Rollercoaster has a dip that bottoms out in a vertical circle of radius r.

Passenger feels seat of car pushing upward on her with a force twice her weight as she goes through the dip.

If r = 20.0 m, how fast is rollercoaster going at bottom of the dip?

Potential energy at the topmost point = mg(2r)

Kinetic energy at dip bottommost point= 1/2 m (v^2)

so conservation of energy means

mg 2r = 1/2 m(v^2)

or 4rg= v^2
or v = 2sqrt(rg)

To find the speed of the rollercoaster at the bottom of the dip, we can use the concept of circular motion and centripetal force.

In this case, the passenger feels the seat of the car pushing upward on her with a force twice her weight. This means that the net force acting on the passenger at the bottom of the dip is equal to three times her weight (weight + upward force).

The net force at the bottom of the dip is given by the centripetal force, which is the force required to keep an object moving in a circular path. It is given by the equation:

F = (m * v^2) / r

Where F is the net force, m is the mass of the object, v is the velocity, and r is the radius of the circular path.

In this case, since the net force is three times the weight (3W), we can equate the two:

3W = (m * v^2) / r

We know that the weight of an object is equal to its mass times the acceleration due to gravity:

W = m * g

Substituting this into the equation above:

3(m * g) = (m * v^2) / r

We can cancel out the mass:

3g = (v^2) / r

Now we can rearrange the equation to find the velocity v:

v^2 = 3g * r

v = sqrt(3g * r)

Given that r = 20.0 m, we need the value of acceleration due to gravity (g). On the surface of the Earth, g is approximately equal to 9.8 m/s^2.

Plugging in the values:

v = sqrt(3 * 9.8 * 20.0)

v = sqrt(588.0)

v ≈ 24.2 m/s

Therefore, the rollercoaster is going at approximately 24.2 m/s at the bottom of the dip.