Question Type 6: Determining speed and velocity for trajectories with varying velocities Exercises

A particle follows a trajectory defined by the position vector $\mathbf{r}(t) = \begin{pmatrix} 3 \\ 4 \end{pmatrix} + t \begin{pmatrix} 9 - t^2 \\ 11 + t \end{pmatrix}$, where $t$ is the time in seconds.

Find the position vector of the particle at $t = -2$.

For the trajectory $\mathbf{r}(t) = (1, 0) + t (8 + t, 2 - t^2)$, find the velocity vector $\mathbf{v}(t)$ and the speed at $t = 2$.

A particle moves such that its position vector $\mathbf{r}$ at time $t$ is given by $\mathbf{r}(t) = \begin{pmatrix} 1 \\ 0 \end{pmatrix} + t \begin{pmatrix} 8 + t \\ 2 - t^2 \end{pmatrix}$.

Find all values of $t$ for which the motion of the particle is purely vertical.

(Note: Motion is purely vertical when the horizontal component of velocity is zero, $x'(t) = 0$.)

Find the position vector at $t = -1$ for the trajectory $\mathbf{r}(t) = \begin{pmatrix} 1 \\ 0 \end{pmatrix} + t \begin{pmatrix} 8 + t \\ 2 - t^2 \end{pmatrix}$.

For $\mathbf{r}(t) = (1, 0) + t \bigl( 8 + t, 2 - t^2 \bigr)$, determine whether there is any real time $t$ when the velocity components are equal, i.e.

$2 - 3t^2 = 8 + 2t$

Find the position vector at $t=3$ for the trajectory $\mathbf{r}(t) = (1, 0) + t \bigl( 8 + t, 2 - t^2 \bigr)$.

A particle moves such that its position at time $t$ seconds, for $t \ge 0$, is given by the vector $\mathbf{r}(t) = \begin{pmatrix} 3 \\ 4 \end{pmatrix} + t \begin{pmatrix} 9 - t^2 \\ 4 - t \end{pmatrix}$.

Find the speed of the particle at the time when its motion is purely horizontal.

For the trajectory defined by $\boldsymbol{r}(t) = (1,0) + t (8 + t, 2 - t^2)$, find the velocity vector $\boldsymbol{v}(t)$ and compute the speed at $t = 1$.

Find the position vector at $t = 2$ for the trajectory $\mathbf{r}(t) = \begin{pmatrix} 3 \\ 4 \end{pmatrix} + t \begin{pmatrix} 9 - t^2 \\ 11 + t \end{pmatrix}$.

A particle moves such that its position vector at time $t$ seconds is given by $\mathbf{r}(t) = \begin{pmatrix} 3 \\ 4 \end{pmatrix} + t \begin{pmatrix} 9 - t^2 \\ 12 - t \end{pmatrix}$, for $t \geq 0$.

Find the value of $t$ for which the motion of the particle is purely horizontal.

For the position vector $\mathbf{r}(t) = \begin{pmatrix} 1 \\ 0 \end{pmatrix} + t \begin{pmatrix} 8 + t \\ 2 - t^2 \end{pmatrix}$, find all times $t$ when the motion is purely horizontal. (Motion is purely horizontal when $y'(t)=0$.)

For $\mathbf{r}(t) = (3, 4) + t(9 - t^2, 11 + t)$, find all times $t$ when the motion is purely vertical. (Solve $x'(t) = 0$.)