Problem 114 - Fundamentals of Physics

Question

A car can be braked to a stop from the autobahn-like speed of \(200 \mathrm{~km} / \mathrm{h}\) in \(170 \mathrm{~m}\). Assuming the acceleration is constant, Find its magnitude in (a) SI units and (b) in terms of g. (c) How much time \(T_{b}\) is required for the braking? Your reaction time \(T,\) is the time you require to perceive an emergency, move your foot to the brake. and begin the braking. If \(T_{r}=400 \mathrm{~ms}\), then (d) what is \(T_{b}\) in terms of \(T_{n}\) and \((\mathrm{c})\) is most of the full time required to stop spent in reacting or braking" Dark sunglasses delay the visual signals sent from the cyes to the visual cortex in the brain, increasing \(T_{r-}\) (f) In the extreme case in which \(T_{r}\) is increased by \(100 \mathrm{~ms}\), how much farther does the car travel during your reaction time?

Step-by-Step Solution

Answer

(a) Acceleration is \(9.09 \mathrm{~m/s^2}\). (b) Acceleration is \(0.928g\). (c) Braking takes \(6.11\mathrm{~s}\). (d) Most time is spent braking. (f) Extra distance is \(5.56 \mathrm{~m}\).

Step 1: Convert Initial Speed to Meters per Second

The initial speed is given as \(200 \mathrm{~km/h}\). We need to convert this to meters per second (\(\mathrm{m/s}\)) using the conversion factor \(1\mathrm{~km/h} = \frac{1}{3.6} \mathrm{~m/s}\). Thus, the speed is \( v_i = \frac{200}{3.6} \mathrm{~m/s} \approx 55.56 \mathrm{~m/s} \).

Step 2: Use Kinematic Equation to Find Acceleration

We use the equation \(v_f^2 = v_i^2 + 2a x\), where \(v_f = 0\), \(v_i = 55.56 \mathrm{~m/s}\), and \(x = 170 \mathrm{~m}\). Solving for \(a\), we have:\[a = \frac{v_f^2 - v_i^2}{2x} = \frac{0 - (55.56)^2}{2 \times 170} \approx -9.09 \mathrm{~m/s^2}.\]

Step 3: Convert Acceleration to g

To find the magnitude of acceleration in terms of \(g\), use the conversion \(g = 9.81 \mathrm{~m/s^2}\). Thus:\[a = \frac{9.09 \mathrm{~m/s^2}}{9.81 \mathrm{~m/s^2}} \approx 0.928g.\]

Step 4: Calculate Braking Time \(T_b\)

Use the kinematic equation \(v_f = v_i + a T_b\), where \(v_f = 0\), \(v_i = 55.56 \mathrm{~m/s}\), and \(a = -9.09 \mathrm{~m/s^2}\), solve for \(T_b\):\[T_b = \frac{0 - 55.56}{-9.09} \approx 6.11\mathrm{~s}.\]

Step 5: Calculate Total Stopping Time

The total time \(T_n\) is the sum of the reaction time \(T_r = 0.4 \mathrm{~s}\) and the braking time \(T_b = 6.11 \mathrm{~s}\). Thus, \(T_n = T_r + T_b = 0.4 + 6.11 = 6.51 \mathrm{~s}.\)

Step 6: Compare Reaction and Braking Times

Compare \(T_r\) and \(T_b\). Since \(T_b = 6.11 \mathrm{~s}\) and \(T_r = 0.4 \mathrm{~s}\), most of the time is spent braking.

Step 7: Calculate Additional Distance due to Increased Reaction Time

If the reaction time \(T_r\) increases by \(0.1 \mathrm{~s}\), it becomes \(0.5 \mathrm{~s}\). Using the speed \(v_i = 55.56 \mathrm{~m/s}\), the additional distance \(d\) is:\[d = v_i \times 0.1 = 55.56 \times 0.1 = 5.56 \mathrm{~m}.\]

Key Concepts

Acceleration

In kinematics, acceleration refers to the rate at which an object changes its velocity. When we say a car accelerates, it could mean it is speeding up, slowing down, or changing direction. In the context of our exercise involving a car braking, the car experiences a negative acceleration, also known as deceleration, as it comes to a stop.
Acceleration can be calculated using the formula \(a = \frac{v_f - v_i}{t}\), but in our scenario, we utilized another kinematic equation due to given distance, initial velocity, and final velocity (which is zero since the car stops).
To compute the car's acceleration while braking, we used the formula \(v_f^2 = v_i^2 + 2ax\). This equation relates the initial velocity \(v_i\), final velocity \(v_f\), acceleration \(a\), and distance \(x\). Solving these values gives the magnitude of the car's acceleration as \(-9.09\,\mathrm{m/s^2}\). The negative sign indicates a reduction in speed.
Understanding acceleration helps us know how quickly an object can change its speed, which is crucial in real-world scenarios like driving, where it directly affects stopping distances and safety.

Kinematic Equations

Kinematic equations describe the motion of objects without considering the forces that cause this motion. They are foundational in classical mechanics and are used to solve various problems related to moving objects, like the car example we explored.
Three commonly used kinematic equations are:

• \(v_f = v_i + at\)
• \(v_f^2 = v_i^2 + 2ax\)
• \(x = v_i t + \frac{1}{2}at^2\)

In our exercise, we applied the second equation. This choice was due to the need to incorporate distance \(x\) and the known speeds to determine acceleration \(a\). Another equation, \(v_f = v_i + aT_b\), was used to find the braking time \(T_b\), providing us with the clear steps needed to solve such questions.
Each kinematic equation is tailored to different known quantities, helping to piece together unknown variables. They are powerful tools for students to predict and analyze the motion of anything from toys to vehicles, fostering a deeper understanding of motion.

Reaction Time

Reaction time is the interval between perceiving a stimulus and responding to it. In driving, this time involves the delay between noticing an emergency and engaging the brakes.
For our problem, \(T_r\) is given as 0.4 seconds. This represents the time taken by the driver to react before braking actually begins. In high-speed scenarios like driving at 200 km/h, even a slight delay in reaction time can significantly affect the stopping distance of a vehicle.
If reaction time increases, say due to external factors like wearing dark sunglasses—alluded to in the exercise as possibly extending \(T_r\) by 0.1 seconds—the car travels an extra distance before braking starts. Using the initial speed of 55.56 m/s, an increased reaction time of 0.5 seconds results in the car covering an additional 5.56 meters.
Understanding and improving reaction time is critical in contexts such as driving, where quick responses can avert accidents and enhance safety. Highlighting the impact of even small reaction delays emphasizes why drivers must be attentive and focused.