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Sagot :
The question is incomplete. The complete question is :
A high-speed bullet train accelerates and decelerates at the rate of 4 ft/s^2. Its maximum cruising speed is 90 mi/h. What is the maximum distance the train can travel if it accelerates from rest until it reaches its cruising speed and then runs at that speed for 15 minutes?
Solution :
Given :
Speed of the bullet train, v = 90 mi/h
= [tex]$90 \times \frac{5280}{3600}$[/tex]
= 132 ft/s
Time = 15 minutes
= 15 x 60
= 900 s
Acceleration from rest,
[tex]$a(t) = 4 \ ft/s^2$[/tex]
[tex]$v(t) = 4t + C$[/tex]
Since, v(0) = 0, then C = 0, so velocity is
v(t) = 4t ft/s
Then find the position function,
[tex]$s(t) = \frac{4}{2}t^2 + C$[/tex]
[tex]$=2t^2+C$[/tex]
It is at position 0 when t = 0, so C = 0, and the final position function for only the time it is accelerating is :
[tex]$s(t) = 2t^2$[/tex]
Time to get maximum cruising speed is :
4t = 132
t = 33 s
Distance travelled (at cruising speed) by speed to get the remaining distance travelled.
[tex]$900 \ s \times 132 \ \frac{ft}{s} = 118800 \ ft$[/tex]
Total distance travelled, converting back to miles,
[tex]$2178 + 118800 = 120978\ ft . \ \frac{mi}{5280 \ ft}$[/tex]
= 22.9125 mi
Therefore, the distance travelled is 22.9125 miles
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