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. $$ \mathbf{r}(t)=(6-t) \mathbf{i}+t \mathbf{j},(3,3) $$

Short answer

The position vector \( r(t) = (6-t)i + tj \) represents a straight line path on the xy-plane from (6,0) to (0,6). The velocity vector at any point is \( r'(t) = -i + j \), a constant vector pointing left and up. The acceleration of the object is \( r''(t) = 0 \), meaning the velocity does not change at any point.

Step-by-step solution

Understand the Position Vector

The given position vector \( r(t) = (6-t)i + tj \) represents the position of the object at any time \( t \). By observing the equations, it can be seen that as time increases, the x-position decreases from 6 while the y-position increases from 0.

Determine the Velocity Vector

The velocity of the particle is the first derivative of the position function. The derivative of \( r(t) = (6-t)i + tj \) is \( r'(t) = -i + j \). This represents the velocity vector at any point in time, showing that the x-component of velocity is -1 and the y-component is 1.

Determine the Acceleration Vector

The acceleration vector can be found by taking the derivative of the velocity vector. So the derivative of \( r'(t) = -i + j \) is \( r''(t) = 0 \). This indicates that the acceleration is constant and zero for all points in time, implying that the object is moving in a straight line.

Calculate the Velocity and Acceleration Vectors at a Certain Point

At the point (3,3) which corresponds to time \( t = 3 \), put \( t \) into \( r'(t) \) and \( r''(t) \) to find the velocity and acceleration vectors. The velocity at \( t = 3 \) is still \( r'(3) = -i + j \) and the acceleration is still \( r''(3) = 0 \).

Plot the Path and Vectors

Now, plot the path of the object's movement (the line (6-t, t)) on the xy-plane from the origin (0,0) to (6,6). At the point (3,3), draw the velocity vector \( v = -i + j \) as an arrow pointing left one unit and up one unit from (3,3). Whereas, the acceleration vector being zero, does not present any arrow on the graph.