Abox is moving along the x-axis and its position varies in time according to the expression:
x = (6.00 m/s^2)t^2,
where x is in meters and t is in seconds.
(a) determine its position (in m) at t = 3.20 s.
(b) determine its position (in m) at t = (3.20 s) + δt.
(c) evaluate the limit of δx δt as δt approaches zero to find the velocity (in m/s) at t = 3.20 s.

38.4 m/s

Explanation:

a) at t = 3.2s.

b) at t = 3.2 + Δt.

c) As Δt approaches 0. We can find the velocity by the ratio of Δx/Δt

As Δt approaches 0, v = 38.4 + 0 = 38.4 m/s

it has to be b, about 97 percent.

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .