91Ó°ÊÓ

To understand how a particle moves, we can use the concept of the position vector. A position vector tells us the exact location of a particle in space at any given moment in time.
In our exercise, the position vector is given by \( \vec{r}(t)=(1 \mathrm{~m}) \hat{\mathrm{i}}+\left[\left(4 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2}\right] \hat{\mathrm{j}} \). This tells us the particle's location on the x-axis and y-axis for any time t.
Key points about the position vector in this problem:

• It has two components: the x-component and the y-component.
• The x-component (1 m) remains constant over time.
• The y-component (\left(4 \mathrm{~m} /\mathrm{s}^{2}\right) t^{2}\hat{\mathrm{j}}) changes with time.

The x-component being constant means that the particle does not move in the x-direction.
Meanwhile, as time progresses, the y-component grows quadratically, indicating that the particle moves faster as time goes on in the y-direction.

Velocity gives us information about how fast the position changes over time. It is the time derivative of the position vector.
In mathematical terms, velocity \( \vec{v}(t) \) is given by the derivative of \( \vec{r}(t) \).
For our exercise,

• Start with \( \vec{r}(t)=(1 \mathrm{~m}) \hat{\mathrm{i}} +\left[\left(4 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2}\right] \hat{\mathrm{j}} \).
• Take its derivative with respect to time: \left[ \vec{v}(t) = \frac{d \vec{r}(t)}{dt} \right].

Working it out gives:

• The x-component derivative: \( \frac{d}{dt}(1 \mathrm{~m}) = 0 \hat{\mathrm{i}} \).
• The y-component derivative: \( \frac{d}{dt}\left[\left(4 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2}\hat{\mathrm{j}}\right] = 8t \hat{\mathrm{j}} \).

Thus, the velocity vector \( \vec{v}(t) = 8t \hat{\mathrm{j}} \). This means speed changes linearly with time in the y-direction while remains zero in the x-direction.
This also confirms that the particle's speed in the y-direction is increasing as time goes on.

Acceleration tells us how fast the velocity of a particle changes over time. It is the time derivative of the velocity vector.
In this problem, acceleration \left( \vec{a}(t) \right) is obtained by differentiating the velocity vector with respect to time.
For our exercise:

• Start with the velocity vector \left( \vec{v}(t) = 8t \hat{\mathrm{j}} \right).
• Take its derivative with respect to time: \left( \vec{a}(t) = \frac{d \vec{v}(t)}{dt} \right).

Working it out gives:

• The derivative of \left( 8t \hat{\mathrm{j}} \right) with respect to time is \left( 8 \hat{\mathrm{j}} \right).

Thus, the acceleration vector \( \vec{a}(t) = 8 \hat{\mathrm{j}} \).
This tells us that the particle has a constant acceleration in the y-direction. The constant value (8) means that the rate of change of velocity remains the same throughout the motion.