Question

The position of a particle moving in an \(x y\) plane is given by \(\overrightarrow{\mathbf{r}}=\left[\left(2 \mathrm{~m} / \mathrm{s}^{3}\right) t^{3}-(5 \mathrm{~m} / \mathrm{s}) t\right] \hat{\mathbf{i}}+\left[(6 \mathrm{~m})-\left(7 \mathrm{~m} / \mathrm{s}^{4}\right) t^{4}\right] \hat{\mathbf{j}} . \quad\) Cal- culate \((a) \overrightarrow{\mathbf{r}},(b) \overrightarrow{\mathbf{v}}\), and \((c) \overrightarrow{\mathbf{a}}\) when \(t=2 \mathrm{~s}\).

Short answer

After taking the steps, we obtain the values for: \n(a) \(\overrightarrow{\mathbf{r}}\) at t=2s \n(b) \(\overrightarrow{\mathbf{v}}\) at t=2s \n(c) \(\overrightarrow{\mathbf{a}}\) at t=2s.

Step-by-step solution

Find Position Vector at t=2s

To find the position vector at \( t=2 s \), we substitute \( t=2 \) into the position vector. \n \[ \overrightarrow{\mathbf{r}}=\left[\left(2 \mathrm{~m} / \mathrm{s}^{3}\right) t^{3}-(5 \mathrm{~m} / \mathrm{s}) t\right] \hat{\mathbf{i}}+\left[(6 \mathrm{~m})-\left(7 \mathrm{~m} / \mathrm{s}^{4}\right) t^{4}\right] \hat{\mathbf{j}} \]

Differentiate position vector to find velocity

Differentiate the position vector with respect to time to find the velocity vector. Remember to apply the power rule for differentiation, d/dx[x^n] = nx^{n-1}. Apply this rule to each term in the position vector to find the velocity vector.

Evaluate velocity vector at t=2s

Substitute \( t = 2 s \) into the equation for the velocity vector to find the velocity of the particle at \( t = 2 s \).

Differentiate velocity vector to find acceleration

Differentiate the velocity vector with respect to time to find the acceleration vector. Use the chain rule for differentiation, d/dx[f(g(x))] = f'(g(x)) * g'(x). Apply this rule to each term in the velocity vector to get the acceleration vector.

Evaluate acceleration vector at t=2s

Substitute \( t = 2 s \) into the equation for the acceleration vector to find the acceleration of the particle at \( t = 2 s \).

Key concepts

Position Vector

In physics, the position vector describes the location of a particle in space relative to a reference point, often the origin of a coordinate system. Think of it as an arrow pointed from the origin to the particle's position. In our exercise, the particle moves in an \(xy\)-plane, and its position vector is given by:
\[\overrightarrow{\mathbf{r}}=[(2 \mathrm{~m} / \mathrm{s}^{3}) t^{3} - (5 \mathrm{~m} / \mathrm{s}) t] \hat{\mathbf{i}} + [(6 \mathrm{~m}) - (7 \mathrm{~m} / \mathrm{s}^{4}) t^{4}] \hat{\mathbf{j}}. \]
This vector consists of two parts:

• \(\hat{\mathbf{i}}\)-component: This reveals how the position changes along the x-axis.
• \(\hat{\mathbf{j}}\)-component: This indicates position changes along the y-axis.

To find the position at a specific time \(t = 2\) seconds, you simply substitute \(t = 2\) into this equation and solve for \(\overrightarrow{\mathbf{r}}\). You'll see how the particle is positioned relative to the origin at that given time, providing the exact coordinates in the \(xy\)-plane.

Velocity Vector

The velocity vector is a crucial concept because it tells us both the speed and direction of a particle's motion. In this exercise, we derive the velocity vector by differentiating the position vector with respect to time. Differentiation helps us find the rate of change, which in this context is the rate of positional change per unit time.
The formula for our velocity vector \(\overrightarrow{\mathbf{v}}\) is derived by applying the standard power rule for differentiation on each term of the position vector. For example, if you have a term like \(t^3\), it becomes \(3t^2\) when differentiated.
After differentiating the position vector, you get the velocity vector in component form, which too, will have \(\hat{\mathbf{i}}\) and \(\hat{\mathbf{j}}\) components. Evaluating this at \(t = 2 \text{ seconds}\) will give you the velocity at that moment, helping you understand how fast and in what direction the particle moves at that exact time point.

Acceleration Vector

To delve into the acceleration vector, think of it as the change in velocity over time. Acceleration indicates how the velocity of a particle changes, which includes changes in the speed as well as the direction of motion. For this exercise, deriving the acceleration vector involves differentiating the velocity vector with respect to time.
By applying the rules of differentiation, again, particularly the power rules, we obtain a new vector equation that represents acceleration. Differentiation translates changes in the velocity vector's \(\hat{\mathbf{i}}\) and \(\hat{\mathbf{j}}\) components into an acceleration vector. Each component now reflects how the rate of change of position—velocity, in this sense—is itself changing.
Substituting \(t = 2\) seconds into this equation gives you the acceleration at that exact time, describing whether the particle is speeding up or slowing down, and how the direction of its motion might be shifting.

Differentiation

Differentiation is a powerful mathematical tool used to determine rates of change. In this context, it's used extensively to move from position to velocity, and from velocity to acceleration. Understanding differentiation can demystify much of motion analysis.
The basic rule you encounter first is the power rule which is expressed as \(\frac{d}{dt}[t^n] = nt^{n-1}\). This becomes useful when identifying how each part of the position changes over time to give you velocity. For combining terms, another vital rule is the chain rule: \(\frac{d}{dt}[f(g(t))] = f'(g(t)) \cdot g'(t)\).
In our exercise, differentiation lets us transform the position vector, which outlines the particle's location at any time, into a velocity vector that explains its speed and path, and then further into an acceleration vector which details how its speed and path are changing.