Question

A meteor this past year was tracked while it re-entered the earth's atmosphere going 18,000 mph. The enormous air friction on it caused it to lose 90% of its velocity in about 10 seconds. How far did it travel in kilometers during this time?

Answer 1

44257m

Step-by-step explanation:

first we convert the initial velocity to m / s

Vo=initial velocity=18000Mph*(0.44m/s)/1mph=8046.7m/s

As the meteorite lost 90% of its speed, it means that it is moving with 10% of the initial speed

Vf=final velocity=(0.1)(8046.7)=804.67m/s

As the meteorite has a uniformly accelerated movement we can use the following equation

A=aceleration=(Vf-Vi)/t

A=(804.67-8046.7)/10=-724.2m/S^2

finally to calculate the displacement we use the following equation

(Vf^2-Vo^2)/2A=X=displacement

X=(804.67^2-8046.7^2)/(2*-724.2)=44257m