Question

The position vector \(r\) describes the path of an object moving in the \(x y\) -plane. Sketch a graph of the path and sketch the velocity and acceleration vectors at the given point. $$ \mathbf{r}(t)=3 t \mathbf{i}+(t-1) \mathbf{j},(3,0) $$

Short answer

The object moves in a straight line along the line \(y=x-3\) at a speed of \(\sqrt{10}\) units per time unit. There is no acceleration vector for this object at the point (3,0) as the acceleration is zero.

Step-by-step solution

Understanding the Position Vector

A position vector describes the object's position in space at any given point of time. Since this is a position vector in a two-dimensional plane (x,y), we know the x-coordinate is described by \(3t\) and the y-coordinate by \(t-1\). It means that the point moves 3 units in the x direction and 1 unit in the y direction for every 1 unit of time.

Graphing the Position Vector

Using information about the direction in the x and y axes, we can sketch the path by plotting the x and y coordinates at different time points t. We plot these coordinates and then draw the path by combining them. At the point (3,0), it indicates \(t=1\) because when we plug \(t=1\) into the equations for x and y coordinates we get \(x=3\) and \(y=0\). We also label this point on the graph.

Calculating and drawing velocity vector

The velocity vector can be obtained by taking the derivative of the position vector. This gives us \(\mathbf{v}(t)= \mathbf{r}'(t)=3\mathbf{i}+\mathbf{j}\). At the point (3,0) (which happens at \(t=1\)), the velocity vector is thus \(\mathbf{v}(1)=3\mathbf{i}+\mathbf{j}\). This is represented as an arrow pointing from the position point. The direction of the arrow represents the direction of the motion, while the magnitude (length) of the arrow represents the speed.

Calculating and drawing acceleration vector

The acceleration vector can be calculated by taking the derivative of the velocity vector. This gives us \(\mathbf{a}(t)=\mathbf{v}'(t)=0\mathbf{i}+0\mathbf{j}\) indicating that acceleration vector is zero, hence, we do not have an acceleration vector to draw at point (3,0).