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A spaceship travels [tex]3.00 \times 10 \, \text{km}[/tex] vertically to transport scientists from one station to another. After the transportation, the captain decides to go horizontally [tex]2.00 \times 10 \, \text{km}[/tex] to visit an Orie.

Calculate its:

a. Distance

b. Displacement​

Answer :

Answer:

a).Don't know if you meant to write 3x10^10km and 2x10^10km

I'll solve with this... If they aren't the values... You can then input the real values.

Distance = Total length traveled.

D= 3x10^10 + 2x10^10

D= 5x10^10Km

b) Displacement:

Since it went vertically... Then went horizontally in the second phase... It is forming a right angled Triangle and the Displacement would be the Hypotenuse of this triangle.

Disp.= √ (3x10^10)²+(2x10^10)²

Disp.=3.61x10^10km.
Letter B, displacement

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