Question

You and a friend are driving to the beach during spring break. You travel 14.5 km east and then 66.5 km south in a total time of 50 minutes. (a) What is the average speed of the trip? (b) What is the magnitude of the average velocity? -In this question for part B, why isnt it necessay to find the unit vectors for displacement?

Answer 1

Answer:

(a). The average speed is 97.23 km/hr.

(b). The average velocity is 81.70 km/min.

Explanation:

Given that,

Distance in east = 14.5 km

Distance in south = 66.5 km

Time = 50 min = 0.833 hr

(a). We need to calculate the average speed

Using formula of average speed

$v_{avg}=\dfrac{D}{T}$

Where, D = total distance

T = total time

Put the value into the formula

$v_{avg}=\dfrac{14.5+66.5}{0.833}$

$v_{avg}=97.23\ Km/hr$

(b). We need to calculate the displacement

Using Pythagorean theorem

$d=\sqrt{(d_{e})^2+(d_{s})^2}$

Put the value into the formula

$d=\sqrt{(14.5)^2+(66.5)^2}$

$d=68.06\ km$

(b). We need to calculate the average velocity

Using formula of average velocity

$v_{avg}=\dfrac{d}{t}$

Where, d = displacement

t = time

Put the value into the formula

$v_{avg}=\dfrac{68.06}{0.833}$

$v_{avg}=81.70\ km/min$

Here, We can not necessary to find the unit vectors for displacement because we need to the displacement for find the average velocity.

Hence, (a). The average speed is 97.23 km/hr.

(b). The average velocity is 81.70 km/hr.