Problem 51 - Calculus: Early Transcendental Functions

Question

Suppose that a car can accelerate from 30 mph to 50 mph in 4 seconds. Assuming a constant acceleration, find the acceleration (in miles per second squared) of the car and find the distance traveled by the car during the 4 seconds.

Step-by-Step Solution

Answer

The car's acceleration is \(2.2352 ms^{-2}\) and it traveled a distance of approximately \(0.044448 miles\) during the 4 seconds.

Step 1: Finding the Acceleration

Firstly, convert the speeds from miles per hour to miles per second because the time is given in seconds. 1 mph is approximately 0.44704 m/s. So, 30 mph is \(30 \times 0.44704 ms^{-1} = 13.4112 ms^{-1}\) and 50 mph is \(50 \times 0.44704 ms^{-1} = 22.352 ms^{-1}\). The acceleration \(a\) can be found using the formula \[a = \frac{{v_f - v_i}}{t}\] where \(v_i\) is the initial velocity (30 mph), \(v_f\) is the final velocity (50 mph) and \(t\) is the time (4 seconds).

Step 2: Calculation of Acceleration

Plugging in the given velocities (after converting them to m/s) and time into the formula will yield: \[a = \frac{{22.352 - 13.4112}}{4} = 2.2352 ms^{-2}\]. Therefore, the acceleration of the car is \(2.2352 ms^{-2}\).

Step 3: Finding the Distance Traveled

The distance \(s\) traveled by the car can be found using the second equation of motion: \[s = v_it + \frac{1}{2}at^2\]. Plugging in the given values: \[s = 13.4112 \times 4 + \frac{1}{2} \times 2.2352 \times (4^2) \].

Step 4: Calculation of Distance

Solving the equation will give: \[s = 53.6448 + 17.8816 = 71.5264 m\]. The distance traveled by the car during the 4 seconds is 71.5264 meters. However, since the question specified 'miles ', we need to convert meters back to miles. As 1 meter is approximately 0.000621371 miles, the final answer is \(0.000621371 \times 71.5264 = 0.044448 miles \).

Key Concepts

Velocity Conversion

To solve problems involving speed, acceleration, and time, it's often necessary to convert units. This is particularly important when you're working with different systems like miles per hour (mph) and meters per second (m/s). In many physics problems, we deal with meters per second because our equations are in SI units.
When converting velocity, remember:

• 1 mph is approximately 0.44704 m/s.
• To convert mph to m/s, multiply by 0.44704.
• To convert m/s back to mph, divide by 0.44704.

Let's apply this conversion. Suppose you're given velocities in mph, like 30 mph and 50 mph. Converting them to m/s gives you 13.4112 m/s and 22.352 m/s, respectively. This step is essential before plugging these values into constant acceleration equations.

Distance Calculation

Calculating the distance traveled when an object is under constant acceleration is a frequent need in physics problems. We use the equation of motion for this: \[ s = v_i t + \frac{1}{2} a t^2 \]where:

• \(s\) is the distance traveled.
• \(v_i\) is the initial velocity.
• \(t\) is the time.
• \(a\) is the acceleration.

Start by inserting your known values—initial velocity, time, and acceleration—into this equation.
In our example, with an initial velocity of 13.4112 m/s, a time of 4 seconds, and an acceleration of 2.2352 m/s², the solution steps through the formula and delivers a distance traveled as 71.5264 meters. Often, your final step will be to convert meters to miles if asked for distance in miles.

Equations of Motion

Understanding the set of equations known as the "equations of motion" is key to solving physics problems involving constant acceleration. These equations relate velocity, time, distance, and acceleration, making it possible to find an unknown value when three are known. Let's focus on a primary equation:The first equation of motion connects initial velocity \(v_i\), final velocity \(v_f\), acceleration \(a\), and time \(t\):\[ v_f = v_i + at \]In our problem, using this helps calculate the acceleration since we know the initial and final velocities and the time.
The second equation of motion, sneaked into our distance calculation, helps find distance when you know initial velocity, time, and acceleration as follows:\[ s = v_i t + \frac{1}{2} a t^2 \]These formulas are part of a toolkit that allows you to dissect almost any motion-based question involving constant acceleration. Remember, these equations assume constant acceleration—an essential detail when applying them to solve real-world and theoretical problems.