Student drove to the university from her home and noted that the odometer on her car increased by 14.0 km. the trip took 18.0 min. (a) what was her average speed? (b) if the straight-line distance from her home to the university is 10.3 km in a direction 25.0° south of east, what was magnitude of her average velocity?

a) Her average speed was 0.778 km/min

b) Her average velocity was 0.572 km/min

Explanation:

a) The average speed is given by the traveled distance over time:

average speed = distance / time

average speed = 14.0 km / 18.0 min = 0.778 km/min

b) The average velocity is given by the displacement over time. It is calculated as the variation of the position over time relative to the frame of reference. In this case, the origin of the frame of reference is located at the home of the student. The final position will be the university. Then:

velocity = (final position - initial position)/ time

velocity = 10.3 km / 18.0 min = 0.572 km/min

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day

your answer is c! hope this : )