x(t) = e ^ t - 2t
x(3) = e^3 - 6
x(1) = e - 2
avg speed = (e^3 - 6 - e + 2)/(3-1)
= (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) .
The position of a particle moving along the x-axis 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.
1 answer
1 answer