- #1
mindcircus
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A boat with initial speed v is launched on a lake. The boat is slowed by the water by a force, F=-ke^(bv). Find the expression for speed v(t).
I've done the problem, but my answer seems too odd to be right...it may be my calculus.
I've drawn a FBD, with the normal force and weight cancelling each other out. The net force is the resisting force, F=-ke^(bv), which I've then set equal to F=ma. I've used dv/dt for a.
-ke^(bv)=m*(dv/dt)
Rearranging to get like terms together gives me
dv/(e^(bv))=-(k/m)dt
(e^-(bv))dv=-(k/m)dt
Setting up the integrand using limits 0 to t and initial v to v gives me:
-b((e^-(bv.initial))-(e^-(bv)))=-(k/m)t
Simplified to:
(kt/bm)=e^(v.initial/v)
ln (kt/bm)=(v.inital/v)
So, v=(v.initial)/ln(kt/bm)
The answer seems too messy...any help would be much appreciated!
I've done the problem, but my answer seems too odd to be right...it may be my calculus.
I've drawn a FBD, with the normal force and weight cancelling each other out. The net force is the resisting force, F=-ke^(bv), which I've then set equal to F=ma. I've used dv/dt for a.
-ke^(bv)=m*(dv/dt)
Rearranging to get like terms together gives me
dv/(e^(bv))=-(k/m)dt
(e^-(bv))dv=-(k/m)dt
Setting up the integrand using limits 0 to t and initial v to v gives me:
-b((e^-(bv.initial))-(e^-(bv)))=-(k/m)t
Simplified to:
(kt/bm)=e^(v.initial/v)
ln (kt/bm)=(v.inital/v)
So, v=(v.initial)/ln(kt/bm)
The answer seems too messy...any help would be much appreciated!