Question

A family on vacation in San Francisco drives from Golden Gate Park due south on 19th Avenue for 2.2 miles and then turns west on Sloat Boulevard and drives an additional 1.1 mile to go to the zoo. The driving time for this trip is 18 minutes. What is the family’s net displacement for this trip? What is the average speed for the trip? What is the average velocity? (Remember to specify the direction of the net displacement and the velocity vectors. This requires giving a reference, like “of the north”, and a direction the angle is measured, like “east” in the direction “20.0° east of north”.)

Answer 1

a.)

Given, d1 = 2.2 miles south = 0 i - 2.2 j

d2 = 1.1 mile west = -1.1 i + 0 j

then, net displacement vector will be,

d = d1 + d2 = (0 i - 2.2 j) + (-1.1 i + 0 j)

d = -1.1 i - 2.2 j

net displacement = |d| = sqrt(1.1^2 + 2.2^2)

|d| = 2.5 mile

For, the direction of the net displacement vectors:

direction = arctan(d_y/d_x) = arctan(2.2/1.1) = 63.43 deg South of west

So, net displacement = 2.5 mile at 63.43 deg south of west

b.)

average speed will be:

v = (total distance)/(total time)

here, total distance = |d1| + |d2| = 2.2 + 1.1 = 3.3 mile

total time = 18 minutes = 18/60 = 0.3 hr

So,

v = 3.3/0.3

v = 11 mph

c.)

average velocity will be:

v = (total displacement)/(total time)

here, total displacement = |d| = 2.5 mile

total time = 18 minutes = 18/60 = 0.3 hr

So,

v = 2.5/0.3

v = 8.33 mph

For, the direction of the average velocity vectors = direction of the net displacement vectors

So, average velocity = 8.33 mph at 63.43 deg south of west

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