Problem 87 - Calculus and its applications

Question

A car accelerates at a constant rate from 0 to \(60 \mathrm{mph}\) in \(30 \mathrm{sec} .\) How far does the car travel during that time?

Step-by-Step Solution

Answer

The car travels 402.336 meters.

Step 1: Identify Known Variables

The car starts from rest, so the initial velocity \( u = 0 \) mph. The final velocity \( v = 60 \) mph, which needs to be converted to \( \frac{m}{s} \). Meanwhile, the time \( t = 30 \) seconds. To convert \( v \) to \( \frac{m}{s} \), use the conversion factor \( 1 \text{ mph} = \frac{0.44704}{\text{m/s}} \).

Step 2: Convert Velocity

Convert \( 60 \text{ mph} \) to meters per second: \[v = 60 \times 0.44704 = 26.8224 \text{ m/s}.\]

Step 3: Find Acceleration

Use the formula for constant acceleration: \[a = \frac{v - u}{t}\]Substitute the values:\[a = \frac{26.8224 \text{ m/s} - 0 \text{ m/s}}{30 \text{ s}} = 0.89408 \text{ m/s}^2.\]

Step 4: Use the Formula for Distance

Use the formula for distance under constant acceleration: \[s = ut + \frac{1}{2} a t^2\]With \( u = 0 \):\[s = 0 \cdot 30 + \frac{1}{2} \cdot 0.89408 \cdot 30^2.\]

Step 5: Calculate Distance

Calculate the distance traveled:\[s = \frac{1}{2} \times 0.89408 \times 900 = 402.336 \text{ meters}.\]

Step 6: Conclusion

The car travels 402.336 meters in 30 seconds while accelerating from 0 to 60 mph.

Key Concepts

Distance Calculation

When calculating the distance traveled by an object under constant acceleration, understanding the formula is crucial. The distance equation we use is:\[ s = ut + \frac{1}{2} a t^2 \]In this formula, the variables are:

• \( s \): the distance traveled.
• \( u \): the initial velocity of the object. In this exercise, it's \(0 \ \text{m/s}\). This simplifies our calculation because any term multiplied by zero will be zero.
• \( a \): the acceleration of the object. This value must be calculated before using the formula.
• \( t \): the time over which the acceleration occurs.

This formula gives us the distance over which the car travels while accelerating. Because the car starts from rest, the formula simplifies to:\[ s = \frac{1}{2} a t^2 \]This step is about application and understanding that, under constant acceleration, the distance is directly dependent on both the square of the time and the acceleration experienced.

Velocity Conversion

Velocity conversion is essential when working with cars because speeds are often given in miles per hour (mph), but we need meters per second (m/s) for scientific calculations. In our original exercise, the car's velocity is given as 60 mph. To convert this, we apply the conversion factor:\[ 1 \ \text{mph} = 0.44704 \ \text{m/s} \]Multiply the velocity by this factor:\[ 60 \ \text{mph} \times 0.44704 \ \text{m/s per mph} = 26.8224 \ \text{m/s} \]Why do we convert to m/s? It ensures we can accurately perform calculations using standard units of measurement. Physics often relies on coherent units like meters, seconds, and kilograms, which align with the international system (SI). Remembering this formula helps simplify and integrate calculations across different fields of physics and engineering.

Acceleration Formula

Acceleration is the rate of change of velocity. For cases involving constant acceleration, we use the formula:\[ a = \frac{v - u}{t} \]Here’s what each symbol means:

• \( a \): the acceleration.
• \( v \): the final velocity (in m/s).
• \( u \): the initial velocity (in m/s).
• \( t \): the time period over which the acceleration occurs.

Applying this to our scenario where the car accelerates from rest, we set \( u = 0 \), and the formula becomes:\[ a = \frac{v}{t} \]For our specific exercise, substituting in the converted final velocity of 26.8224 m/s and the time of 30 seconds gives:\[ a = \frac{26.8224 \ \text{m/s}}{30 \ \text{s}} = 0.89408 \ \text{m/s}^2 \]Understanding these principles enables us to find how fast an object is speeding up and comprehend how forces result in changes to an object’s motion.