Motion problem-solving question

During the final braking stage of a lunar landing, a lander is $48.0\,\mathrm{m}$ above the Moon's surface and moving vertically downwards at $12.0\,\mathrm{m\,s^{-1}}$. Its engines produce a constant resultant acceleration of $1.50\,\mathrm{m\,s^{-2}}$ upwards. Take upwards as the positive direction. (a) Calculate the time taken for the lander to come to rest. [2] (b) Calculate the lander's displacement during this time. [3] (c) State, with a reason, whether this motion produces a soft landing at the lunar surface. [1]

Take upwards as positive. The initial velocity and acceleration are therefore $$u=-12.0\,\mathrm{m\,s^{-1}}$$ and $$a=+1.50\,\mathrm{m\,s^{-2}}.$$ (a) When the lander comes to rest, $v=0$. Using $v=u+at$: $$0=-12.0+(1.50)t.$$ Therefore $$t=8.00\,\mathrm{s}.$$ (b) Using $s=ut+\frac{1}{2}at^2$: $$s=(-12.0)(8.00)+\frac{1}{2}(1.50)(8.00)^2$$ $$s=-96.0+48.0=-48.0\,\mathrm{m}.$$ The lander's displacement is $48.0\,\mathrm{m}$ downwards. (c) The lander begins $48.0\,\mathrm{m}$ above the surface and comes to rest after a displacement of $48.0\,\mathrm{m}$ downwards. It therefore reaches the lunar surface with zero velocity, so this idealised motion produces a soft landing.

(a) (M1) Uses v = u + at with a consistent sign convention, including u = -12.0 m s^-1, v = 0 and a = +1.50 m s^-2. (a) (A1) Obtains t = 8.00 s. [2] (b) (M1) Uses a valid constant-acceleration equation for displacement. (b) (A1) Substitutes values with a consistent sign convention. (b) (A1) Obtains s = -48.0 m, or states 48.0 m downwards, with a unit. [3] (c) (C1) States that the lander reaches the surface with zero velocity because it starts 48.0 m above the surface and moves 48.0 m downwards, so the idealised landing is soft. [1]

Take upwards as positive, so the initial downward velocity is negative. At rest, the final velocity is zero. Compare the displacement with the starting height.