An Amtrak going 250m/s comes to a stop in 12s. What is the acceleration?
1 answer:
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Answer:
(a) With a short line, the A,B,C,D parameters are:
A = 1pu B = 1.685∠60.8°Ω C = 0 S D = 1 pu
(b) The sending-end voltage for 0.9 lagging power factor is 35.96
(c) The sending-end voltage for 0.9 leading power factor is 33.40
Explanation:
(a)
Considering the short transition line diagram.
Apply kirchoff's voltage law to the short transmission line.
Write the equation showing the relations between the sending end and the receiving end quantities.
Compare the line equations with the A,B,C,D parameter equations.
(b)
Determine the receiving-end current for 0.9 lagging power factor.
Determine the line-to-neutral receiving end voltage.
Determine the sending end voltage of the short transition line.
Determine the line-to-line sending end voltage which is the sending end voltage.
(c)
Determine the receiving-end current for 0.9 leading power factor.
Determine the sending-end voltage of the short transition line.
Determine the line-to-line sending end voltage which is the sending end voltage.
Answer:
25.52 m
Explanation:
Relative speed between the person and I would be the difference of our speeds
Time taken by me to walk up to the person = 14.5 s
Distance is given by
The person was 25.52 m ahead of me when I started running