Question

NASA's Langley Research Center has been experimenting with the use of air bags to soften the landings of crew exploration vehicles (CEV) on land. What stopping time will be required in order to safely stop a 7250 kg CEV moving at 7.65 m/s with an average force of 426000 N (an average force of 6 Gs)?

Answer 1

Answer:

0.130 s

Explanation:

a=F/M.

a=58.76 m/s².

t=v/a=7.65/58.76=0.130 s

Answer 2

Answer:

stopping time is 0.13s

Explanation:

Given the mass of the vehicle,  $m=7250 kg$

velocity,  $u= 7.65 m/s$

and force $F= 426000 N$

Acceleration is given by $a=\frac{F}{m}$

                                             $=\frac{426000}{7250}=58.76m/s^{2}$

From the equation of motion, $v=u+at$

As $v=0$ therefore $u=-at$

negative sign shows deceleration  

and stopping time, $t=\frac{u}{a}$

                                  $=\frac{7.65}{58.76}=0.13s$