The Physics Behind an Automobile Accident Analysis

Analyze the Speed of Car A Before Braking

Because of your expert physics skills, you have been hired as an analyst in a court case involving an automobile accident. The accident involved a 3,000 kg car (car A) which approached a stationary car of mass 1,841 kg (car B). The driver of car A slammed on his brakes (locking his wheels) 14 m before he/she crashed into car B. After the collision, car A slid 19 m forward while car B slid 26 m forward (Note: the cars did not stick together upon colliding). The coefficient of kinetic friction between the locked wheels and the road was measured to be 0.8. Calculate the speed of car A, in mph, at the moment he began braking. What is the speed of car A, in mph, at the moment he began braking?

The speed of car A at the moment he began braking is 91.8 Km/h. Explanation: We must start this exercise at the end. Let's look for the lighter car acceleration (B). For this, we use Newton's second law: fr = m * a a = fr / m fr = μ * N N-W = 0 Let's replace: a = μ * m * g / m a = μ * g a = 0.8 * 9.8 a = 7.84 m/s² Since car B reached an initial velocity vo₀₂ and at the end the speed was zero, let's use kinematics: v₀₂ = 2 * a * x₂ v₀₂ = √(2 * 7.84 * 26) v₀₂ = 20.19 m/s Let's perform the same procedure for car A. The acceleration is the same as it does not depend on the mass of the vehicles: v₀₁ = √(2 * a * x₁) v₀₁ = √(2 * 7.84 * 19) v₀₁ = 17.36 m/s Now, let's use momentum conservation where the system is the two vehicles: Initial momentum before the crash: p₀ = M * v₁ + 0 After the crash: p_{f} = M * v₀₁ + m * v₀₂ p₀ = p_{f} M * v₁ = M * v₀₁ + m * v₀₂ v₁ = v₀₁ + m / M * v₀₂ v₁ = 17.36 + 1841/3000 * 20.19 v₁ = 20.75 m/s Lastly, to find the speed of car A when the brakes were applied (v): v₁² = v² - 2 * a * x₁ v = √(v₁² + 2 * a * x₁) v = √(20.75² + 2 * 7.84 * 14) v = 25.50 m/s v = 25.50 m/s (1 km/1000 m) * (3600 s/1 h) v = 91.8 Km/h

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