Math Problem Statement

Step by step

A 945 kg car traveling rightward at 22.6 m/s slams on the brakes and skids to a stop (with locked wheels). If the coefficient of friction between tires and road is 0.972, determine the distance required to stop.

Solution

To solve this problem, we need to determine the stopping distance of a car with the following given values:

Given Data:

  • Mass of the car, m=945kgm = 945 \, \text{kg}
  • Initial speed of the car, v=22.6m/sv = 22.6 \, \text{m/s}
  • Coefficient of kinetic friction between the tires and the road, μk=0.972\mu_k = 0.972
  • The car comes to a stop, so the final velocity, vf=0m/sv_f = 0 \, \text{m/s}

Solution Steps:

  1. Determine the Frictional Force:

    The frictional force ff can be calculated using the formula: f=μkNf = \mu_k \cdot N where NN is the normal force. Since the car is on a flat surface, N=mgN = m \cdot g, where g=9.8m/s2g = 9.8 \, \text{m/s}^2 (acceleration due to gravity).

    Substituting the values: f=μkmg=0.9729459.8f = \mu_k \cdot m \cdot g = 0.972 \cdot 945 \cdot 9.8

    Calculating ff: f=9006.036Nf = 9006.036 \, \text{N}

  2. Calculate the Deceleration (Negative Acceleration):

    Since this frictional force is the only horizontal force acting on the car, it provides the deceleration aa: f=maf = m \cdot a Solving for aa: a=fm=9006.036945a = \frac{f}{m} = \frac{9006.036}{945}

    Calculating aa: a=9.528m/s2a = 9.528 \, \text{m/s}^2 This deceleration is in the opposite direction of motion, so we take it as 9.528m/s2-9.528 \, \text{m/s}^2.

  3. Calculate the Stopping Distance Using Kinematic Equation:

    We use the kinematic equation: vf2=v2+2adv_f^2 = v^2 + 2 \cdot a \cdot d Rearranging to solve for dd: d=vf2v22ad = \frac{v_f^2 - v^2}{2 \cdot a} Substituting vf=0v_f = 0, v=22.6m/sv = 22.6 \, \text{m/s}, and a=9.528m/s2a = -9.528 \, \text{m/s}^2: d=0(22.6)22(9.528)d = \frac{0 - (22.6)^2}{2 \cdot (-9.528)}

    Calculating dd: d=510.7619.05626.8md = \frac{-510.76}{-19.056} \approx 26.8 \, \text{m}

Final Answer:

The distance required to stop the car is approximately 26.8m26.8 \, \text{m}.

Would you like further clarification on any step or additional details?

Here are five related questions to consider:

  1. How would the stopping distance change if the coefficient of friction were lower?
  2. What would happen to the stopping distance if the car's mass were doubled?
  3. How does an incline affect the stopping distance if the road is sloped?
  4. How would a higher initial speed affect the stopping distance?
  5. What role does tire condition play in determining the stopping distance?

Tip: Remember, the stopping distance increases with both speed and reduced friction, making it crucial to consider both factors when driving conditions change.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Newton's Laws of Motion
Friction

Formulas

Frictional force: f = μ * N
Normal force: N = m * g
Kinematic equation: v_f^2 = v^2 + 2 * a * d

Theorems

Newton's Second Law of Motion

Suitable Grade Level

Grades 10-12