Problem 35 - College Physics

Question

An electric drag racer is much like its piston engine counterpart, but instead it is powered by an electric motor running off of onboard batteries. These vehicles are capable of covering a \(\frac{1}{4}\) mile straight-line track in \(10 \mathrm{~s}\). (a) Determine the acceleration of the drag racer in units of \(\mathrm{m} / \mathrm{s}\). (Assume that the acceleration is constant throughout the race.) (b) How does the value you get compare with the acceleration of gravity? (c) Calculate the final speed of the drag racer in \(\mathrm{mi} / \mathrm{h}\).

Step-by-Step Solution

Answer

(a) 8.045 m/s², (b) Less than 9.81 m/s², (c) 179.95 mi/h.

Step 1: Convert Distance to Meters

The track covers a distance of \(\frac{1}{4}\) mile. First, convert this distance into meters. Since 1 mile is approximately 1609 meters, we calculate:\[\text{Distance in meters} = \frac{1}{4} \times 1609 = 402.25 \text{ meters}\]

Step 2: Determine the Acceleration

Using the formula for constant acceleration, \(s = ut + \frac{1}{2}at^2\), where the initial speed \(u = 0\), we solve for acceleration \(a\):\[s = \frac{1}{2}at^2\]Substitute \(s = 402.25 \text{ meters}\) and \(t = 10 \text{ seconds}\) to get:\[402.25 = \frac{1}{2}a(10)^2\]\[402.25 = 50a\]\[a = \frac{402.25}{50} = 8.045 \text{ m/s}^2\]

Step 3: Compare with Gravity

The acceleration due to gravity is \(9.81 \text{ m/s}^2\). Compare the drag racer's acceleration with gravity:\[a = 8.045 \text{ m/s}^2 < 9.81 \text{ m/s}^2\]The racer's acceleration is slightly less than gravitational acceleration.

Step 4: Calculate Final Speed

Use the final speed formula: \(v = u + at\), with \(u = 0\):\[v = 0 + 8.045 \times 10 = 80.45 \text{ m/s}\]To convert \(v\) to miles per hour (since 1 \(\text{m/s} = 2.237 \text{mi/h}\)):\[v = 80.45 \times 2.237 = 179.95 \text{ mi/h}\]

Key Concepts

Understanding Constant Acceleration

Electric drag racers rely on constant acceleration to achieve high speeds over a short distance. Constant acceleration means the velocity of the drag racer increases steadily over time. In our exercise, the acceleration remains the same from start to finish, regardless of the racer's speed at any given moment. This simplifies calculations and makes it easier to predict the elapsed time over a certain distance.
The formula we'll use to explore this is \[ s = ut + \frac{1}{2}at^2 \] where:

• \( s \) is the distance traveled,
• \( u \) is the initial velocity (zero in our example),
• \( a \) is the acceleration,
• \( t \) is the time.

This equation helps us solve for acceleration \( a \) when time and distance are known, establishing a foundation for understanding motion in physics.

Comparing Acceleration with Gravity

In physics, comparing different forms of acceleration can offer valuable insight into their relative strength. Here, we compare the electric drag racer's acceleration to the acceleration due to gravity.
The standard acceleration due to gravity on Earth is \( 9.81 \text{ m/s}^2 \). In our case, we calculated the racer's acceleration to be \( 8.045 \text{ m/s}^2 \). This means that the racer accelerates a bit more slowly than free-falling objects drawn by gravity.
These comparisons help us understand how rapid the change in velocity is for electric drag racers, placing it in the context of a natural force all of us are familiar with—gravity. It also highlights just how powerful these vehicles are, coming close to a natural force that influences everything on Earth.

The Importance of Unit Conversions

When solving physics problems, converting units is often necessary to make your calculations consistent and meaningful. In this exercise, we needed to convert the distance from miles to meters to use the standard units in physics equations.
Since 1 mile roughly equals 1609 meters, converting a quarter-mile distance gives us approximately 402.25 meters. This allows us to employ consistent units when using equations for motion, such as \( s = \frac{1}{2}at^2 \).
Later, we had to convert the final speed from meters per second (\text{m/s\}) to miles per hour (\text{mi/h\}). This conversion used the factor \( 1 \text{ m/s} = 2.237 \text{ mi/h} \).
Accurate unit conversion is crucial as it ensures that our calculations reflect real-world measurements, providing us with results that are universally understood.

Calculating the Final Speed

To determine how fast the electric drag racer is traveling at the quarter-mile mark, we calculate its final speed. This uses the equation \[ v = u + at \] where:

• \( v \) is the final velocity,
• \( u \) is the initial velocity (zero in this scenario),
• \( a \) is the acceleration,
• \( t \) is the time elapsed.

Plugging in our values, we find \( v = 80.45 \text{ m/s} \).
We convert this speed into miles per hour for better relatability, resulting in approximately \( 179.95 \text{ mi/h} \). This final speed demonstrates the incredible power of electric drag racers, capable of reaching such elevated velocities in just ten seconds.