Aparticle moves along the x-axis with velocity given by v(t)=3t2+6t for time tâ¥0. if the particle at position x=2 at time t=0, what is the position of the particle at time t=1?
a) 4
b) 6
c) 0
d) 11
e) 12

b) 6

Explanation:

Given

v(t)=3t²+6t

X(0) = 2

X(1) = ?

Knowing that

v(t)=3t²+6t = dX/dt

⇒ ∫dX = ∫(3t²+6t)dt

⇒ X - X₀ = t³ + 3t²

⇒ X(t) = X₀ + t³ + 3t²

If  X(0) = 2

⇒  X(0) = X₀ + (0)³ + 3(0)² = 2

⇒  X₀ = 2

then we have

X(t) = t³ + 3t² + 2

when

t = 1

X(1) = (1)³ + 3(1)² + 2

X(1) = 6

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 .