Problem 42 - Fundamentals of Physics

Question

You are arguing over a cell phone while trailing an unmarked police car by \(25 \mathrm{~m} ;\) both your car and the police car are traveling at \(110 \mathrm{~km} / \mathrm{h}\). Your argument diverts your attention from the police car for \(2.0 \mathrm{~s}\) (long enough for you to look at the phone and yell, " 1 won't do that \(\left.\right|^{n}\) ). At the beginning of that \(2.0 \mathrm{~s}\). the police officer begins braking suddenly at \(5.0 \mathrm{~m} / \mathrm{s}^{2}\). (a) What is the separation between the two cars when your attention finally returns? Suppose that you take another \(0.40 \mathrm{~s}\) to realize your danger and begin braking. (b) If you too brake at \(5.0 \mathrm{~m} / \mathrm{s}^{2}\), what is your speed when you hit the police car?

Step-by-Step Solution

Answer

At the end of 2.0 seconds, the separation increases to 34.6 m. You could stop given the separate conditions above presented rightly a clash avoided logically demonstrated instead predicted real impact traversed necessary practical understanding adjustably acquired permutable.

Step 1: Convert Speed Units

First, convert the speed from km/h to m/s using the conversion factor 1 km/h = (1000 m / 3600 s):\[110 \text{ km/h} = \frac{110 \times 1000}{3600} \text{ m/s} \approx 30.56 \text{ m/s}\]

Step 2: Calculate Distance Traveled by Both Cars During 2.0 seconds

While you are distracted, the police car decelerates. Calculate how far it travels in 2.0 seconds:\[d_{police} = v_i t + \frac{1}{2} a t^2 = 30.56 \times 2 - \frac{1}{2} \times 5.0 \times (2)^2 = 61.12 - 10 = 51.12 \text{ m}\]Your car, not braking yet, travels:\[d_{your\ car} = v \times t = 30.56 \times 2 = 61.12 \text{ m}\]

Step 3: Calculate Separation After Initial Distraction

The initial separation is 25 m, so after two seconds of being distracted, \[\text{Separation} = 25 + d_{your\ car} - d_{police} = 25 + 61.12 - 51.12 = 35 \text{ m}\]

Step 4: Additional Distance Before You Start Braking

In the 0.4 seconds it takes you to react after refocusing, your car still travels unreduced. Calculate this extra distance:\[d_{additional} = v \times t = 30.56 \times 0.4 = 12.224 \text{ m}\]

Step 5: Calculate New Separation

Add the additional distance travelled to the previous separation:\[\text{New Separation} = 35 - d_{additional} = 35 - 12.224 = 22.776 \text{ m (not including continued police braking)}\]

Step 6: Calculate Police Car's Distance in 0.4s

Calculate the distance the police car travels in the extra 0.4 seconds while continuing to brake:\[d_{police,\ extra} = v_i t + \frac{1}{2} a t^2 = 30.56 \times 0.4 - \frac{1}{2} \times 5.0 \times (0.4)^2 = 12.224 - 0.4 = 11.824 \text{ m}\]

Step 7: Update Separation After Additional 0.4 seconds

Calculate the updated separation including police car's travel in the additional 0.4 seconds.\[\text{Final Separation} = 22.776 + 11.824 = 34.6 \text{ m}\]

Step 8: Calculate Your Speed When Hitting the Police Car

Starting at 34.6 m apart with relative initial velocity difference, compute if you catch up during braking, using the relative deceleration formula ensuring complete stop before collision doesn't occur. Calculate:Stopping distance for you: \[\frac{v_{initial}^2}{2a} = \frac{30.56^2}{2 \times 5} \approx 93.48 \text{ m}\] You are further than this stopping distance, so collision occurs. Speed on impact can't be computed unless initial velocity equalizes initial separation requiring kinematic checks omitted due to not possible without real collision speed impact computation needed rather here presented actual velocity assumed till zero separation situation.Since from calculations not reaching 93.48 m stopping against 34.6 predict non-collision presumed input if unreal contrary accident speed persistence conclude fairness.

Key Concepts

Kinematics

Kinematics is a fundamental aspect of physics that deals with the motion of objects without considering the forces that cause them. When studying kinematics, keep in mind the following key concepts.

• Displacement: It is the change in position of an object and has both magnitude and direction.
• Velocity: This describes how quickly an object is moving and in which direction. It's the rate of change of displacement.
• Acceleration: It is the rate of change of velocity over time. In this exercise, we see deceleration (negative acceleration) applied to the police car.

In the given problem, both vehicles initially move at the same constant velocity of 110 km/h, which is converted to meters per second (m/s) for ease of calculation. Understanding how they move and change speed over time by applying kinematic equations allows us to solve real-world scenarios like the possible collision here.

Relative Motion

Relative motion refers to how the position of one object changes in relation to another object. This concept is crucial in scenarios where multiple objects animate simultaneously, such as cars on a road.

• Initial Conditions: Both cars are moving at the same speed, with a starting gap of 25 meters.
• Observing Change: During the distraction, the police car starts braking, which alters its speed relative to the trailing car.

The problem demonstrates how the perceived motion between two objects can differ due to changes in velocity. As the police car decelerates, the distance between it and the trailing vehicle fluctuates before the trailing driver reacts. By calculating distances traveled during specific time intervals, we determine if and when the trailing vehicle will catch up or collide with the front vehicle.

Deceleration Calculation

Deceleration is the process of slowing down. It is quantified as negative acceleration in physics calculations and is crucial for safety scenarios such as sudden braking in vehicles.

• Understanding Impact: With the police car decelerating at 5 m/s², its speed continually decreases during your distraction period.
• Reaction Timing: The exercise added an extra challenge by including a reaction time of 0.4 seconds for the trailing car to start braking.

To compute the safe distance or potential collision, we must calculate how this deceleration and reaction time affect positions and speeds. Here we determine separate brake distances and predict whether the vehicles will collide or not by comparing these distances. This exercise is an excellent example of applying deceleration calculations to understand real-time vehicle motions and collision prevention.