Question

An engineer is considering possible trajectories to use for emergency descent of a lunar module from low moon orbit to the lunar surface and decides it to) to investigate one for which the vertical component of velocity as a function of time is described by vy(t) = voe 2:2 where vo = 0.49 km/s, to = 390 s, and o = 65 s. Part F = to-o= 325 s? Recall that acceleration is For this trajectory, what would the vertical component of acceleration for the module be at time tm the derivative of velocity with respect to time.

Answer 1

Answer:

4.57 m/s

Explanation:

We are given that

$v_y(t)=v_0e^{-\frac{(t-t_0)^2}{2\sigma^2}}$

$v_0=0.49 km/s=490m/s$

1km=1000 m

$t_0=390 s$

$\sigma=65$

$t=325 s$

$v'_y(t)=v_0\times (-2\frac{(t-t_0)}{2\sigma^2})e^{-\frac{(t-t_0)^2}{2\sigma^2}}$

Substitute the values

$a_y=v'_y=2\times 490\times \frac{-(325-390)}{2(65)^2}e^{-\frac{(325-390)^2}{2(65)^2}}$

$a_y=v'_y=4.57m/s$