Answer to Question #103948 in Mechanics | Relativity for havefun7741

Question

A car of negligible size is travelling from A to B with 5 s, from B to C with 4 s, from C to D with 3 s, then the car rolls off at D in projectile motion to cross the pit EF. The angles of AB and CD with horizontal are given as 30° and 50° respectively.(a) (i) Find the average speed from A to D.(ii) Find the displacement from A to D in unit vector form.(iii) Find the average velocity from A to D in unit vector form.(iv) Find the magnitude and angle (with horizontal) of the average velocity from A to D.

Answer 1

Draw the trajectory (as if we are in a helicopter or satellite):

i. If the car moves at a constant speed, the average speed is the total distance over total time: the distance is

$v(5+4+3)=12v,$

the time is

$t=12 \text{ s}.$

$$v_{av}=\frac{12v}{12}=v.$$

ii. Find the magnitudes:

$\vec{r}_{ABj}=AB_y=vt_{AB}\text{ sin}30^\circ=2.5v\textbf{j},\\ \vec{r}_{BCj}=BC_y=0,\\ \vec{r}_{CDj}=CD_y=vt_{CD}\text{ sin}50^\circ=2.3v\textbf{j}.\\ \vec{r}_{ADj}=AD_y=AB_y+BC_y+CD_y=4.8v\textbf{j}.\\ \space\\ \vec{r}_{ABi}=AB_x=vt_{AB}\text{ cos}30^\circ=4.3v\textbf{i},\\ \vec{r}_{BCi}=BC_x=vt_{bc}=4v\textbf{i},\\ \vec{r}_{CDi}=CD_x=vt_{CD}\text{ cos}50^\circ=1.9v\textbf{i}.\\ \vec{r}_{ADi}=AD_x=AB_x+BC_x+CD_x=10.3v\textbf{i}.\\ \space\\ |\vec{r}|=AD=\sqrt{AD_x^2+AD_y^2}=11.4v.$

Therefore:

$$\vec{r}=\vec{r}_{AB}+\vec{r}_{BC}+\vec{r}_{CD}=10.3v\textbf{i}+4.8v\textbf{j}.$$

iii, iv. Calculate the angle:

$$\alpha=\text{atan}\frac{\vec{r}_{ADj}}{\vec{r}_{ADi}}=17.6^\circ.$$

The magnitude is v. Therefore:

$$\vec{v}=v(\text{sin}\alpha\textbf{i}+\text{ cos}\alpha\textbf{j})=0.3v\textbf{i}+0.95v\textbf{j}.$$