No, it only does when entering an atmosphere
The car’s velocity at the end of this distance is <em>18.17 m/s.</em>
Given the following data:
- Initial velocity, U = 22 m/s
- Deceleration, d = 1.4

To find the car’s velocity at the end of this distance, we would use the third equation of motion;
Mathematically, the third equation of motion is calculated by using the formula;

Substituting the values into the formula, we have;

<em>Final velocity, V = 18.17 m/s</em>
Therefore, the car’s velocity at the end of this distance is <em>18.17 m/s.</em>
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Read more: brainly.com/question/8898885
Explanation:
for v vs t graph for t=9 to 11,v=15kmph
similarly v2=0,v3=60kmph&v4= -40kmph
Image from a far away object formed by a concave mirror
I have no idea but this is my best guess as a sophomore in college
What your saying doesn't make sense.