Problem 30 - Calculus Single Variable

Question

An automobile is cruising at a constant speed of \(55 \mathrm{mi} / \mathrm{hr}\). To pass another vehicle, the car accelerates at a constant rate. In the course of one minute the car covers 1.3 miles. What is the rate at which the car is accelerating? What is the speed of the car at the end of this minute?

Step-by-Step Solution

Answer

Acceleration is 0.7666 miles/min²; final speed is 101 mph.

Step 1: Identify Initial Conditions

The car is initially traveling at a constant speed of 55 mi/hr, which we need to convert into miles per minute. Since there are 60 minutes in an hour, the initial speed in miles per minute is \( \frac{55}{60} \approx 0.9167 \) miles per minute.

Step 2: Establish Total Distance Travelled

We know the car covers a total of 1.3 miles in one minute while accelerating.

Step 3: Use Kinematic Equation for Distance

The kinematic equation relating distance \( d \), initial speed \( v_0 \), time \( t \), and acceleration \( a \) is:\[ d = v_0 t + \frac{1}{2} a t^2\]Substitute the known values: \( d = 1.3 \) miles, \( v_0 = 0.9167 \) miles/min, and \( t = 1 \) minute.

Step 4: Substitute Values into Equation

Substitute the known values into the equation:\[ 1.3 = 0.9167 \times 1 + \frac{1}{2} a \times 1^2\]Simplify to find the acceleration \( a \).

Step 5: Solve for Acceleration

Rearrange the equation to solve for \( a \):\[1.3 - 0.9167 = \frac{1}{2} a\]\[0.3833 = \frac{1}{2} a\]\[a = 2 \times 0.3833 = 0.7666 \text{ miles/min}^2\]

Step 6: Determine Final Speed

Use the equation for final speed \( v = v_0 + at \). Substitute the known values: \( v_0 = 0.9167 \) miles/min, \( a = 0.7666 \) miles/min², and \( t = 1 \) min.\[v = 0.9167 + 0.7666 \times 1\]\[v = 1.6833 \text{ miles/min}\]

Step 7: Convert to Miles per Hour

Since 1 mile/min equals 60 mph, convert the final speed from miles/min to miles/hr:\[v = 1.6833 \times 60 = 101 \text{ mph}\]

Key Concepts

Understanding Acceleration

Imagine pressing the gas pedal when driving. The car goes faster. This change in speed over time is called acceleration. It tells us how quickly a car (or anything in motion) is speeding up or slowing down. Acceleration is measured in terms of speed per unit of time. For example, in this exercise, it was measured in miles per minute squared (\(\text{mi/min}^2\)). The exercise shows how even a small acceleration like \(0.7666 \text{ mi/min}^2\) can significantly impact speed. This number indicates how much the velocity increases every minute.Knowing how to calculate acceleration through kinematic equations is important: it allows us to predict future vehicle behavior during speed changes.

Concept of Motion

Motion is all about change - moving from one position to another. When we talk about a car moving, we describe its motion using parameters like speed and acceleration. In this exercise, the car initially moves at a steady pace but begins to accelerate to pass another vehicle. To calculate the motion precisely, we use kinematic equations. These equations relate speed, time, distance, and acceleration. For instance, if a car accelerates, we can calculate how far it travels in a certain period and its new speed after that time. Understanding motion helps in various real-world scenarios, from planning a journey to more complex physics-related tasks. It's the difference between just watching a car drive by and knowing what's happening under the hood.

Calculating Distance Using Kinematics

Knowing how far something travels in a given time while it's accelerating is crucial in kinematic equations. Distance calculation becomes vital when assessing motion problems, like knowing how much distance a vehicle covers while speeding up.In this exercise, we utilized the equation\[d = v_0 t + \frac{1}{2} a t^2\]where:

• \( d \)is the total distance traveled,
• \( v_0 \)is the initial velocity,
• \( t \)is the time,
• \( a \)is the acceleration.

To find how far the car traveled while it was accelerating, we plugged in the known values into this equation and solved it. This step is essential as it lays down the foundation to understand both the current and future positions of objects in motion.