A particle is moving with the given data. Find the position of the particle. a(t

Question

A particle is moving with the given data. Find the position of the particle. a(t) = t^2 - 7t + 9, s(0) = 0, s(1) = 20 s(t) = A high-speed bullet train accelerates and decelerates at the rate of 10 ft/s^2. Its maximum cruising speed is 75 mi/h. (Round your answers to three decimal places.) 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? mi Suppose that the train starts from rest and must come to a complete stop in 15 minutes. What is the maximum distance it can travel under these conditions? mi Find the minimum time that the train takes to travel between two consecutive stations that are 37.5 miles apart. min The trip from one station to the next takes at minimum 37.5 minutes. How far are the stations? mi

Explanation / Answer

Solution: (14)

a(t) = dv/dt = t^2 - 7t + 9

v = (t^2 - 7t + 9) dt = (1/3)t^3 - (7/2) t^2 + 9t + C = ds/dt

Again intregate

s(t) = (1/12)t^4 - (7/6)t^3 + (9/2)t^2 + Ct + D

s(0) = 0 - 0 + 0 + 0 + D = 0, so D = 0

s(1) = (1/12) - (7/6) + (9/2) + C = 20

multiply by 12,

=> 1 - 14 + 54 + 12C = 240

=> 12C = 201
=> C = 16.75 or 67/4

s(t) = (1/12)t^4 - (7/6)t^3 + (9/2)t^2 + (67/4)t

Solution: (15)

Distance before maximum speed S=at^2/2
time to reach: v=at=>v/a=t

a) S=a(v/a)^2/2
S2=Vmaxt, t= 15 min=1/4 h
S2=75/4
S1=10(5280*75/(10*3600)^2/2=605/5280 ml
S=75/4+605/5280ml=18.865 ml

b) t0=15 min
Tacc=v/a=11 s
S1=at^2/2=605/5280 ml
S2=Vmax(t0-Tacc)=5280*75/3600(15*60-11)=18.52 ml
S=18.52+605/5280=18.635

c)
S=2S1+S2
2S1=1210/5280
S2=37.5-2S1=37.271
T1=Tacc=22 s
T2=S2/Vmax=30 min 11 s

d)
S1=1210/5280
S2=46.417
S=46.646 ml

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