SOLUTION: A train leaves a station, moving along a straight line. During the next 3 hours, its velocity (in mph) is given by:
v(t) = 2t^2 - (t^3)/2, for 0≤t≤3
where t is it
Algebra ->
Equations
-> SOLUTION: A train leaves a station, moving along a straight line. During the next 3 hours, its velocity (in mph) is given by:
v(t) = 2t^2 - (t^3)/2, for 0≤t≤3
where t is it
Log On
Question 996204: A train leaves a station, moving along a straight line. During the next 3 hours, its velocity (in mph) is given by:
v(t) = 2t^2 - (t^3)/2, for 0≤t≤3
where t is its time (in hours) since its departure from the station.
(a). Determine the position function f(t) of the train from its velocity.
(b). Where is the train 3 hours after its departure. Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! A train leaves a station, moving along a straight line. During the next 3 hours, its velocity (in mph) is given by:
v(t) = 2t^2 - (t^3)/2, for 0≤t≤3
where t is its time (in hours) since its departure from the station.
(a). Determine the position function f(t) of the train from its velocity.
-------------
v is the 1st derivative of distance
f(t) = INT(2t^2 - t^3/2) = 2t^3/3 - t^4/8
----------------
(b). Where is the train 3 hours after its departure.
f(t) = 2t^3/3 - t^4/8
f(3) = 2*9/3 - 81/8 = 6 - 10.125
f(3) = -4.125
--> 4.125 miles from the station.
The minus sign indicates the direction.
