???Help Please???

Question

A small space probe of mass 220 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will eventually land. At a time 22.9 seconds after it is launched, the probe is at location < 4100, 7500, 0 > m, and at this same instant its momentum is < 41000, -8100; 0 > kg·m/s. At this instant, the net force on the probe due to the gravitational pull of Mars plus the air resistance acting on the probe is < -4850, -730, 0 > N. Assuming that the force is nearly constant during the time interval from 22.9 seconds after the probe is launched to 23.3 seconds after the launch, so that the velocity is changing at a constant rate, what is the change of the position of the probe during this time interval?
I get = < 74.5 ,-14.73 , 0 > m.
I don't understand why the 74.5 is wrong.

Answer 1

position = initial position + v0 t + 1/2at^2

initial position is given
initial velocity is the given momentum divided by the mass--v0 = p0/m
acceleration is the given force divided by the mass--a=F/m
time is the difference betwen those two times

change in position = s - s0
= p0/m(t2-t1) + 1/2 (F/m)(t2-t1)^2

Do the problem in 2 pieces. You do the exact same thing for the x part and the y part. The z-part, of course, is trivial--you got that right already.

Answer 2

Ok. you have the starting position and starting velocity (the momentum divided by the mass) and also the acceleration (force divided by mass). Then the position as a funciton of time is:

X = 1/2 A*t^2 +V*t + X0 where the capital letters are vector quantities of acceleration (A), Velocity (V), and initial position (X0)

V = <1863.6, -36.82, 0>
A = <-22.05, -3.32, 0>

Now t = 23.3-22.9 = 0.4 sec

Then X = <4843.5, 7485, 0>

The initial speed in x is so large it will take a while before you see the x direction decrease in value.