Question

Question: A golf ball is hit off a tee at the edge of a cliff. Itsx and y coordinates as functions of time are givenby x = 18.0tand y = 4.00t -4.90t²where x and y are in meters and t is in seconds.

(a) Write a vector expression for the ball's position as a function of time, using the unit vectors $\hat{\mathbf{i}}$and $\hat{\mathbf{j}}$. By taking derivatives, obtain expressions for

(b) The velocity vector $\overrightarrow{\mathbf{v}}$as a function of time and

(c) The acceleration vector $\overrightarrow{\mathbf{a}}$as a function of time.

(d) Next use unit-vector notation to write expressions for the position, the velocity, and the acceleration of the golf ball at t = 3.00 s.

Short answer

(a)The vector expression for ball is$\left(\left(18t\right)\text{m}\right)\hat{i}+\left(\left(4t-4.9t^2\right)\text{m}\right)\hat{j}$ .

(b)The velocity vector is $(18\mathrm{m}/\mathrm{s})\hat{\mathit{i}}+((4-9.8t\mathit{}\mathit{)}\mathit{}\mathrm{m}/\mathrm{s})\hat{\mathrm{j}}$.

(c)The acceleration vector is $(-9.8\mathrm{m}/{\mathrm{s}}^2)\hat{\mathrm{j}}$.

(d) The expressions for position, velocity and acceleration are,
$$\begin{gathered} \overrightarrow{\mathrm{r}}=(54m)\hat{\mathrm{i}}-(32.1m)\hat{\mathrm{j}} \\ \overrightarrow{v}=(18m/\mathrm{s})\hat{\mathrm{i}}-(29.4\mathrm{m}/\mathrm{s})\hat{\mathrm{j}} \\ \overrightarrow{\mathrm{a}}=(-9.8\mathrm{m}/{\mathrm{s}}^2)\hat{\mathrm{j}} \end{gathered}$$

Step-by-step solution

Definition of velocity vector and Acceleration vector.

The rate of change of an object's position with respect to time is represented by a velocity vector. The angle between the velocity vector and the acceleration is$\mathbf{90}\mathbf{°}$, if the motion is circular.

Given Data

Coordinates are$x=18.0t\text{and}y=4.00t-4.90t^2$

Determine the expression for r→.

(a)

The position vector is expressed in unit-vector notation as:
$$\begin{gathered} \overrightarrow{\mathrm{r}}=x\hat{\mathrm{i}}+y\hat{\mathrm{j}} \\ =(18.0t\mathit{}\mathit{)}\mathit{}\hat{\mathit{i}}\mathit{}\mathit{+}\mathit{}(4.00t\mathit{}\mathit{-}\mathit{}\mathit{4}\mathit{.}\mathit{90}{t}^2\mathit{})\hat{\mathrm{j}} \end{gathered}$$
Thus, the expression for $\overrightarrow{r}$is$(18.0t\mathit{})\hat{\mathrm{i}}+(4.00t\mathit{}\mathit{-}\mathit{}\mathit{4}\mathit{.}\mathit{90}{\mathrm{t}}^2)\hat{\mathrm{j}}$ .

Differentiate the r→ with respect to time.

(b)

In order to find the velocity vector, we derive the components of position vectorand with respect to time. So, x and y components of velocity will be-
$$\begin{gathered} v_x=\frac{dx}{dt} \\ =\frac{d\left(18.0t\right)}{dt} \\ =18\text{m/s}··································\text{(1)} \end{gathered}$$
$$\begin{gathered} {\mathrm{v}}_{\mathrm{y}}=\frac{\mathrm{dy}}{\mathrm{dy}} \\ =\frac{\mathrm{d}(4.00t\mathit{}\mathit{-}\mathit{}4.90t^{\mathit{2}}\mathit{}\mathit{)}}{\mathrm{dt}} \\ =(4-9.8t\mathit{})\mathit{}\mathrm{m}/\mathrm{s}·································\left(2\right) \end{gathered}$$

Then, we substitute the values of v_x and v_y from Eq.(1) and Eq.(2) into the following formula:

$$\begin{gathered} \overrightarrow{\mathrm{v}}=v_x\hat{\mathrm{i}}+v_y\hat{\mathrm{j}} \\ \overrightarrow{v}=(18\mathrm{m}/\mathrm{s})\hat{\mathrm{i}}+\left(\right(4-9.8\mathrm{t})\mathrm{m}/\mathrm{s})\hat{\mathrm{j}} \end{gathered}$$

Thus, the velocity vector for the time is$(18\mathrm{m}/\mathrm{s})\hat{\mathrm{i}}+\left(\right(4-9.8t\mathit{})\mathit{}\mathrm{m}/\mathrm{s})\hat{\mathrm{j}}$.

Differentiate the v→ with respect to time.

(c)

In order to find the acceleration vector, we derive the components of velocity vectorand with respect to time. The x and y components of acceleration are given as-
$$\begin{gathered} a_x=\frac{d{v}_x}{dt} \\ =\frac{d\left(18\text{m/s}\right)}{dt} \\ =0{\text{m/s}}^2························\text{(3)} \end{gathered}$$
$$\begin{gathered} a_y=\frac{d{v}_y}{dt} \\ =\frac{d\left((4-9.8t)\text{m/s}\right)}{dt} \\ =(0-9.8){\text{m/s}}^2····································\text{(4)} \end{gathered}$$
Then, we substitute the values of a_x and a_y from Eq.(3) and Eq.(4) into the following formula:
$$\begin{gathered} \overrightarrow{\mathrm{a}}=a_x\hat{\mathrm{i}}+a_y\hat{\mathrm{j}} \\ \overrightarrow{a}=(-9.8\mathrm{m}/{\mathrm{s}}^2)\hat{\mathrm{j}} \end{gathered}$$

Thus, the acceleration vector for the time is $(-9.8\mathrm{m}/{\mathrm{s}}^2)\hat{\mathrm{j}}$.

Substitute the value t = 3s in the position, the velocity, the acceleration vector.

(d)

We substitute t = 3s in the expressions for $\overrightarrow{\mathrm{r}},\overrightarrow{v}and\overrightarrow{a}$

Thus, the expressions for position, velocity and acceleration are,
$$\begin{gathered} \overrightarrow{\mathrm{r}}=(54m)\hat{\mathrm{i}}-(32.1\mathrm{m})\hat{\mathrm{j}} \\ \overrightarrow{\mathrm{v}}=(18\mathrm{m}/\mathrm{s})\hat{\mathrm{i}}-(29.4\mathrm{m}/\mathrm{s})\hat{\mathrm{j}} \\ \hat{\mathrm{a}}=(-9.8\mathrm{m}/{\mathrm{s}}^2)\hat{\mathrm{j}} \end{gathered}$$