Question

The position of an object is given by x=2t^(3)+0.2t^(5), where t is greater than or equal to 0. Find the position and acceleration when the velocity is 40.

Answer 1

To get the equation of the velocity, take the derivate on both sides.
d/dt (x = 2t^(3) + 0.2t^(5))
dx/dt = 6t^(2) + t^(4)

Since dx/dt = 40, that will be
40 = 6t^(2) + t^(4)

You need to solve the value of t.
Let u = t^2
40 = 6u + u^2
u^2 + 6u - 40 = 0
Using the quadratic formula to get the roots, that is
u = 4, u = -10

Absurd u = -10. So, use u = 4. Going back to the roots,
4 = t^2
t = 2

The value of the position (which is the original equation) is
x = 2[(2)^3] + 0.2[(2)^5]
x = 22.4

For the acceleration, you take the second order derivative, that is
d/dt [dx/dt = 6t^(2) + t^(4)]
d^2x/dt^2 = 12t + 4t^3
Finally,
d^2x/dt^2 = 12(2) + 4[(2^3)]
d^2x/dt^2 = 56

Summary: position = 22.4, velocity = 40, acceleration = 56