Problem 30 - Fundamentals of Physics

Question

The brakes on your car can slow you at a rate of \(5.2 \mathrm{~m} / \mathrm{s}^{2}\). (a) If you are going \(137 \mathrm{~km} / \mathrm{h}\) and suddenly see a state trooper, what is the minimum 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 keep your high speed from being detected with a radar or laser gun.) (b) Graph \(x\) versus \(t\) and \(v\) versus \(t\) for such a slowing.

Step-by-Step Solution

Answer

It takes a minimum of approximately 2.51 seconds to decelerate to 90 km/h.

Step 1: Convert units to standard format

First, convert the given speeds from kilometers per hour to meters per second. 1 kilometer per hour is equal to \( \frac{1000}{3600} \) meters per second.Initial speed: \[ 137 \frac{km}{h} = 137 \times \frac{1000}{3600} = 38.06 \frac{m}{s} \] Final speed (speed limit): \[ 90 \frac{km}{h} = 90 \times \frac{1000}{3600} = 25 \frac{m}{s} \]

Step 2: Use the equation for motion

The equation to find time when the acceleration and initial and final velocities are known is given by:\[ v = u + at \]where:- \(v\) is the final velocity (25 m/s),- \(u\) is the initial velocity (38.06 m/s),- \(a\) is the acceleration (-5.2 m/s², negative since it's deceleration),- \(t\) is the time we need to find.

Step 3: Solve for time (t)

Substituting the known values in the equation:\[ 25 = 38.06 - 5.2t \]Rearrange to solve for \(t\):\[ 5.2t = 38.06 - 25 \]\[ 5.2t = 13.06 \]\[ t = \frac{13.06}{5.2} \approx 2.51 \, ext{seconds} \]

Step 4: Analyze the conclusions of urgency

It would take approximately 2.51 seconds to slow down to the speed limit. The quick deceleration but still visible over-speeding highlights the futility of attempting to brake abruptly to avoid detection.

Step 5: Graphing x vs t and v vs t

To graph these, consider the relationship:- Distance (x) travelled over time can be calculated from the initial velocity and the acceleration using:\[ x = ut + \frac{1}{2}at^2 \]For the graphs:- **x versus t**: plot the calculated distance over the timespan, representing a parabolic curve since acceleration affects displacement quadratically.- **v versus t**: plot a straight line with a negative slope given by the deceleration rate.

Step 6: Calculation of distance (x) for graphing

Substitute values into the equation:\[ x = 38.06 \times t + \frac{1}{2}(-5.2)t^2 \]\[ x = 38.06 \times 2.51 - 2.6 \times (2.51)^2 \]\[ x = 95.4976 - 2.6 \times 6.3001 \]\[ x = 95.4976 - 16.38026 \]\[ x \approx 79.12 \, ext{meters} \]

Key Concepts

Kinematic Equations

Kinematic equations are very helpful when you want to understand and predict motion. These equations relate the five main variables of motion: displacement, initial velocity, final velocity, acceleration, and time. They enable you to solve various types of problems involving motion, including deceleration like when a car brakes.

• The most common kinematic equation is: \[ v = u + at \]This equation shows how velocity changes over time under constant acceleration.


• To find the displacement over time, use:\[ x = ut + \frac{1}{2}at^2 \]


• Another form connects velocity and displacement without time:\[ v^2 = u^2 + 2ax \]

In our exercise, these equations were key in calculating the time needed for a car to decelerate. The initial speed, rate of deceleration, and desired final speed were known, and the kinematic equations helped solve for the unknown time.

Velocity Conversion

Velocity conversion is essential when solving problems using kinematic equations that require standardized units. In physics, velocity is often measured in meters per second (m/s) rather than kilometers per hour (km/h).

• This conversion is crucial to attain consistent units across all calculations. For example, to convert 137 km/h to m/s:\[ 137 \times \frac{1000}{3600} = 38.06 \frac{m}{s} \]


• Similarly, to convert 90 km/h to m/s:\[ 90 \times \frac{1000}{3600} = 25 \frac{m}{s} \]

Using the converted values ensures that the calculations using kinematic equations result in accurate, reliable solutions. It also helps in understanding how fast an object is moving in relation to standard scientific measures.

Graphing Motion

Graphing motion involves plotting key motion characteristics like distance and velocity against time, offering a visual comprehension of how an object's motion evolves.

• **x versus t Graph**: This represents the distance travelled over time. Given that the equation \( x = ut + \frac{1}{2}at^2 \) includes acceleration, the graph is a curve. At the start, the initial velocity contributes to the motion, but as time advances, the deceleration causes the distance curve to flatten.


• **v versus t Graph**: A linear downward slope signifies the constant rate of deceleration. It starts at the initial velocity (38.06 m/s) and decreases linearly to the final velocity (25 m/s) over the calculated deceleration time (2.51 seconds).

These graphs enable you to visualize motion, helping to interpret the problem's physical context beyond mere numbers and equations.