Question

part 2
your answer is partially correct.
(b) if the ball continued to travel at the same average velocity during the next second, where would it be at the end of that second? that is, where would it be at time t = 2 s? (express your answer in vector form.)
(vec{r}_f=lt38.6,36.1,0gt m
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Explanation:

Step1: Assume the average - velocity vector $\vec{v}$

Let the average - velocity vector be $\vec{v}=v_x\hat{i}+v_y\hat{j}+v_z\hat{k}$. If the ball continues to travel with the same average velocity for the next second, and its position at time $t = 2s$ is to be found, and we know that the displacement $\Delta\vec{r}=\vec{v}\Delta t$. Since $\Delta t = 1s$, the new position $\vec{r}_{new}=\vec{r}_{old}+\vec{v}\Delta t$. But we are not given the initial position vector $\vec{r}_{old}$ and the velocity vector $\vec{v}$ components clearly in the problem statement. However, if we assume the position vector at a certain time is $\vec{r}=\langle38.6,36.1,0
angle$ and the average - velocity remains constant, for a time interval $\Delta t = 1s$, the new position vector $\vec{r}_{new}$ at $t = 2s$ (assuming the given position is at $t = 1s$) is just the sum of the position vector at $t = 1s$ and the displacement vector (which is $\vec{v}\Delta t$, and since $\Delta t = 1s$, it's the same as the average - velocity vector). But since no other information about acceleration or change in velocity is given, we assume a constant - velocity motion.

Step2: Calculate the new position

If the position vector at $t = 1s$ is $\vec{r}=\langle38.6,36.1,0
angle$ and the ball moves with a constant average velocity for the next second, the position vector at $t = 2s$ is $\vec{r}_{2}=\langle38.6 + v_x,36.1 + v_y,0 + v_z
angle$. Since no information about velocity change is given, if we assume the average velocity is such that there is no acceleration, and the position at $t = 1s$ is $\vec{r}=\langle38.6,36.1,0
angle$, and the ball moves with a constant average velocity for one more second, the position vector at $t = 2s$ is $\langle38.6 + 0,36.1+ 0,0 + 0
angle=\langle38.6,36.1,0
angle$ (assuming zero acceleration and thus constant velocity).

Answer:

$\langle38.6,36.1,0
angle$