The position of a particle moving along the xaxis at time t > 0 seconds is given by the function x(t) = e ^ t  2t feet.
a) Find the average velocity of the particel over the interval [1,3].
b) In what direction and how fast is the particle moving at t= 1 seconds?
c) For what values of t is the particle moving to the right?
d) Find the postition of the particle when its velocity is 0.
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1 answer

x(t) = e ^ t  2t
x(3) = e^3  6
x(1) = e  2
avg speed = (e^3  6  e + 2)/(31)
= (e^3  e  4)/2 ft/s
x'(t) = e^t  2
x'(1) = e  2 = about .72 ft/s
looks like it is moving to the right
moving to the right when e^t  2 > 0
e^t > 2 , let's take ln of both sides
t ln e > ln2 ,
t > ln2
it stops when t = ln2
now find x(ln2) . 👍
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answered by mathhelper
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