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.

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