# Particle undergoes a displacement

• sherlockjones
In summary, the conversation discusses the concept of displacement of a particle, which refers to the change in position from its initial to final position. It is different from distance, which measures the total length of the path taken. Various factors such as initial velocity, acceleration, and external forces can affect displacement. It can be calculated by subtracting the initial position from the final position. Real-life examples of particle displacement include everyday movements such as walking and driving, as well as its importance in understanding fluid and gas dynamics.
sherlockjones
Lets say we have $$\vec{r} = x\vec{i} + y\vec{j} + z\vec{k}$$. If the particle undergoes a displacement $$\Delta \vec{r}$$ in time $$\Delta t$$ then we know that $$\Delta r \doteq \frac{d\vec{r}}{dt} \Delta t = \vec{v}\Delta t$$.

How is $$\vec{v} \Delta t$$ tangent to the particles trajectory, when we take $$\Delta t \rightarrow 0$$? Wouldn't the expression become 0? I can see how $$\Delta \vec{r} \rightarrow 0$$ as $$\Delta t \rightarrow 0$$.

Also let's say we have the following:

$$\vect{r} = A(e^{\alpha t}\vec{i} + e^{-\alpha t}\vec{j})$$

I know that $$\vec{v} = A(\alpha e^{\alpha t}\vec{i} + e^{-\alpha t}\vec{j})$$

As $$t\rightarrow \infty, e^{\alpha t} \rightarrow \infty$$ and $$e^{-\alpha t} \rightarrow 0$$. From this how do we come to the conclusion that $$\vec{r} \rightarrow Ae^{\alpha t}\vec{i}$$? I thought that $$\vec{r} \rightarrow \infty$$.

Similarily, $$\vec{v} \rightarrow \alpha Ae^{\alpha t}\vec{i}$$ when it seems like it should approach $$\infty$$. Perhaps it has to do something with the x-components?

$$\alpha$$ is a constant.

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sherlockjones said:
How is $$\vect{v} \Delta t$$ tangent to the particles trajectory, when we take $$\Delta t \rightarrow 0$$? Wouldn't the expression become 0? I can see how $$\Delta \vect{r} \rightarrow 0$$ as $$\Delta t \rightarrow 0$$.
The expression $$v\Delta t$$ would approach zero. It's the vector $$v=\lim_{t\to 0}\Delta r/Delta t$$ that is tangential to the particle's trajectory.

Also let's say we have the following:

$$\vect{r} = A(e^{\alpha t}\vect{i} + e^{-\alpha t}\vect{j})$$

I know that $$\vect{v} = A(\alpha e^{\alpha t}\vect{i} + e^{-\alpha t}\vect{j})$$

As $$t\rightarrow \infty, e^{\alpha t} \rightarrow \infty$$ and $$e^{-\alpha t} \rightarrow 0$$. From this how do we come to the conclusion that $$\vect{r} \rightarrow Ae^{\alpha t}\vect{i}$$? I thought that $$\vect{r} \rightarrow \infty$$.

Similarily, $$\vect{v} \rightarrow \alpha Ae^{\alpha t}\vect{i}$$ when it seems like it should approach $$\infty$$. Perhaps it has to do something with the x-components?

I`m not sure what you want to do here. There's an error in the expression for v (not sure if that's a typo). Then (assuming alpha>0) you take the limit as t-> infinity (what for?).

It's true that r-> infinity if t-> infinity, but If you want to approximate the trajectory for large t, you can drop the second term since it is vanishingly small in comparison with the first term.

They wanted me to sketch the trajectory. But $$\vec{r} = Ae^{\alpha t }\vec{i}} + Ae^{-\alpha t}\vec{j}$$?
Then the second term approaches 0 and the first term approaches $$\infty$$. So then doesn't $$\vec{r} \rightarrow \infty$$?

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Yes, but that doesn't help in sketching the trajectory. You want the asymptotic behaviour of r, not the limit as t-> infinity.

## 1. What is a displacement of a particle?

A displacement of a particle refers to the change in position of the particle from its initial position to its final position. It is a vector quantity that takes into account both the distance and direction of the movement.

## 2. How is displacement different from distance?

Displacement and distance are two different measurements of movement. Distance is the total length of the path taken by a particle, while displacement only considers the straight-line distance between the initial and final positions.

## 3. What factors affect the displacement of a particle?

The displacement of a particle can be affected by various factors such as the initial velocity, acceleration, and direction of the particle. External forces, such as friction and gravity, can also impact the displacement of a particle.

## 4. How is displacement calculated?

To calculate the displacement of a particle, you need to know its initial and final positions. The displacement can be calculated by subtracting the initial position vector from the final position vector.

## 5. What are some real-life examples of particle displacement?

Particle displacement can be observed in many everyday situations, such as a car moving from one location to another, a person walking from their house to the store, or a ball being thrown and caught. In the field of science, it is also a crucial concept in understanding the movement of particles in fluids and gases.

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