Question

At an elevation of 56.7 km above the surface of Mars, your spacecraft is dropping vertically at a speed of 265 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 56.7 km until it reaches the surface of Mars. What is the magnitude of the acceleration?

Answer 1

0.62 m/s²

Step-by-step explanation:

Consider the motion of spacecraft in vertical direction

`y` = vertical displacement of the spacecraft = elevation oabove the surface of mars = 56.7 km = 56700 m

`v_(oy)` = initial speed of spacecraft at the given elevation = 265 m/s

`v_(fy)` = final speed of spacecraft on the surface of mars = 0 m/s

a = acceleration experienced by the spacecraft

Using the kinematics equation based on the data available, we have

`v_(fy)^(2) = v_(oy)^(2) + 2 a y\\0^(2) = 265^(2) + 2 (56700) a\\0 = 70225 + 113400 a \\a = - 0.62 ms^(-2)`

hence the magnitude of acceleration is 0.62