IB Mathematics Applications & Interpretation Question Type 6: Determining Speed and Velocity for Trajectories with Varying Velocities Exercises

A particle follows a trajectory defined by the position vector $\mathbf{r}(t)=\left(\begin{array}{c}3\\ 4\end{array}\right)+t\left(\begin{array}{c}9-t^2\\ 11+t\end{array}\right)$, 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 $\mathbf{r}(t)=\left(\begin{array}{c}1\\ 0\end{array}\right)+t\left(\begin{array}{c}8+t\\ 2-t^2\end{array}\right)$.

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, $x^{\prime }(t)=0$.)

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

For $\mathbf{r}(t)=(1,0)+t(8+t,2-t^2)$, 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 $\mathbf{r}(t)=(1,0)+t(8+t,2-t^2)$.

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

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

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

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

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

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

For the position vector $\mathbf{r}(t)=\left(\begin{array}{c}1\\ 0\end{array}\right)+t\left(\begin{array}{c}8+t\\ 2-t^2\end{array}\right)$, find all times $t$ when the motion is purely horizontal. (Motion is purely horizontal when $y^{\prime }(t)=0$.)

For $\mathbf{r}(t)=(3,4)+t(9-t^2,11+t)$, find all times $t$ when the motion is purely vertical. (Solve $x^{\prime }(t)=0$.)