# help with physics?

A cyclist travels with the velocity of 6.0m/s[W] for 45 minutes. she then heads south with a speed of 4.0m/s for 30.0 minutes.

a) calculate the displacement of the cyclist from her starting point

b) determine the average velocity for the trip

### 3 Answers

- electron1Lv 71 year agoFavorite Answer
Since the unit of the velocity is m/s, the time must be in seconds.

t1 = 45 * 60 = 2,700 seconds

t2 = 30 * 60 = 1,800 seconds

d1 = 6 * 2,700 = 16,200 meters west

d2 = 4 * 1,800 = 7,200 meters south

Since west is perpendicular to south, the following equation is used to determine the displacement.

d = √(d1^2 + d2^2)

d = √(16,200^2 + 7,200^2) = √3.1428 * 10^8

This is approximately 17,700 meters.

The following equation is used to determine the average velocity.

v = d ÷ t

t = 2,700 + 1,800 = 4,500 seconds

v = √3.1428 * 10^8 ÷ 4,500

This is approximately 3.94 m/s. To determine the angle south of west, use the following equation.

Tan θ = South ÷ West = 7,200 ÷ 16,200

This is approximately 24˚.

OR

Tan θ = West ÷ South = 16,200 ÷ 7,200 = 2.25

This is approximately 66˚. These two angles are complementary. It is good when you receive to answers that are the same. I hope this is helpful for you.

- Andrew SmithLv 71 year ago
the time for going W is 45 * 60 s and the displacement in this direction is then 6 * 45 * 60 m

( 16200 m)

The time going s is 30 * 60 s so the displacement in this direction is 4 * 30 * 60 m

(7200 m)

The MAGNITUDE of the net displacement is given by Pythagoras = sqrt( 16200^2 + 7200^2)

~= 17.7 km

and the magnitude of the net velocity ( average velocity) is displacement / time

= 17.7 * 10 ^ 3 / ( ( 45+30)*60) ~= 3.94 m/s

The DIRECTION is found by tan ( theta) = opposite / adjacent

theta = atan ( 7200/16200) south of West