The speed of a bird is related to the time (t) since it leaves the nest by S = 20 - t^2 + 3t, where t is time in seconds. What is the distance traveled by the bird in 1 hour?

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The velocity of a bird flying away from its nest is given as a function of the time since it leaves the nest. If x is the displacement of the bird, its instantaneous velocity is the first derivative of x with respect to t, `dx/dt` .

`dx/dt = 20 - t^2 + 3t`

`dx = 20 - t^2 + 3t dt`
`x = int 20 - t^2 + 3t dt`
The displacement of the bird in 1 hour is the definite integral `int_0^3600 20 - t^2 + 3t dt`

= `[20t - t^3/3 + 3t^2/2]_0^3600`

= `20*(3600 - 0) - (3600 - 0)^3/3 + 3*(3600 - 0)^2/2`

= `20*(3600 - 0) - (3600 - 0)^3/3 + 3*(3600 - 0)^2/2`

= `72000 - 1.5552*10^10 + 19440000`

`~~` -1.5532488*10^10

The velocity of the bird after a certain duration of time speed of the bird after flying for some time becomes negative. It travels a distance equal to 1.5532488*10^10 in the direction opposite to that it which it initially started to fly.

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