Problem 31 - Calculus of a Single Variable

Question

Stopping Distance The total stopping distance \(T\) of a vehicle is $$T=2.5 x+0.5 x^{2}$$ where \(T\) is in feet and \(x\) is the speed in miles per hour. Approximate the change and percent change in total stopping distance as speed changes from \(x=25\) to \(x=26\) miles per hour.

Step-by-Step Solution

Answer

After performing the steps, first the stopping distance at 25 mph will be found, then at 26 mph. The difference between them will give the change in stopping distance. Dividing this by the initial stopping distance at 25 mph and multiplying by 100 will give the percent change.

Step 1: Calculate the Stopping Distance at 25 mph

Substitute \(x = 25\) into the equation \(T = 2.5x + 0.5x^2\). This would give the stopping distance at 25 mph. Perform the necessary calculations to find the numerical value.

Step 2: Calculate the Stopping Distance at 26 mph

Next, substitute \(x = 26\) into the equation \(T = 2.5x + 0.5x^2\). This would give the stopping distance at 26 mph. Calculate it as in the previous step.

Step 3: Calculate the Change in Stopping Distance

Having the stopping distance at both 25 mph and 26 mph, the change in stopping distance can be found by subtracting the stopping distance at 25 mph from that at 26 mph.

Step 4: Find the Percent Change

Finally, find the percent change. For this, divide the change in stopping distance (which was found in step 3) by the initial stopping distance at 25 mph, then multiply the result by 100 to get the percent change.

Key Concepts

Calculus in Physics

Calculus is a branch of mathematics that deals with rates of change and sums of changes. In the realm of physics, calculus enables us to solve problems related to motion and forces, which typically involve variables that change continuously. For example, when looking at stopping distances, calculus allows us to derive equations that can determine how distance varies with speed.

By using a function for stopping distance that incorporates speed, such as in our exercise where the total stopping distance is a function of speed with the equation \( T = 2.5x + 0.5x^2 \), we apply calculus concepts to compute precise changes. This equation is a quadratic function, a fundamental concept in algebra, showing calculus's close relationship with other areas of mathematics. Calculus also permits us to calculate the exact point at which a vehicle will come to a halt by considering continuous acceleration, or in this case, deceleration.

Quadratic Equations

Quadratic equations take the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) is the unknown variable. In our stopping distance scenario, \( x \) represents speed and the equation \( T = 2.5x + 0.5x^2 \) is a quadratic because of the \( x^2 \) term.

The shape of a quadratic equation's graph is a parabola, which means that our stopping distances will increase exponentially, not linearly, with speed. Quadratic equations are useful for modeling situations where there is acceleration, such as the increase in stopping distance relative to the speed of a vehicle.

Percent Change

The percent change is a mathematical calculation that shows how much a quantity has changed in proportion to its initial value, expressed as a percentage. It is calculated by taking the difference (change) in the value of a quantity, dividing it by the original value, and then multiplying by 100 to convert it into a percentage.

For instance, in the context of our stopping distance problem, once we determine the stopping distances at 25 mph and 26 mph, we can calculate the absolute change. To then find the percent change, we divide this change by the stopping distance at 25 mph and multiply by 100. This gives insights into how sensitive stopping distance is to changes in speed - crucial for understanding safety dynamics in vehicle motion.

Motion-related Calculus

Motion-related calculus is all about understanding how objects move and change over time. This involves analyzing rates of change - a foundational concept in differential calculus. For example, when a vehicle is slowing down to a stop, we are interested in its stopping distance and how it varies with respect to its speed.

To solve motion-related problems in calculus, we often use derivatives to find rates of change and integrals to find total changes over intervals. These tools help us quantify how motion is affected by various factors, such as acceleration, time, and in this case, speed. In the stopping distance example, using calculus allows us to predict how much additional distance is required to stop if a vehicle's speed increases - a practical application of calculus in everyday life.