Saved Bookmarks
| 1. |
15. A train starts from rest and accelerates uniformly at a rate of 2 m s-2 for 10 s. It then maintains aconstant speed for 200 s. The brakes are thenapplied and the train is uniformly retarded andcomes to rest in 50 s. Find : (i) the maximumvelocity reached, (ii) the retardation in the last50 s, (iii) the total distance travelled, and (iv) theaverage velocity of the train.-1 G 0.4 |
Answer» ANSWER:
GIVEN:
TO FIND:
EXPLANATION:1) The maximum velocity reached:The maximum velocity should be reached after 10 seconds, because after that it travels at constant speed and again retards. u = 0 a = 2 m/s² t = 10 s v = 0 + 2(10) v = 20 m/s The maximum velocity will be 20 m/s. 2) The retardation in the last:The initial velocity is what we found above because after 10 s the train travels at a constant speed, then starts retarding and at last it comes to rest(v = 0). u = 20 m/s v = 0 t = 50s 0 = 20 + a(50) 50A = - 20 a = - 2/5 a = - 0.4 m/s² It retards uniformly at a rate of 0.4 m/s². 3) The total distance travelled:Lets split the distance as THREE parts A,B,C First part: A = 0(10) + 1/2(2)(10²) A = 1/2(2 × 100) A = 100 m Second part: Here velocity is constant and hence accleration is zero B = 20 × 200 B = 4000 m Third part: C = 20(50) + 1/2(- 0.4)(50)(50) C = 1000 - (2500)/5 C = 1000 - 500 C = 500 m Total distance = 100 + 4000 + 500 Total distance = 4600 m. 4) The average velocity of the train:Avg velocity = 17.69 m/s. |
|