91Ó°ÊÓ

When we talk about instantaneous velocity, we're diving into the exact speed and direction of an object at a specific moment in time. To visualize this, think of a video showing a dot moving across a screen. If you pause the video at a particular second, you're looking at the dot's instantaneous velocity at that point. How do we determine this? We take the derivative of the position vector. In the exercise, the position was expressed as \[ \vec{r}(t) = [4.0 \, \text{cm} + (2.5 \, \text{cm/s}^2) \, t^2] \hat{\imath} + (5.0 \, \text{cm/s}) \, t \hat{\jmath} \]The derivative gives us the rate of change of the position, which tells us the instantaneous velocity:\[ \vec{v}(t) = \frac{d\vec{r}}{dt} = (5.0t) \hat{\imath} + 5.0 \hat{\jmath} \]At specific times, we calculate it by substituting the time values. This is why you get different velocities at various times like \(t = 0\), \(t = 1\), and \(t = 2\) seconds.
Remember, instantaneous velocity includes both magnitude (how fast) and direction (where to), making it a vector quantity.

The trajectory is the path traced by a moving object, such as a dot on a screen. Imagine throwing a ball and watching its arc; that's the trajectory of the ball. In this exercise, the dot's movement from one point to another over time creates its trajectory. Each time point, like \(t = 0\), \(t = 1\), and \(t = 2\), corresponds to a specific position of the dot.

• At \(t = 0\): the dot is at \( (4.0, 0) \).
• At \(t = 1\): the dot moves to \( (6.5, 5.0) \).
• At \(t = 2\): it reaches \( (14.0, 10.0) \).

By plotting these points, you can see the trajectory as a connected line or curve. It's the visual path the dot takes over the time interval. The shape of this path depends on the specifics of the position vector equation. For example, a quadratic term \((2.5t^2)\) in the position vector suggests a parabolic trajectory when you plot it on a graph over time.

The position vector is essentially the coordinate system that describes where an object is located in a space relative to an origin point, using vectors. In the animation problem we encountered, the position vector \(\vec{r}(t)\) tells us where the dot is at any time \(t\) in a 2-dimensional plane, being represented in the form:\[\vec{r} = [4.0 \, \text{cm} + (2.5 \, \text{cm/s}^2)t^2] \hat{\imath} + (5.0 \, \text{cm/s}) t \hat{\jmath}\]This vector is made of two components:

• \(\hat{\imath}\): representing movement along the x-axis.
• \(\hat{\jmath}\): representing movement along the y-axis.

The scalar coefficients in front of these unit vectors indicate the position in centimeters at any given time. By multiplying these coefficients by time, we see exactly how the dot's position changes. The position vector equation, as seen above, allows us to calculate exact coordinates for specific times, thereby painting a clear picture of the dot's journey across the screen. Knowing the position vector aids in calculating velocities, both average and instantaneous, as these depend on changes in position over time.