Question

A motorist travels due North at 90 km/h for 2 hours. She changes direction and travels West at 60 km/for 1 hour.

Answer 1

Answer:

a) S =  63.2 km/h

b) V =  63.2 km/h*(-0.316 , 0.949)

Explanation:

Let's define:

North as the positive y-axis

East as the positive x-axis.

Also, remember the relation:

Distance = Time*Speed

Let's assume that she starts at the position (0km, 0km)

Then she travels due North at 90km/h for two hours, then the displacement is

90km/h*2h = 180km to the north

Then the new position is:

(0km, 180km)

Then she travels West at 60km/h for one hour.

Then the distance traveled to the West (negative x-axis) is:

60km/h*1h = 60km to the west

Then the new position is:

(-60km, 180km).

a) The average speed is defined as the quotient between the displacement and the time.

We know that the total time traveled is 3 hours.

And the displacement is the difference between the final position and the initial position.

this is:

D = √( -60km - 0km)^2 + (180km - 0km)^2)=

D = √( (60km)^2 + (180km)^2) = 189.7 km

Then the average speed is:

S = (189.7 km)/(3 h) = 63.2 km/h

b) Now we want to find the average velocity, this will be equal to the average speed times a versor that points from the origin to the direction of the final position.

So, if the final position is (-60km, 180km)

We need to find a vector that represents the same angle, but that is on the unit circle.

Then, if the module of the final position is 189.7 km (as we found above), then the versor is just given by:

(-60km/ 189.7 km, 180km/ 189.7 km)

(-60/189.7 , 180/189.7)

We can just check that the module of the above versor is 1.

$module = \sqrt{(\frac{-60}{189.7} )^2 + (\frac{180}{189.7} )^2} = \frac{1}{189.7}* \sqrt{(-60 )^2 + (180 )^2} = 1$

Then the average velocity is:

V = 63.2 km/h*(-60/189.7 , 180/189.7)

We can simplify our versor so the velocity equation is easier to read:

V = 63.2 km/h*(-0.316 , 0.949)