Question

A motorist drives north for 35.0 minutes at 85.0 kph. He then stops for 15.0 minutes. The motorist then drives 130.0 km in 2.0 hours. Calculate the average velocity of the motorist if the second leg of the trip
A. Was north
B. Was south

Answer 1

Answer:

A. 216.36 $\frac{km}{h}  $

B. 96.56 $\frac{km}{h}  $

Explanation:

Let $s_{1}  $ be the distance in first part.

$s_{1}  $ = velocity × time

$s_{1}  $ = 85 × $\frac{35}{60}  $

$s_{1}  $ = 49.58 km

Let $s_{2}  $ be the distance in first part.

$s_{2}  $ = 130 km

Average velocity = $\frac{Total displacement}{Total time}  $

When second leg of the trip is

A. Toward north

Average velocity = $\frac{[tex] s_{1} + s_{2}  $}{Total time}  [/tex]

Average velocity = $\frac{130+49.58}{0.25+0.58}  $

Average velocity =216.36 $\frac{km}{h}  $

B. Toward south

Average velocity = $\frac{[tex] s_{1} - s_{2}  $}{Total time}  [/tex]

Average velocity = $\frac{130-49.58}{0.25+0.58}  $

Average velocity =96.56 $\frac{km}{h}  $