A student drove to the university from her home and noted that the odometer reading of her car increased by 7 miles. The trip took 20 min. (Note: 1 mile = 1.6 km). If the straight-line distance from her home to the university is 10 km, and the university is directly south of her home, what is the vector of her average velocity?

Respuesta :

Answer:

Vector of the average velocity = 3.33 km\hr towards south.

Explanation:

Given:

Distance covered by the student = [tex]7[/tex] miles = [tex]7\times 1.6 =11.2[/tex] km

Total displacement = [tex]10[/tex] km in [tex]South[/tex]

Total time taken = [tex]20[/tex] mins = [tex]\frac{20}{60} = 0.33[/tex] hr

Note:

Odometer reading is the distance covered while straight-line distance is the displacement which is the shortest route.

Distance is scaler where as displacement is a vector quantity.

Scaler has only magnitude while vector has magnitude and direction.

We have to find the average velocity.

Average velocity = Ratio of total displacement and total time.

⇒ Average velocity ([tex]v[/tex]) = [tex]\frac{displacement \ (x)}{time\ (t)}[/tex]

⇒ Plugging the values.

⇒ ([tex]v[/tex]) = [tex]\frac{10}{0.33}[/tex]

⇒ ([tex]v[/tex]) = [tex]3.33[/tex] km\hr

So the vector of the average velocity = 3.33 km\hr towards south.