At 20 sec velocity is 50 ft/sec. t = 30 – 20 = 10 s S 3 = ut + 1/2 at 2 = 50 × 10 + (1/2)(–5)(10) 2 = 250 m Total distance travelled is 30 sec = S 1 + S 2 + S 3 = 250 + 500 + 250 = 1000 ft. 8. a) Initial velocity u = 2 m/s. final velocity v = 8 m/s time = 10 sec, acceleration = 10 2 8 ta u v ? ? ? = 0.6 m/s 2 b) v 2 – u 2 = 2aS ? Distance S = a 2 u v 2 2 ? = 6 . 0 2 2 8 2 2 ? ? = 50 m. c) Displacement is same as distance travelled. Displacement = 50 m. 9. a) Displacement in 0 to 10 sec is 1000 m. time = 10 sec. V ave = s/t = 100/10 = 10 m/s. b) At 2 sec it is moving with uniform velocity 50/2.5 = 20 m/s. at 2 sec. V inst = 20 m/s. At 5 sec it is at rest. V inst = zero. At 8 sec it is moving with uniform velocity 20 m/s V inst = 20 m/s At 12 sec velocity is negative as it move towards initial position. V inst = – 20 m/s. 10. Distance in first 40 sec is, ? OAB + ?BCD = 2 1 × 5 × 20 + 2 1 × 5 × 20 = 100 m. Average velocity is 0 as the displacement is zero. 11. Consider the point B, at t = 12 sec At t = 0 ; s = 20 m and t = 12 sec s = 20 m So for time interval 0 to 12 sec Change in displacement is zero. So, average velocity = displacement/ time = 0 ? The time is 12 sec. 12. At position B instantaneous velocity has direction along BC . For average velocity between A and B. V ave = displacement / time = ) t / AB ( t = time 10 t 5 2 4 6 8 t 10 (slope of the graph at t = 2 sec) 2.5 50 100 0 t 5 7.5 15 40 t (sec) 20 5 m/s O A B C D 20 B 10 10 12 20 4 B 2 C 4 2 x 6 y Chapter-3 3.3 We can see that AB is along BC i.e. they are in same direction. The point is B (5m, 3m). 13. u = 4 m/s, a = 1.2 m/s 2 , t = 5 sec Distance = s = 2 at 2 1 ut ? = 4(5) + 1/2 (1.2)5 2 = 35 m. 14. Initial velocity u = 43.2 km/hr = 12 m/s u = 12 m/s, v = 0 a = –6 m/s 2 (deceleration) Distance S = ) 6 ( 2 u v 2 2 ? ? = 12 m Chapter-3 3.4 15. Initial velocity u = 0 Acceleration a = 2 m/s 2 . Let final velocity be v (before applying breaks) t = 30 sec v = u + at ? 0 + 2 × 30 = 60 m/s a) S 1 = 2 at 2 1 ut ? = 900 m when breaks are applied u ? = 60 m/s v ? = 0, t = 60 sec (1 min) Declaration a ? = (v – u)/t = = (0 – 60)/60 = –1 m/s 2 . S 2 = a 2 u v 2 2 ? ? ? ? = 1800 m Total S = S 1 + S 2 = 1800 + 900 = 2700 m = 2.7 km. b) The maximum speed attained by train v = 60 m/s c) Half the maximum speed = 60/2= 30 m/s Distance S = a 2 u v 2 2 ? = 2 2 0 30 2 2 ? ? = 225 m from starting point When it accelerates the distance travelled is 900 m. Then again declarates and attain 30 m/s. ? u = 60 m/s, v = 30 m/s, a = –1 m/s 2 Distance = a 2 u v 2 2 ? = ) 1 ( 2 60 30 2 2 ? ? = 1350 m Position is 900 + 1350 = 2250 = 2.25 km from starting point. 16. u = 16 m/s (initial), v = 0, s = 0.4 m. Deceleration a = s 2 u v 2 2 ? = –320 m/s 2 . Time = t = 320 16 0 a u v ? ? ? ? = 0.05 sec. 17. u = 350 m/s, s = 5 cm = 0.05 m, v = 0 Deceleration = a = s 2 u v 2 2 ? = 05 . 0 2 ) 350 ( 0 2 ? ? = –12.2 × 10 5 m/s 2 . Deceleration is 12.2 × 10 5 m/s 2 . 18. u = 0, v = 18 km/hr = 5 m/s, t = 5 sec a = 5 0 5 t u v ? ? ? = 1 m/s 2 . s = 2 at 2 1 ut ? = 12.5 m a) Average velocity V ave = (12.5)/5 = 2.5 m/s. b) Distance travelled is 12.5 m. 19. In reaction time the body moves with the speed 54 km/hr = 15 m/sec (constant speed) Distance travelled in this time is S 1 = 15 × 0.2 = 3 m. When brakes are applied, u = 15 m/s, v = 0, a = –6 m/s 2 (deceleration) Chapter-3 3.5 S 2 = ) 6 ( 2 15 0 a 2 u v 2 2 2 ? ? ? ? = 18.75 m Total distance s = s 1 + s 2 = 3 + 18.75 = 21.75 = 22 m.Read More

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