You are driving your 1800 kg car at 21 m/s over a circular hill that has a radius of 150 m. A deer running across the road causes you to hit the brakes hard while right at the summit of the hill, and you start to skid. The coefficient of kinetic friction between your tires and the road is 0.75. What is the magnitude of your horizontal acceleration as you begin to slow? Express your answer with the appropriate units.

To solve for the magnitude of your horizontal acceleration, we need to consider the forces acting on the car as it starts to skid at the top of the hill.

### Step 1: Analyze the forces involved

When you are at the summit of a circular hill and hit the brakes, the following forces act on the car:
- **Gravitational force (\( F_g \)):** This is the weight of the car, which is \( F_g = m \cdot g \), where \( m = 1800 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \).
- **Normal force (\( F_n \)):** At the top of the hill, the normal force is reduced because part of the gravitational force is used to provide the centripetal force to keep the car moving in a circular path.
- **Frictional force (\( F_f \)):** This is the kinetic friction that opposes the motion and slows down the car. The frictional force is given by \( F_f = \mu_k \cdot F_n \), where \( \mu_k \) is the coefficient of kinetic friction.

Since the car is at the summit of the hill, the normal force is less than the full weight of the car due to the centripetal force requirement. The net vertical force provides the centripetal force to keep the car moving in the circular path.

### Step 2: Calculate the normal force

At the top of the hill, the normal force \( F_n \) is reduced by the amount of force required for centripetal motion. The centripetal force \( F_c \) is:

\[
F_c = \frac{m \cdot v^2}{r}
\]

where:
- \( m = 1800 \, \text{kg} \) (mass of the car),
- \( v = 21 \, \text{m/s} \) (speed of the car),
- \( r = 150 \, \text{m} \) (radius of the hill).

Substitute the values:

\[
F_c = \frac{1800 \cdot (21)^2}{150} = \frac{1800 \cdot 441}{150} = 5292 \, \text{N}
\]

The gravitational force \( F_g \) is:

\[
F_g = m \cdot g = 1800 \cdot 9.8 = 17640 \, \text{N}
\]

Now, the normal force at the top of the hill is the difference between the gravitational force and the centripetal force:

\[
F_n = F_g - F_c = 17640 - 5292 = 12348 \, \text{N}
\]

### Step 3: Calculate the frictional force

The frictional force \( F_f \) is given by:

\[
F_f = \mu_k \cdot F_n = 0.75 \cdot 12348 = 9261 \, \text{N}
\]

### Step 4: Calculate the horizontal acceleration

The frictional force is what causes the car to decelerate. Using Newton's second law, \( F = m \cdot a \), the horizontal acceleration \( a \) is:

\[
a = \frac{F_f}{m} = \frac{9261}{1800} \approx 5.14 \, \text{m/s}^2
\]

### Final Answer:
The magnitude of your horizontal acceleration as you begin to slow down is approximately:

\[
\boxed{5.14 \, \text{m/s}^2}
\]

Would you like more details or have any questions?

---

Here are 5 related questions you might consider:
1. How would the acceleration change if the radius of the hill were larger?
2. What if the coefficient of friction were lower—how would that affect your deceleration?
3. How does the mass of the car impact the centripetal force at the summit?
4. What role does centripetal force play when driving over a circular hill?
5. How does the gravitational force contribute to the deceleration?

**Tip:** When calculating forces on inclines or curves, always account for how forces decompose based on the direction of motion and the shape of the path.

You are driving your 1800 kg car at 21 m/s over a circular hill that has a radius of 150 m. A deer running across the road causes you to hit the brakes hard while right at the summit of the hill, and you start to skid. The coefficient of kinetic friction between your tires and the road is 0.75. What is the magnitude of your horizontal acceleration as you begin to slow? Express your answer with the appropriate units.

Learn how to calculate the horizontal acceleration of an 1800 kg car skidding at 21 m/s over a circular hill with a radius of 150 m. This problem involves Newton's laws, centripetal force, and kinetic friction. Suitable for Grades 11-12 physics students.

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