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It is 2058 and you are taking your grandchildren to Mars. At an elevation of 34.7 km above the surface of Mars, your spacecraft is dropping vertically at a speed of 293 m/s. The spacecraft is to make a soft landing -- that is, at the instant it reaches the surface of Mars, its velocity is zero. Assume the spacecraft undergoes constant acceleration from the elevation of 34.7 km until it reaches the surface of Mars. What is the magnitude of the acceleration

Sagot :

Answer: [tex]1.23\ m/s^2[/tex]

Explanation:

Given

At an elevation of [tex]y=34.7\ km[/tex], spacecraft is dropping vertically at a speed of [tex]u=293\ m/s[/tex]

Final velocity of the spacecraft is [tex]v=0[/tex]

using equation of motion i.e. [tex]v^2-u^2=2as[/tex]

Insert the values

[tex]\Rightarrow 0-(293)^2=2\times a\times (34.7\times 10^3)\\\\\Rightarrow a=-\dfrac{293^2}{2\times 34.7\times 10^3}\\\\\Rightarrow a=-1.23\ m/s^2[/tex]

Therefore, magnitude of acceleration is [tex]1.23\ m/s^2[/tex].

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