Question

show that acceleration is given by a=10-(1/4)(v) if velocity is 40(1-e^-t/4)

Answer 1

\[v(t) = 40 - 40 e^{-t/4}\]\[a(t) = v'(t) = 10 - \frac {1}{4} v(t)\]Okay, so we take the derivative of the velocity function to get the acceleration function.\[\frac {d}{dt} \left[ v(t) = 40 - 40 e^{-t/4} \right] \rightarrow \frac {dv}{dt} = -40e^{-t/4} \times \frac {d}{dt}\left[ \frac {-t}{4} \right]\]\[\frac {dv}{dt} = a(t) = 10e^{-t/4}\]And that is equivalent to the other form of a(t)\[10e^{-t/4} = 10 - \frac {1}{4} v(t) \]\[10e^{-t/4} = 10 - (10 - 10e^{-t/4})\]\[10e^{-t/4} = 10e^{-t/4}\]So the acceleration is that.

Answer 2

yea, i got upto here, but now i dont know what to do...

Answer 3

That's all you have to do, use the derivative to show that a = 10-(1/4)(v)

Answer 4

oh okay

Answer 5

okay one more question, if you dont mind.

Answer 6

for what values of m does the function y=Ae^(mt) satisfy the following equation?