作业帮求问一道数学题!急!
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求问一道数学题!急!
求问一道数学题!急!
求问一道数学题!急!
Average acceleration of train A over 2≤t≤8 is a = [v(8)-v(2)] / (8-2) = (-120-100)/6 = - 110/3 meters/minute²
Yes,it does.
v(t) is continuous,so for any value V satisfying v(5)<V<v(8),there exist a time 5<T<8 such that V=v(T).
v(2)=100>0 as the train is moving east,so a positive velocity v(t) corresponds to the train moving east,then 
Given v(2)=100,v(5)=40,v(8)=-120,v(12)=-150,d can be estimated as the trapezoidal sum 
d = -150,so the train is approxiamately 150m to the west of the Origin Station.
Let X(t) be the distance of train A to the east of Origin Station,Y(t) be the distance of train B to the north of the Origin Station,then the distance between train A,B is D=√(X²+Y²).
Differentiate with respect to time t to get the rate of change of the distance:
(Uses the chain rule)
At t=2,X(2)=300,Y(2)=400,X'=vA(2)=100,Y'=vB(2) = -5(2)²+60*2+25=125.
Substitute the numbers in D',we have D'(2)=0.45m
Hope this helps.
那也要能看得懂才行