Chapter 2: Problem 29
The brakes on your automobile are capable of slowing down your car at a rateof \(5.2 \mathrm{~m} / \mathrm{s}^{2} .(\mathrm{a})\) If you are going \(137\mathrm{~km} / \mathrm{h}\) and suddenly see a state trooper, what is theminimum time in which you can get your car under the \(90 \mathrm{~km} /\mathrm{h}\) speed limit? The answer reveals the futility of braking to keepyour high speed from being detected with a radar or laser gun.) (b) Graph \(x\)vs. \(t\) and \(v \mathrm{vS}, t\) for such a deceleration.
Short Answer
Expert verified
The minimum time required to reduce the speed from 137 km/h to 90 km/h is 2.51 seconds
Step by step solution
01
- Convert Speeds to m/s
First, convert both speeds from km/h to m/s. Use the conversion factor: 1 km/h = 0.27778 m/s.Initial speed ( v_i n) = 137 km/h * 0.27778 = 38.06 m/sFinal speed ( v_f ) = 90 km/h * 0.27778 = 25 m/s
02
- Use the Deceleration Formula
The deceleration formula is given by: a = (v_f - v_i) / t, where a is the acceleration (or deceleration in this case).Rearrange it to solve for time ( t ): t = (v_f - v_i) / a
03
- Substitute Values
Substitute the given values into the rearranged formula:t = (25 - 38.06) / (-5.2)= -13.06 / -5.2 = 2.51 seconds
04
- Graph x vs. t and v vs. t
Plot the graphs of both distance ( x ) versus time ( t ) and velocity ( v ) versus time ( t ):For velocity vs. time:the line will be a straight line with a negative slope starting at 38.06 m/s at t = 0 and ending at 25 m/s at t = 2.51 s.For distance vs. time:, use the formula x = v_i * t + 0.5 * a * t^2. Plug in the values and calculate the distance as a function of time.
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Kinematics
In physics, kinematics is the branch that deals with the motion of objects without considering the forces that cause the motion. When analyzing the movement of a car decelerating, we need to apply kinematic equations. These equations relate the initial and final velocities, acceleration (or deceleration), time, and displacement. Here’s a breakdown of key terms:
- Velocity (\textbf{v}): The rate at which an object changes its position. Measured in meters per second (m/s).
- Acceleration (\textbf{a}): The rate at which an object changes its velocity. When a car slows down, this is termed deceleration and usually has a negative value. Measured in meters per second squared (m/s\textsuperscript{2}).
- Time (\textbf{t}): The duration over which the change occurs. Measured in seconds (s).
We use these components in the formula:\[ a = \frac{v_f - v_i}{t} \]Rearranging it to find time (t) gives:\[ t = \frac{v_f - v_i}{a} \]Applying these principles helps in determining how quickly a car can decelerate to a lower speed.
Conversion of Units
When solving physics problems, it’s crucial to ensure all units are consistent. In the given problem, the speeds are provided in kilometers per hour (km/h), but we need them in meters per second (m/s). This is done using the conversion factor: 1 km/h = 0.27778 m/s. Here’s a detailed conversion:
Initial speed (137 km/h) conversion:
- Multiply by the conversion factor: 137 * 0.27778 = 38.06 m/s.
Final speed (90 km/h) conversion:
- Multiply by the conversion factor: 90 * 0.27778 = 25 m/s.
These conversions are essential for accurately applying the kinematic formulas, which require consistent units of measure. Converting units ensures that you don’t end up with incorrect values due to mismatched units.
Graphical Analysis
Graphing is a vital part of physics as it provides visual insight into the motion described by the kinematic equations. For this problem, we consider two types of graphs: velocity (\textbf{v}) vs. time (\textbf{t}) and displacement (\textbf{x}) vs. time (t):
- **Velocity vs. Time:** This graph for deceleration is a straight line with a negative slope. Starting from 38.06 m/s at t = 0 and ending at 25 m/s at t = 2.51 s. The slope represents the deceleration rate.
- **Distance vs. Time:** To plot this, we use the formula for displacement:\[ x = v_i * t + 0.5 * a * t^2 \]Plugging in the values:
\[ x = 38.06 * t + 0.5 * (-5.2) * t^2 \]This gives a parabolic curve, starting at the origin (0,0) and curving upwards before slowing down due to deceleration.
Graphing these equations helps in understanding how the vehicle's speed decreases over time and how far it travels during deceleration. Visual tools like these are very helpful in making abstract concepts more concrete.
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