Question

A particle moves so that its position as a function of time is $$ \overrightarrow{\mathbf{r}}(t)=A \hat{\mathbf{i}}+B t^{2} \hat{\mathbf{j}}+C t \hat{\mathbf{k}} $$ where \(A=1.0 \mathrm{~m}, B=4.0 \mathrm{~m} / \mathrm{s}^{2}\), and \(C=1.0 \mathrm{~m} / \mathrm{s}\). Write ex- pressions for \((a)\) its velocity and \((b)\) its acceleration as functions of time. \((c)\) What is the shape of the particle's trajectory?

Short answer

The velocity of the particle is \(0\hat{\mathbf{i}}+ 8.0t\hat{\mathbf{j}}+1.0\hat{\mathbf{k}}\) m/s, the acceleration of the particle is \(0\hat{\mathbf{i}}+ 8.0\hat{\mathbf{j}}+0\hat{\mathbf{k}}\) m/s², and the shape of the particle's trajectory is a parabolic curve.

Step-by-step solution

Calculation of velocity

The velocity of the particle is given by the derivative of the position. Thus, \(\overrightarrow{\mathbf{v}}(t) = d\overrightarrow{\mathbf{r}} / dt = 0\hat{\mathbf{i}}+ (2Bt)\hat{\mathbf{j}}+C\hat{\mathbf{k}} = 0\hat{\mathbf{i}}+ (2(4.0)t)\hat{\mathbf{j}}+(1.0)\hat{\mathbf{k}} = 0\hat{\mathbf{i}}+ 8.0t\hat{\mathbf{j}}+1.0\hat{\mathbf{k}}\)

Calculation of acceleration

The acceleration is given by the derivative of the velocity. Thus, \(\overrightarrow{\mathbf{a}}(t) = d\overrightarrow{\mathbf{v}} / dt = 0\hat{\mathbf{i}}+ (2B)\hat{\mathbf{j}}+0\hat{\mathbf{k}} = 0\hat{\mathbf{i}}+ (2(4.0))\hat{\mathbf{j}}+0\hat{\mathbf{k}} = 0\hat{\mathbf{i}}+ 8.0\hat{\mathbf{j}}+0\hat{\mathbf{k}}\)

Shape of Trajectory

For the shape of the trajectory, observe that only the y-component (vertical component) of the position depends on time, and in particular, it depends quadratically on time, which is characteristic of a parabolic trajectory. The x-component (horizontal component) and z-component do not change with time, indicating that the particle moves along a line in these directions. Therefore, the shape of the trajectory in 3D space would be a parabolic curve.

Key concepts

Velocity

To understand the velocity of a particle, we look at how its position changes with time. Velocity is a vector, which means it has both a magnitude and a direction. In physics, the velocity of a particle is obtained by differentiating its position with respect to time. In this case, the position is given as \[ \overrightarrow{\mathbf{r}}(t) = A \hat{\mathbf{i}} + B t^{2} \hat{\mathbf{j}} + C t \hat{\mathbf{k}} \]. By applying differentiation, we find the velocity function:

• The derivative of the constant component \(A\) in \(\hat{\mathbf{i}}\) equals 0.
• The derivative of \(B t^2\) becomes \(2Bt\) in \(\hat{\mathbf{j}}\).
• The derivative of \(C t\) is simply \(C\) in \(\hat{\mathbf{k}}\).

This results in the velocity as \(\overrightarrow{\mathbf{v}}(t) = 0\hat{\mathbf{i}}+ 8.0t\hat{\mathbf{j}}+1.0\hat{\mathbf{k}}\).
This equation tells us that the particle's velocity changes in the \(\hat{\mathbf{j}}\) direction, which increases linearly with time, while it remains constant in the \(\hat{\mathbf{k}}\) direction.

Acceleration

Acceleration further explains how the velocity of the particle changes with time. It is the rate of change of velocity and can also be defined as the second derivative of the position function.
To find the acceleration, take the derivative of the velocity function \(\overrightarrow{\mathbf{v}}(t)\), which we derived as \(0\hat{\mathbf{i}} + 8.0t\hat{\mathbf{j}} + 1.0\hat{\mathbf{k}}\).

• The derivative of zero in \(\hat{\mathbf{i}}\) is still zero.
• The derivative of \(8.0t\) is \(8.0\) in \(\hat{\mathbf{j}}\).
• The derivative of the constant 1.0 in \(\hat{\mathbf{k}}\) is zero.

Thus, the acceleration is given by \(\overrightarrow{\mathbf{a}}(t) = 0\hat{\mathbf{i}} + 8.0\hat{\mathbf{j}} + 0\hat{\mathbf{k}}\).
This means the particle experiences a constant acceleration in the \(\hat{\mathbf{j}}\) direction.

Particle Trajectory

The trajectory of a particle is the path it follows as it moves through space. In this problem, trajectory is determined by analyzing the position function \[ \overrightarrow{\mathbf{r}}(t) = A \hat{\mathbf{i}} + B t^{2} \hat{\mathbf{j}} + C t \hat{\mathbf{k}} \]. Notably, the only time-dependent component is the \(B t^2\) in the \(\hat{\mathbf{j}}\) direction, which produces a characteristic parabolic curve.

• The \(A\) term in \(\hat{\mathbf{i}}\) is constant, indicating no movement along the x-axis.
• The \(B t^2\) in \(\hat{\mathbf{j}}\) results in a quadratic time dependence, shaping a parabolic motion along the y-axis.
• The \(C t\) in \(\hat{\mathbf{k}}\) contributes a linear progression along the z-axis.

The trajectory, therefore, reveals that the particle moves along a line in the x and z directions, but displays a parabolic motion due to its \(y\)-component. This mixture creates a 3D parabolic curve in space.