Understanding Distance and Displacement through Integration

Integration is an important idea in calculus. It is widely used in physics to learn about distance and displacement. To understand these ideas better, we need to know the difference between distance and displacement and how to express them mathematically.

Distance vs Displacement

1. Displacement:

   Displacement is a term that describes how far an object has moved from its starting point. It includes both how far it went and in which direction.

   For example, if an object starts at point $x_1$ and moves to point $x_2$, we can find displacement ($\Delta x$) using this simple formula:

   $$\Delta x = x_2 - x_1$$

2. Distance:

   Distance is a measure of how far an object has traveled in total, no matter which direction it moved. For instance, if an object goes around in a circle and comes back to where it started, its displacement is zero. However, the distance it traveled is the entire length of the circle.

How to Represent These Concepts Mathematically

In physics, we often look at how distance and displacement relate to velocity, which is how fast something is moving. We can describe an object's position over time with a function, like $s(t)$ for position at time $t$.

• The velocity function $v(t)$ shows how position changes over time: $$v(t) = \frac{ds}{dt}$$

To find the displacement over a certain time period, from time $t_0$ to $t_1$, we can use an integral of the velocity function:

$$\Delta s = \int_{t_0}^{t_1} v(t) \, dt$$

Using Integration to Calculate Distance

When we want to figure out the total distance traveled, we need to think about the absolute value of velocity. This way, we account for when the object changes direction:

$$\text{Distance} = \int_{t_0}^{t_1} |v(t)| \, dt$$

This measure adds up all the small bits of distance covered during every tiny moment, no matter the direction.

Let's Work through an Example

Imagine we have an object moving with a velocity expressed by the function $v(t) = t^2 - 4t + 3$, where $t$ is in seconds and $v(t)$ is in meters per second.

1. Finding the Displacement:

   To calculate the displacement from time $t = 1$ to $t = 4$, we set up this integral:

   $$\Delta s = \int_{1}^{4} (t^2 - 4t + 3) \, dt$$

   After evaluating, we get:

   $$= \left[ \frac{t^3}{3} - 2t^2 + 3t \right]_{1}^{4} = \left[ \frac{64}{3} - 32 + 12 \right] - \left[ \frac{1}{3} - 2 + 3 \right] = \frac{43}{3} \text{ meters}$$

2. Finding the Distance:

   First, we need to find when the object changes direction. We do this by setting $v(t) = 0$:

   $$t^2 - 4t + 3 = 0$$

   Factoring gives us $t = 1$ and $t = 3$.

   Now we can find the total distance by adding the sections where the object moves:

   $$\text{Distance} = \int_{1}^{3} |v(t)| \, dt + \int_{3}^{4} |v(t)| \, dt$$

   We need to keep track of the sign of $v(t)$ to calculate each part correctly and determine the total distance traveled.

In Conclusion

Integration is a helpful tool in physics for figuring out both distance and displacement. By using integrals, we can summarize the total change in position over certain times. This way, we can better understand how objects move in a clear and mathematical way.