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The problem involves calculating the apparent weight of a driver as a car travels over a hill and through a dip at constant speed, with the actual weight of the driver given as 680 N. The forces involved are based on circular motion.

### Part 1: Apparent Weight at the Top of the Hill
The hill has a radius of curvature \( r_h = 107 \, \text{m} \). At the top of the hill, the centripetal force is directed downward, reducing the normal force that provides the driver's apparent weight.

The apparent weight at the top of the hill, \( W_h \), is given by:
\[
W_h = W - \frac{m v^2}{r_h}
\]
Where:
- \( W = 680 \, \text{N} \) (actual weight),
- \( v = 25 \, \text{m/s} \) (speed),
- \( r_h = 107 \, \text{m} \) (radius of curvature).

First, we calculate the mass of the driver:
\[
m = \frac{W}{g} = \frac{680 \, \text{N}}{9.8 \, \text{m/s}^2} \approx 69.39 \, \text{kg}
\]

Now, calculate the centripetal force:
\[
\frac{m v^2}{r_h} = \frac{69.39 \times 25^2}{107} \approx 405.47 \, \text{N}
\]

So, the apparent weight at the top of the hill is:
\[
W_h = 680 \, \text{N} - 405.47 \, \text{N} = 274.53 \, \text{N}
\]
The result given in the image is approximately \( 270.6 \, \text{N} \), which aligns closely with this calculation.

### Part 2: Apparent Weight at the Bottom of the Dip
The dip has a radius of curvature \( r_d = 77 \, \text{m} \). At the bottom of the dip, the centripetal force is directed upward, adding to the normal force.

The apparent weight at the bottom of the dip, \( W_d \), is given by:
\[
W_d = W + \frac{m v^2}{r_d}
\]

Calculate the centripetal force at the bottom:
\[
\frac{m v^2}{r_d} = \frac{69.39 \times 25^2}{77} \approx 563.31 \, \text{N}
\]

So, the apparent weight at the bottom of the dip is:
\[
W_d = 680 \, \text{N} + 563.31 \, \text{N} = 1243.31 \, \text{N}
\]
However, the image shows a calculated \( W_d \approx 738 \, \text{N} \). This suggests either a different speed or another factor affecting the given solution.

### Clarification
Would you like further details about how the apparent weights were computed or want to discuss any corrections to the provided answers?

### Related Questions:
1. How does increasing the car's speed affect the apparent weight at the hilltop and the dip?
2. What happens to the apparent weight at the top of the hill if the hill's radius increases?
3. Why is the apparent weight less than the actual weight at the top of the hill?
4. How does the concept of centripetal force relate to the feeling of "weightlessness" in this scenario?
5. Can the driver's apparent weight ever be zero, and under what conditions?

### Tip:
The apparent weight is always influenced by the net forces acting on an object, including gravitational force and any additional forces due to circular motion.

A car travels down a straight country road that leads over hills and through valleys. On one particular stretch of road, the car encounters a hill that can be approximated as the top of a circle with a radius rh = 107 m. Later, the car comes to a dip with a radius of curvature rd = 77 m. Assume that the car maintains a constant speed of v = 25 m/s as it goes over the hill and through the dip. The actual weight of the driver, as measured on a flat stretch of road, is 680 N. What is the apparent weight of the driver at the top of the hill? What is the apparent weight of the driver at the bottom of the dip?

Calculate the apparent weight of a driver as a car moves over a hill and through a dip with radii 107 m and 77 m respectively. Given a speed of 25 m/s and an actual weight of 680 N, determine the driver's apparent weight at both locations using circular motion and centripetal force principles. Suitable for Grades 10-12.

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