Need help with solving for t value and plugging back into equation?

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To find the position of an electron described by the equation x=16*te^-t when it momentarily stops, the derivative x' is calculated as x'=e^-t - t*e^-t. Setting the derivative to zero leads to the equation -te^-t = -e^-t, simplifying to t = 1. The discussion confirms that this value of t is correct for determining when the electron stops. The thread emphasizes the importance of correctly applying calculus to solve for the t value and subsequently plugging it back into the original equation.
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1. An electron moving along the x-axis has a position given by x=16*te^-t m, where t is in seconds. How far is the electron from the origin when it momentarily stops?


2. Homework Equations
uv=u'v+uv'


3. The Attempt at a Solution
x'=e^-t - t*e^-t
e^-t - t*e^-t = 0
-te^-t = -e^-t
t = 1?
 
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hi itsgabriella! welcome to pf! :smile:

(try using the X2 button just above the Reply box :wink:)
itsgabriella said:
x'=e^-t - t*e^-t
e^-t - t*e^-t = 0
-te^-t = -e^-t
t = 1?

yup! :smile:

what's worrying you about that? :confused:​
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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