Speed-time graph question Watch

This discussion is closed.
faa
Badges: 0
Rep:
?
#1
Report Thread starter 14 years ago
#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.

are these answers right?
0
Jonny W
Badges: 8
Rep:
?
#2
Report 14 years ago
#2
Your answers are right.

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
faa
Badges: 0
Rep:
?
#3
Report Thread starter 14 years ago
#3
(Original post by Jonny W)
Your answers are right.

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

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • University of the Arts London
    MA Performance Design and Practice Open Day Postgraduate
    Thu, 24 Jan '19
  • Coventry University
    Undergraduate Open Day Undergraduate
    Sat, 26 Jan '19
  • Brunel University London
    Undergraduate Experience Days Undergraduate
    Sat, 26 Jan '19

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%

Watched Threads

View All