Question #103948

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.

Expert's answer

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}."

ii. Find the magnitudes:

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

The magnitude is v. Therefore:

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