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Keshroom
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Homework Statement
A particle of mass m is subject to a force
F(t) = mae^{-bt}
The initial position and speed are zero. Find x(t)
Homework Equations
The Attempt at a Solution
how would i begin to start this??
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Keshroom said:A particle of mass m is subject to a force
F(t) = mae^{-bt}
Keshroom said:F(t) = m(dv/dt) e^{-bt}
tiny-tim said:You mean "a" is the acceleration?
No, that makes no sense.
If F(t) is the force, and m is the mass, then F(t) = ma,
so F(t) cannot be mae^{-bt}.
What exactly was the question?
Mathoholic! said:I think I get it:
F(t)=ma(t) in which a(t)=a_{0}e^{-bt}
If you integrate you get:
v(t)=v_{0}-(a_{0}/b)e^{-bt}
And again:
x(t)=x_{0}+v_{0}t+(a_{0}/b^{2})e^{-bt}
Since x_{0} and v_{0} are both zero.
x(t)=(a_{0}/b^{2})e^{-bt}
Keshroom said:wait..you can just integrate like that without the dt on the right hand side?
Mathoholic! said:I simply skipped the integration steps
Mathoholic! said:I think I get it:
F(t)=ma(t) in which a(t)=a_{0}e^{-bt}
If you integrate you get:
v(t)=v_{0}-(a_{0}/b)e^{-bt}
And again:
x(t)=x_{0}+v_{0}t+(a_{0}/b^{2})e^{-bt}
Since x_{0} and v_{0} are both zero.
x(t)=(a_{0}/b^{2})e^{-bt}
berkeman said:Please do not do the student's homework for them. That is against the PF rules.
Keshroom said:F(t) = ma_{0}e^{-bt}
m(dv/dt) = ma_{0}e^{-bt}
∫mdv = ∫ma_{0}e^{-bt}dt (from v=0 to v=v and t=0 to t=t)
mv = m(a_{0}e^{-bt})/-b
x(t) refers to the position of an object at a specific time t. It is related to time dependant force because the force acting on an object can cause its position to change over time.
The equation for finding x(t) is x(t) = x_{0} + v_{0}t + 1/2at^{2}, where x_{0} is the initial position, v_{0} is the initial velocity, a is the acceleration, and t is the time.
Yes, x(t) can be negative or zero. This depends on the initial position and the direction of the time dependant force.
The units for x(t) are meters (m) or any other unit of length, while the units for time dependant force are Newtons (N) or any other unit of force.
Yes, x(t) can be determined for all types of time dependant forces as long as the acceleration is constant. If the acceleration is not constant, the equation for x(t) will be more complex and may require calculus to solve.