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Chapter 4: Problem 23

The driver of a 1750 \(\mathrm{kg}\) car traveling on a horizontal road at 110 \(\mathrm{km} / \mathrm{h}\) suddenly applies the brakes. Due to a slippery pavement, the friction of the road on the tires of the car, which is what slown the car, is 25\(\%\) of the weight of the car. (a) What is the acceleration of the car? (b) How many meters does it travel before stopping under these conditions?

Short Answer

Expert verified
The acceleration is \(-2.4525 \, \text{m/s}^2\) and the stopping distance is approximately 190.35 meters.

Step by step solution

01

Calculate the Force of Friction

First, we need to determine the force of friction, which acts as the braking force. The frictional force \( F_f \) is 25% of the weight of the car. The weight of the car \( W \) is equal to the gravitational force, which is the mass \( m \) of the car multiplied by the acceleration due to gravity \( g = 9.81 \, \text{m/s}^2 \). First calculate the weight: \[ W = mg = 1750 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 17167.5 \, \text{N} \]Now, calculate the frictional force:\[ F_f = 0.25 \times 17167.5 \, \text{N} = 4291.875 \, \text{N} \]
02

Calculate the Acceleration Using Newton's Second Law

Use the force of friction to find acceleration. According to Newton's second law, \( F = ma \), where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration. Solving for acceleration \( a \):\[ a = \frac{F_f}{m} = \frac{4291.875 \, \text{N}}{1750 \, \text{kg}} = 2.4525 \, \text{m/s}^2 \]Notice that this is the magnitude of acceleration since the car is slowing down, the actual acceleration would be negative: \( a = -2.4525 \, \text{m/s}^2 \).
03

Convert Initial Speed to Meters per Second

The initial speed of the car is given as 110 km/h. We need to convert this to meters per second using the conversion factor \( 1 \, \text{km/h} = \frac{1}{3.6} \, \text{m/s} \):\[ v_i = 110 \, \text{km/h} \times \frac{1}{3.6} = 30.56 \, \text{m/s} \]
04

Calculate the Stopping Distance Using Kinematic Equation

To find the stopping distance \( d \), use the kinematic equation \[ v_f^2 = v_i^2 + 2ad \]where \( v_f = 0 \, \text{m/s} \) (final velocity), \( v_i = 30.56 \, \text{m/s} \) (initial velocity), and \( a = -2.4525 \, \text{m/s}^2 \). Set the final velocity \( v_f \) to 0 and solve for \( d \):\[ 0 = (30.56)^2 + 2(-2.4525)d \]Solve for \( d \):\[ d = \frac{(30.56)^2}{2 \times 2.4525} = \frac{933.9136}{4.905} \approx 190.35 \, \text{m} \]
05

Conclusion

The car's acceleration is \(-2.4525 \, \text{m/s}^2\) and it travels approximately \(190.35 \, \text{m}\) before stopping.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Frictional Force
Frictional force plays a crucial role in slowing down our car. It's the resistance force that occurs when two surfaces interact. In this problem, the road and the car's tires create friction. The car slows down because of this interaction. When the driver hits the brakes, we calculate friction as 25% of the car's weight. How do we compute this? First, find the car's weight using its mass and gravity. In this case, the mass is 1750 kg, and the gravity is 9.81 m/s虏.
  • Calculate the car's weight: \( W = 1750 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 17167.5 \, \text{N} \)
  • Frictional force is 25% of this weight: \( F_f = 0.25 \times 17167.5 \, \text{N} = 4291.875 \, \text{N} \)
This frictional force is the maximum braking force available to slow the car down.
Kinematic Equations
Kinematic equations help us understand motion. They connect variables like velocity, acceleration, and displacement. In our exercise, we use a kinematic equation to find the stopping distance. This equation relates the initial velocity, final velocity, acceleration, and distance:\[v_f^2 = v_i^2 + 2ad\]Where:
  • \(v_f\) is the final velocity (0 m/s when the car stops)
  • \(v_i\) is the initial velocity
  • \(a\) is the acceleration
  • \(d\) is the stopping distance
In this context, using the kinematic equation allows us to solve for the distance the car travels before it comes to a complete stop, utilizing known values for initial speed and the computed acceleration.
Acceleration
Acceleration is how fast an object's velocity changes. It tells us how quickly the car slows down. In our problem, we find acceleration using Newton's Second Law: \[F = ma\]Where:
  • \( F \) is the force (friction in this case)
  • \( m \) is the car's mass
  • \( a \) is acceleration
Solving for acceleration gives us:\[a = \frac{F}{m} = \frac{4291.875 \, \text{N}}{1750 \, \text{kg}} = -2.4525 \, \text{m/s}^2\]The negative sign indicates the car is decelerating. In other words, it's slowing down.
Conversion of Units
To correctly calculate the stopping distance, we need the car's velocity in meters per second, not kilometers per hour. Conversion of units is essential when units differ from what's required, or when input data is in different formats.To convert from km/h to m/s, use the conversion factor:
  • 1 km/h = \( \frac{1}{3.6} \) m/s
For the car's initial speed:\[v_i = 110 \, \text{km/h} \times \frac{1}{3.6} = 30.56 \, \text{m/s}\]After converting, we ensure the velocity is compatible with the other values in calculations, giving us accurate results.
Stopping Distance
Stopping distance is the distance a vehicle travels before it stops completely after the brakes are applied. In our exercise, it combines the elements of initial speed, friction as the braking force, and acceleration. Using the kinematic equation:\[0 = (30.56)^2 + 2(-2.4525)d\]We solve for \(d\), the stopping distance:\[d = \frac{(30.56)^2}{2 \times 2.4525} = \frac{933.9136}{4.905} \approx 190.35 \, \text{m}\]This result tells us how far the car travels before halting under given conditions: low friction caused by the slippery road surface and the specific deceleration rate.

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