Your car is performing a transformation of energy of:
Chemical energy to Mechanical energy
The chemical is the gasoline which is then converted to fire as the car runs thus creating the movement of the car which is mechanical energy.
1) 3 miles/Hour
The speed is defined as the distance covered divided by the time taken:
![v=\frac{d}{t}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bd%7D%7Bt%7D)
where
d = 1.5 mi is the distance
t = 0.5 h is the time taken
Substituting,
![v=\frac{1.5}{0.5}=3 mi/h](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B1.5%7D%7B0.5%7D%3D3%20mi%2Fh)
2) 1.34 m/s south
Velocity, instead, is a vector, so it has both a magnitude and a direction. We have:
is the displacement in meters
is the time taken in seconds
Substituting,
![v=\frac{2414 m}{1800 s}=1.34 m/s](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B2414%20m%7D%7B1800%20s%7D%3D1.34%20m%2Fs)
And the direction of the velocity is the same as the displacement, so it is south.
Answer:
<u>Distance</u><u> </u><u>between</u><u> </u><u>them</u><u> </u><u>is</u><u> </u><u>4</u><u>,</u><u>2</u><u>0</u><u>0</u><u> </u><u>meters</u><u>.</u>
Explanation:
Consinder car A:
![{ \bf{distance = speed \times time }}](https://tex.z-dn.net/?f=%7B%20%5Cbf%7Bdistance%20%3D%20%20speed%20%5Ctimes%20time%20%7D%7D)
substitute:
![distance = 20 \times (2 \times 60) \\ = 2400 \: m](https://tex.z-dn.net/?f=distance%20%3D%2020%20%5Ctimes%20%282%20%5Ctimes%2060%29%20%5C%5C%20%20%3D%202400%20%5C%3A%20m)
Consider car B:
![distance = 15 \times (2 \times 60) \\ = 1800 \: m](https://tex.z-dn.net/?f=distance%20%3D%2015%20%5Ctimes%20%282%20%5Ctimes%2060%29%20%5C%5C%20%20%3D%201800%20%5C%3A%20m)
since these cars move in opposite directions, distance between them is their summation:
![distance \: between = { \sum(distance \: of \: each \: car)} \\ = 2400 + 1800 \\ = 4200 \: m](https://tex.z-dn.net/?f=distance%20%5C%3A%20between%20%3D%20%7B%20%5Csum%28distance%20%5C%3A%20of%20%5C%3A%20each%20%5C%3A%20car%29%7D%20%5C%5C%20%20%3D%202400%20%2B%201800%20%5C%5C%20%20%3D%204200%20%5C%3A%20m)