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

1. User Answers skyluke89
Acceleration is given by:
[tex]a=\frac{vu}{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/s5361.1 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^2u^2}{2d}=\frac{0^2(89.2 m/s)^2}{2(75 m)}=53.0 m/s^2[/tex]