Physics

Question

In January 2004, NASA landed exploration vehicles on Mars. Part of the descent consisted of the following stages:
Stage A: Friction with the atmosphere reduced the speed from 19300 km/h to 1600 km/h in 4.0 min.
Stage B: A parachute then opened to slow it down to 321 km/h in 94 s.
Stage C: Retro rockets then fired to reduce its speed to zero over a distance of 75 m.
Assume that each stage followed immediately after the preceding one and that the acceleration during each stage was constant.

1- Find the rocket's acceleration (in m/s2) during stage A

2- Find the rocket's acceleration (in m/s2) during stage B

3-Find the rocket's acceleration (in m/s2) during stage C

1 Answer

  • Acceleration is given by:

    [tex]a=\frac{v-u}{t}[/tex]

    where

    v is the final velocity

    u is the initial velocity

    t is the time interval

    Let's apply the formula to the different parts of the problem:

    A) [tex]-20.5 m/s^2[/tex]

    Let's convert the quantities into SI units first:

    [tex]u = 19300 km/h \cdot \frac{1000 m/km}{3600 s/h} = 5361.1 m/s[/tex]

    [tex]v=1600 km/h  \cdot \frac{1000 m/km}{3600 s/h}  =444.4 m/s[/tex]

    t = 4.0 min = 240 s

    So the acceleration is

    [tex]a=\frac{444.4 m/s-5361.1 m/s}{240 s}=-20.5 m/s^2[/tex]

    B) [tex]-3.8 m/s^2[/tex]

    As before, let's convert the quantities into SI units first:

    [tex]u = 444.4 m/s[/tex]

    [tex]v=321 km/h  \cdot \frac{1000 m/km}{3600 s/h}  =89.2 m/s[/tex]

    t = 94 s

    So the acceleration is

    [tex]a=\frac{89.2 m/s - 444.4 m/s}{94 s}=-3.8 m/s^2[/tex]

    C) [tex]-53.0 m/s^2[/tex]

    For this part we have to use a different formula:

    [tex]v^2 - u^2 = 2ad[/tex]

    where we have

    v = 0 is the final velocity

    u = 89.2 m/s is the initial velocity

    a is the acceleration

    d = 75 m is the distance covered

    Solving for a, we find

    [tex]a=\frac{v^2-u^2}{2d}=\frac{0^2-(89.2 m/s)^2}{2(75 m)}=-53.0 m/s^2[/tex]

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