Problem 39 A car travels s meters in the \(... [FREE SOLUTION] (2024)

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Differentiation is a fundamental concept in calculus used to find the rate at which something changes. In this exercise, we need to find the rate of change of the car's position, which gives us the car's instantaneous velocity. The position function of the car is given by \(s(t) = -9.9 t^{2} + 79.2 t\). To find the velocity as a function of time, we take the derivative of \(s(t)\) with respect to \(t\). This derivative gives us the formula for the instantaneous velocity \(v(t) \).

The differentiation process follows these steps:

• Identify the function you want to differentiate, which is \(s(t) = -9.9 t^{2} + 79.2 t\).


• Apply the power rule of differentiation. If you have a term like \(a t^n\), its derivative is \(n a t^{n-1}\).

Following these rules, the derivative of \(s(t)\) becomes
\[ v(t) = \frac{ds}{dt} = -19.8 t + 79.2 \].

This equation tells us how the velocity of the car changes with time after the brakes are applied.

Kinematic equations describe the motion of objects without considering the causes of motion. They are essential tools in physics for analyzing the motion of vehicles, among other things. In this exercise, the kinematic equation provided is the position function of the car \(s(t) = -9.9 t^{2} + 79.2 t\).

This equation has useful information:

• The term \(-9.9 t^{2}\) relates to the decelerative effect after the brakes are applied.


• The term \(79.2 t \) gives the initial velocity component of the car before deceleration significantly affects its motion.

This function helps you track the exact position of the car at any time \(t\).

By differentiating this position function, we convert it into a velocity function that tells us how fast and in which direction the car is moving at any time.

Calculus is widely used in physics to solve problems about motion, force, and energy. This exercise is a perfect example of how calculus helps in understanding physical phenomena. Here, we've used differentiation to move from a position function to a velocity function.

This shift helps solve questions like:

• When will the car stop? To answer this, we set the velocity function to zero and solve for time, giving \(t = 4 \) seconds.


• How far will it travel before stopping? By substituting \( t = 4 \) back into the position function \(s(4)\), we find that the car travels 158.4 meters before coming to a stop.

Integrating these calculations into real-life scenarios provides a keen insight into how initial speeds and deceleration rates can influence travel distances and stopping times. Calculus methods like differentiation and integration play crucial roles in predicting and analyzing these outcomes.