In January 2004, NASA landed exploration vehicles on Mars. Part of the descent c

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

Explanation / Answer

we have an initial and a final velocity and a time, and are asked for an acceleration. Know any formulas with those 4 things in them? ;-)

v = u + at

v-u = at

a = (v-u)t

So for the first question, t = 4 min, but we need it in seconds, = 240 s. The velocities are in km/h and we need them in m/s, so v = 444.4 m/s and u = 5361 m/s

a = (444-5361)/240 = -20.5 m/s/s (or ms^-2)

I'll leave you to do the second question the same way. (t is in seconds but remember to convert velocities to m/s)

The third question is slightly different - we have an initial and a final velocity, but not a time. Instead we have a distance and are asked for an acceleration. The appropriate formula is:

v^2 = u^2 + 2ad (some people use s instead of d for distance)

Rearranging to have a as the subject:

v^2-u^2 = 2ad

a = (v^2-u^2)/2d

The final velocity is zero, and the initial velocity is 89 m/s so

a = (0^2-89^2)/2*75 = -52.8 m/s/s

For the fourth and fifth questions we now know times, initial and final velocities and accelerations, and are looking for distances. One more formula from the toolbox:

d = ut + 1/2 at^2

For the fourth one:

d = 5361*240 + 1/2*(-20.5)*240^2 = 696 240 m = 696.24 km

The 5th one is the same so I'll leave it for you

For the 6th one we don't know the time, but we're actually already given the distance in the question. 75 m = 0.075 km.

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