For the normal force in the drawing to have the same magnitude at all points on the vertical track, the stunt driver must adjust the speed to be different at different points. Suppose, for example, that the track has a radius of 4.84 m and that the driver goes past point 1 at the bottom with a speed of 16.6 m/s. What speed must she have at point 3, so that the normal force at the top has the same magnitude as it did at the bottom?

What drawing? Where is point 3?

We are not clairvoyant.

How do I attach photos?

It is located on the top

I figured it out.

2mg=mv3^2/r-mv1^2/r

V3 being the speed at the top and V1 being the speed given at the bottom

set them equal: V3^2=2gr+V1^2

sqrt(2*9.8*4.84+16.6^2)

V3= 19.2459 m/s

To find the speed that the stunt driver must have at point 3 in order for the normal force at the top to have the same magnitude as it did at the bottom, we can use the concept of conservation of mechanical energy.

Here's how to solve it step by step:

Step 1: Identify the known quantities and variables:
- Radius of the track (r) = 4.84 m
- Speed at point 1 (v1) = 16.6 m/s
- Speed at point 3 (v3) - unknown
- Acceleration due to gravity (g) = 9.8 m/s^2

Step 2: Find the potential energy at point 1:
At point 1 (bottom of the track), all of the mechanical energy is in the form of kinetic energy. Since there is no height, the potential energy is zero (PE1 = 0).

Step 3: Find the potential energy at point 3:
At point 3 (top of the track), all of the mechanical energy is in the form of potential energy. Since the potential energy is equal to the kinetic energy at point 1, we can write: PE3 = KE1.

Potential energy formula: PE = mgh
Since the mass (m) cancels out, we can write: gh3 = 1/2 mv1^2.

Step 4: Find the speed at point 3:
Using the conservation of mechanical energy, we equate the potential energy at point 3 with the kinetic energy at point 3.
PE3 = KE3
mgh3 = 1/2 mv3^2.
Simplifying the equation: gh3 = 1/2 v3^2.

Step 5: Equate the two equations containing h3 and solve for v3:
gh3 = 1/2 mv1^2
gh3 = 1/2 v3^2

Since the height (h) is the same at both points, we can equate the two equations:
1/2 v3^2 = gh3 = 1/2 mv1^2

Simplifying and rearranging the equation, we get:
v3^2 = v1^2

Take the square root of both sides to find v3:
v3 = sqrt(v1^2)

Now we can substitute the given values:
v3 = sqrt((16.6 m/s)^2)
v3 = sqrt(275.56 m^2/s^2)
v3 = 16.6 m/s

Therefore, the stunt driver must have a speed of 16.6 m/s at point 3 in order for the normal force at the top to have the same magnitude as it did at the bottom.