- #1
Saladsamurai
- 3,020
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So I posted this earlier and it got to a point where differentiation is neccessary. I am a little familiar with differentiation, but not to the point where I know how to apply the concepts I have recently learned in Calculus. i would greatly appreciate somebody walking me through the process...it should be fairly simple in this particular problem. Again, this problem is for my own personal practice, not for a class, and I am very interested in finding out how to apply these concepts.
Thanks~Casey
Original Post:
A 1000kg Boat is traveling 90km/h when its engine is cut. The magnitude of the frictional force fk is proportional to the boat's speed v: fk=70v, where v is in m/s and fk is in Newtons. Find the time required for the boat to slow to 45 km/h.
Newton's Second
V^2=Vo^2+2a(X-Xo)
X-Xo=VoT+1/2at^2
V=Vo+aT
Vo=25m/s
V=12.5m/s
fk=70v=1750N
I drew a FBD and it seems that since the engine was cut, there is only fk in the x direction. Thus, fk=ma--->1750=-1000a-->a=-1.75
Then I used V=Vo+at---> t=(V-Vo)/a
-->t=(12.5-25)/-1.75=7.1
But this is not correct...9.9seconds is the correct solution.
Any advice is appreciated.
~Casey
...It was pointed out that "a" is not constant. But I am not sure where to go from here as I have only dealt with problems dealing with constant acceleration...what am I differentiating? I am not sure of the equation...or how to derive one. hollah.
Thanks~Casey
Original Post:
Homework Statement
A 1000kg Boat is traveling 90km/h when its engine is cut. The magnitude of the frictional force fk is proportional to the boat's speed v: fk=70v, where v is in m/s and fk is in Newtons. Find the time required for the boat to slow to 45 km/h.
Homework Equations
Newton's Second
V^2=Vo^2+2a(X-Xo)
X-Xo=VoT+1/2at^2
V=Vo+aT
The Attempt at a Solution
Vo=25m/s
V=12.5m/s
fk=70v=1750N
I drew a FBD and it seems that since the engine was cut, there is only fk in the x direction. Thus, fk=ma--->1750=-1000a-->a=-1.75
Then I used V=Vo+at---> t=(V-Vo)/a
-->t=(12.5-25)/-1.75=7.1
But this is not correct...9.9seconds is the correct solution.
Any advice is appreciated.
~Casey
...It was pointed out that "a" is not constant. But I am not sure where to go from here as I have only dealt with problems dealing with constant acceleration...what am I differentiating? I am not sure of the equation...or how to derive one. hollah.