Aparticle initially located at the origin has an acceleration of = 2.00ĵ m/s2 and an initial velocity of i = 8.00î m/s. find (a) the vector position of the particle at any time t (where t is measured in seconds), (b) the velocity of the particle at any time t, (c) the position of the particle at t = 7.00 s, and (d) the speed of the particle at t = 7.00

(a) The vector position of the particle at any time t; is s(t) = (8i)t + (j)t²

(b) the velocity of the particle at any time is v = 8i + (2j)t

(c) the position of the particle at t = 7.00 s is 56i + 49 j at a distance of 74.4 m

(d) the speed of the particle at t = 7.00 is 16.12 m/s

The given parameters;

acceleration of the particle, a = 2 j m/s²

initial velocity of the particle, v = 8 i m/s

(a) The vector position of the particle at any time t;

(b) the velocity of the particle at any time

(c) the position of the particle at t = 7.00 s

(d) the speed of the particle at t = 7.00

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a. The particle has position vector

b. Its velocity vector is equal to the derivative of its position vector:

c. At , the particle has position

That is, it's 56.0 m to the right and 49.0 m up relative to the origin, a total distance of away from the origin in a direction of relative to the positive axis.

d. The speed of the particle at is the magnitude of the velocity at this time:

Then its speed at this time is

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 .