# Nakafa Learning Content > For AI agents: use [llms.txt](https://nakafa.com/llms.txt) for the site index. Markdown versions are available by appending `.md` to content URLs or sending `Accept: text/markdown`. URL: https://nakafa.com/en/subjects/mathematics/derivative-function/application-of-derivative Source: https://raw.githubusercontent.com/nakafaai/nakafa.com/refs/heads/main/packages/contents/material/lesson/mathematics/derivative-function/application-of-derivative/en.mdx Learn how derivatives solve real-world physics problems. Calculate velocity, acceleration, and maximum heights with worked examples and formulas. --- ## Understanding Motion Through Derivatives Derivatives are not just an abstract concept in mathematics; they help us understand **change**. One of their most tangible applications is in physics, especially for analyzing the motion of an object. If we know the position function of an object at any given time, we can use derivatives to find its velocity and even its acceleration at any moment. ## Instantaneous Velocity from Position Imagine you are driving a car. Your position is constantly changing over time. **Velocity** is the rate of change of that position. If we have a function that describes position, $$s(t)$$, then the instantaneous velocity, $$v(t)$$, at time $$t$$ is the first derivative of the position function. Visible text: Imagine you are driving a car. Your position is constantly changing over time. **Velocity** is the rate of change of that position. If we have a function that describes position, , then the instantaneous velocity, , at time is the first derivative of the position function. ```math v(t) = s'(t) = \frac{ds}{dt} ``` This means we can determine the precise velocity at any instant, not just the average velocity over a journey. ### Determining the Velocity Function For example, the motion of a particle is determined by the position function $$s(t) = t^3 - 6t^2 + 9t$$, where $$s$$ is in meters and $$t$$ is in seconds. To find its velocity function, we simply differentiate the function $$s(t)$$ with respect to $$t$$. Visible text: For example, the motion of a particle is determined by the position function , where is in meters and is in seconds. To find its velocity function, we simply differentiate the function with respect to . Component: MathContainer Children: ```math v(t) = \frac{d}{dt}(t^3 - 6t^2 + 9t) ``` ```math v(t) = 3t^2 - 12t + 9 ``` With this function, we can calculate the particle's velocity at any time. For instance, at $$t = 1 \text{ second}$$, the velocity is $$v(1) = 3(1)^2 - 12(1) + 9 = 0 \text{ m/s}$$. This zero result indicates that the particle is momentarily at rest, perhaps pausing for a moment or even changing direction. Visible text: With this function, we can calculate the particle's velocity at any time. For instance, at , the velocity is . This zero result indicates that the particle is momentarily at rest, perhaps pausing for a moment or even changing direction. ## Acceleration from Velocity Now, what if the velocity itself is changing? Perhaps you're pressing the gas pedal or the brakes. This change in velocity is called **acceleration**. Just as velocity is the derivative of position, acceleration, $$a(t)$$, is the first derivative of the velocity function. This also makes it the second derivative of the position function. Visible text: Now, what if the velocity itself is changing? Perhaps you're pressing the gas pedal or the brakes. This change in velocity is called **acceleration**. Just as velocity is the derivative of position, acceleration, , is the first derivative of the velocity function. This also makes it the second derivative of the position function. ```math a(t) = v'(t) = s''(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2} ``` ### Determining the Acceleration Function Continuing with the previous particle example, we already have the velocity function $$v(t) = 3t^2 - 12t + 9$$. We can find its acceleration function by differentiating the function $$v(t)$$. Visible text: Continuing with the previous particle example, we already have the velocity function . We can find its acceleration function by differentiating the function . Component: MathContainer Children: ```math a(t) = \frac{d}{dt}(3t^2 - 12t + 9) ``` ```math a(t) = 6t - 12 ``` From this, we can find the particle's acceleration at any time. For example, at $$t = 3 \text{ seconds}$$, its acceleration is $$a(3) = 6(3) - 12 = 6 \text{ meters/second}^2$$. Visible text: From this, we can find the particle's acceleration at any time. For example, at , its acceleration is . ## Peak Application in Vertical Motion A ball is thrown straight up. Its height, $$h(t)$$ in meters, after $$t \text{ seconds}$$ is given by the equation $$h(t) = 50t - 5t^2$$. Visible text: A ball is thrown straight up. Its height, in meters, after is given by the equation . First, we can find its velocity and acceleration functions. Velocity function: ```math v(t) = h'(t) = 50 - 10t ``` Acceleration function: ```math a(t) = v'(t) = -10 ``` > The acceleration here is constant and negative. This represents Earth's **gravitational acceleration**, which always pulls objects downward. Its value is approximately $$-9.8 \text{ m/s}^2$$, but it is often rounded to $$-10 \text{ m/s}^2$$ in physics calculations to simplify things. Visible text: > The acceleration here is constant and negative. This represents Earth's **gravitational acceleration**, which always pulls objects downward. Its value is approximately , but it is often rounded to in physics calculations to simplify things. One of the most interesting questions is: **when does the ball reach its maximum height?** The ball reaches its highest point exactly when it stops for a moment to change direction and fall back down. In other words, this occurs when its velocity is exactly zero. Component: MathContainer Children: ```math v(t) = 0 ``` ```math 50 - 10t = 0 ``` ```math 10t = 50 ``` ```math t = 5 ``` So, the ball reaches its peak at $$5 \text{ seconds}$$. To find out what the maximum height is, we substitute $$t=5$$ back into the initial height function. Visible text: So, the ball reaches its peak at . To find out what the maximum height is, we substitute back into the initial height function. ```math h(5) = 50(5) - 5(5)^2 = 250 - 125 = 125 ``` Thus, the maximum height the ball reaches is **$$125 \text{ meters}$$**. Through derivatives, we not only describe motion but can also analyze and determine crucial moments, such as the highest point of a projectile. Visible text: Thus, the maximum height the ball reaches is ****. Through derivatives, we not only describe motion but can also analyze and determine crucial moments, such as the highest point of a projectile.