text: # Speed-time graph questionWatch
  
  This discussion is closed.
  #1
  A straight stretch of railway line passes over a viaduct which is 600m long. An express train on this stretch of line normally travels at a speed of 50ms-1. Some structual weakness in the viaduct is detected and engineers specify that all trains passing over the viaduct must do so at a constant speed of no more than 10ms-1. Approaching the viaduct the train must therefore reduce its speed from 50ms-1 wich constant deceleration 0.5ms-2, reaching a speed of precisely 10ms-1 just as it reaches the viaduct. It then passes over the viaduct at a constant speed of 10ms-1. As soon as it reaches the other end of the viaduct it accelerates to its normal speed of 50ms-1 with constant acceleration 0.5ms-2.
  
  a) sketch a speed time graph from deceleration to acceleration.
  b) Find the total distance the train travelled under 50ms-1
  c) Find the extra time taken for the journey because of the speed restrictions..
  
  I got
  
  b) (40x80) + (10x220) = 5400m
  c) 5400/50 = 108 seconds without speed limits
  
  220 seconds with limits, resulting in extra time of 112 seconds.
  
  0
  14 years ago
  #2
  
  Time taken to slow down = 40/0.5 = 80s
  Average speed during that time = 30m/s
  Distance travelled during that time = 80*30 = 2400m
  
  Time taken to speed up = 40/0.5 = 80s
  Average speed during that time = 30m/s
  Distance travelled during that time = 80*30 = 2400m
  
  (b) 2400 + 600 + 2400 = 5400m
  
  (c)
  Original time to travel 5400m = 5400/50 = 108s
  New time to travel 5400m = 4800/30 + 600/10 = 220s
  Extra time required = 220 - 108 = 112s
  0
  #3
  (Original post by Jonny W)
  
  Time taken to slow down = 40/0.5 = 80s
  Average speed during that time = 30m/s
  Distance travelled during that time = 80*30 = 2400m
  
  Time taken to speed up = 40/0.5 = 80s
  Average speed during that time = 30m/s
  Distance travelled during that time = 80*30 = 2400m
  
  (b) 2400 + 600 + 2400 = 5400m
  
  (c)
  Original time to travel 5400m = 5400/50 = 108s
  New time to travel 5400m = 4800/30 + 600/10 = 220s
  Extra time required = 220 - 108 = 112s
  cheers jonny! that was a question for 11 marks in total!!
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explanation: The document contains a math question related to a speed-time graph and the movement of a train over a viaduct. It also includes an answer to the question, with calculations and explanations provided by a user and verified by another user.
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distilabel_metadata.content: # Speed-time graph questionWatch
  
  This discussion is closed.
  #1
  A straight stretch of railway line passes over a viaduct which is 600m long. An express train on this stretch of line normally travels at a speed of 50ms-1. Some structual weakness in the viaduct is detected and engineers specify that all trains passing over the viaduct must do so at a constant speed of no more than 10ms-1. Approaching the viaduct the train must therefore reduce its speed from 50ms-1 wich constant deceleration 0.5ms-2, reaching a speed of precisely 10ms-1 just as it reaches the viaduct. It then passes over the viaduct at a constant speed of 10ms-1. As soon as it reaches the other end of the viaduct it accelerates to its normal speed of 50ms-1 with constant acceleration 0.5ms-2.
  
  a) sketch a speed time graph from deceleration to acceleration.
  b) Find the total distance the train travelled under 50ms-1
  c) Find the extra time taken for the journey because of the speed restrictions..
  
  I got
  
  b) (40x80) + (10x220) = 5400m
  c) 5400/50 = 108 seconds without speed limits
  
  220 seconds with limits, resulting in extra time of 112 seconds.
  
  0
  14 years ago
  #2
  
  Time taken to slow down = 40/0.5 = 80s
  Average speed during that time = 30m/s
  Distance travelled during that time = 80*30 = 2400m
  
  Time taken to speed up = 40/0.5 = 80s
  Average speed during that time = 30m/s
  Distance travelled during that time = 80*30 = 2400m
  
  (b) 2400 + 600 + 2400 = 5400m
  
  (c)
  Original time to travel 5400m = 5400/50 = 108s
  New time to travel 5400m = 4800/30 + 600/10 = 220s
  Extra time required = 220 - 108 = 112s
  0
  #3
  (Original post by Jonny W)
  
  Time taken to slow down = 40/0.5 = 80s
  Average speed during that time = 30m/s
  Distance travelled during that time = 80*30 = 2400m
  
  Time taken to speed up = 40/0.5 = 80s
  Average speed during that time = 30m/s
  Distance travelled during that time = 80*30 = 2400m
  
  (b) 2400 + 600 + 2400 = 5400m
  
  (c)
  Original time to travel 5400m = 5400/50 = 108s
  New time to travel 5400m = 4800/30 + 600/10 = 220s
  Extra time required = 220 - 108 = 112s
  cheers jonny! that was a question for 11 marks in total!!
  0
  X
  new posts
  Latest
  My Feed
  
  ### Oops, nobody has postedin the last few hours.
  
  Why not re-start the conversation?
  
  see more
  
  ### See more of what you like onThe Student Room
  
  You can personalise what you see on TSR. Tell us a little about yourself to get started.
  
  ### University open days
  
  • University of the Arts London
  MA Performance Design and Practice Open Day Postgraduate
  Thu, 24 Jan '19
  • Coventry University
  Sat, 26 Jan '19
  • Brunel University London
  Sat, 26 Jan '19
  
  ### Poll
  
  Join the discussion
  
  #### Are you chained to your phone?
  
  Yes (78)
  19.35%
  Yes, but I'm trying to cut back (163)
  40.45%
  Nope, not that interesting (162)
  40.2%
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question related to a speed-time graph and the movement of a train over a viaduct. It also includes an answer to the question, with calculations and explanations provided by a user and verified by another user.
  
  Label: question-and-answer