Problem 52 - Calculus: Early Transcendental Functions

Question

Suppose that a car can come to rest from 60 mph in 3 seconds. Assuming a constant (negative) acceleration, find the acceleration (in miles per second squared) of the car and find the distance traveled by the car during the 3 seconds (i.e., the stopping distance).

Step-by-Step Solution

Answer

The car's acceleration is approximately -8.9413 mps^2, and its stopping distance is approximately 40.234 m.

Step 1: Convert speed from mph to mps

In order to solve this problem, all the measurements need to be consistent. So, firstly, convert the speed from miles per hour(mph) to miles per second(mps). 1 mph is approximately equal to 0.44704 mps. So, multiply 60 mph by 0.44704 to get the speed in mps.

Step 2: Calculate acceleration

The formula for acceleration is \(a = \frac{{v_f - v_i}}{{t}}\) where \(v_f\) is the final velocity, \(v_i\) is the initial velocity and \(t\) is the time. Given that the car comes to rest, it means that the final velocity \(v_f\) is 0. Plug the values into the formula to calculate acceleration.

Step 3: Calculate stopping distance

To calculate the distance traveled or stopping distance, use the formula \(d = v_i * t + \frac{1}{2} * a * t^2\). Plug\(v_i\), \(t\), and \(a\) (from step 2) into the formula to calculate the stopping distance.

Key Concepts

Constant Acceleration

When we talk about constant acceleration, it refers to the unchanging rate at which an object's velocity increases or decreases over time. It's crucial to understand that acceleration isn't only about speeding up; it can also mean slowing down, which in physics terms is called deceleration or negative acceleration.

In the context of the exercise, the car is decelerating as it comes to a stop, which means it has a constant negative acceleration. Knowing this is a prerequisite for using the kinematic equations that describe the motion of the car. When the acceleration is constant, calculating variables like final velocity and distance becomes more straightforward because the acceleration doesn't change at different times throughout the process.

Velocity Conversion

Velocity conversion is an essential part of solving many physics problems because equations require consistency in units. In our exercise, the car's initial speed is given in miles per hour (mph), but to use it in kinematic formulas, we need to convert that speed into miles per second (mps).

How to Convert from mph to mps

Since 1 mph is approximately equal to 0.44704 mps, you multiply the speed value by this conversion factor. For the car traveling at 60 mph, you would calculate its speed in mps as follows:

1. Multiply the speed value by the conversion factor:
   \(60 \times 0.44704 \approx 26.8224 \; \text{mps}\)

This step ensures that all subsequent calculations are accurate and consistent. It's crucial for students to remember to convert speeds into the same units their kinematic equations are set up for.

Kinematic Equations

Kinematic equations allow us to predict the future position and velocity of an object moving under constant acceleration. They are crucial tools for solving a wide range of physics problems, including our exercise of calculating stopping distance.

The kinematic equation relevant to this exercise is:
\[d = v_i * t + \frac{1}{2} * a * t^2\]
In this equation:

• \(d\) represents the stopping distance, or how far the car travels before it comes to rest.
• \(v_i\) is the initial velocity of the car (which we've converted to mps).
• \(t\) is the time over which the car decelerates to a stop.
• \(a\) is the constant acceleration (in this case, deceleration).

Using the converted velocity and the time given in the problem, we can plug these values into the kinematic equation to find the car's stopping distance. This equation comes from integrating the mathematical definition of acceleration twice with respect to time, and it assumes a constant acceleration throughout the duration of the car's deceleration, which simplifies the problem significantly.