# 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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The text passage contains a detailed physics problem involving a train's speed, deceleration, and acceleration over a viaduct. It includes a question with multiple parts and a provided answer.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# 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
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### 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%
user
The text passage contains a detailed physics problem involving a train's speed, deceleration, and acceleration over a viaduct. It includes a question with multiple parts and a provided answer.

TYPE: SINGLE_QUESTION_WITH_ANSWER