A spacecraft descends vertically near the surface of Planet X. An upward thrust of 25.0 kN from its engines slows it down at a rate of 1.20 \(m/s^2\), but it speeds up at a rate of 0.80 \(m/s^2\) with an upward thrust of 10.0 kN. (a) In each case, what is the direction of the acceleration of the spacecraft? (b) Draw a free-body diagram for the spacecraft. In each case, speeding up or slowing down, what is the direction of the net force on the spacecraft? (c) Apply Newton’s second law to each case, slowing down or speeding up, and use this to find the spacecraft’s weight near the surface of Planet X.

Solution 46P Step 1: Upward thrust F = 25 kN (Slows Down) 1 Upward thrust F = 10 kN (Speeds Up) 2 When slows down a = 1.2 m/s 2 1 When speeds up a = 0.8 m/s 2 2 Problem (a) To find the direction of the acceleration of the spacecraft Step 1: When the spacecraft slows down, The direction of the acceleration is upward because the spacecraft tries to land.