Dayv O. answered • 12/15/23
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Caring Super Enthusiastic Knowledgeable Calculus Tutor
Looks more like a first order differential equation
distance=d(distance)/dt ,,,,t measured in hours
d'-d=0
multiply by e-t
e-t*d'-e-t*d=0
(e-t*d)'=0 ,,,,product rule for derivatives
e-t*d=C ,,,integrating both sides
d=Cet
and of course d'=Cet, that is premise of problem
call 1pm t=0, then C=100
at 2pm t=1 and d=d'=100e≈272mph
Dayv O.
with d=100e^t, t=0 d=d'=100. d=d'=50, d=100e^ln(1/2) so d=d'=50 when t=ln(1/2). The problem wants solution for d'=d,,,not d=-d'. Thanks for comment.
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12/16/23
Dayv O.
if d=-d'; so d=|-d';|, wouldn't setup be d= -d(d)/dt with solution d=Ce^-t? ,,,, It is implied that the 100 miles is positive from origin.
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12/16/23
Bradford T.
It is not clear if the train is going away from the place or towards the place. If away, then your solution is correct. If it is going towards the place, then 100 mph = 100 miles, 99mph = 99 miles, 98mph = 98 miles, etc., with a varying delta-t. If x = distance traveled since 1PM, dx/dt = 100-x. In that case it would be 100(1-1/e) mph12/15/23