Acceleration vs velocity

Question

I am hoping someone can provide an equation that would enable me to change acceleration value (G force) to yield different results in miles per hour. But the core question I need answered is this: suppose a violent earthquake occurred in which a 160 pound man had his back against a wall. The quake produced an acceleration of 12.9 G's in 0.24 seconds. Assume that the acceleration direction was from behind the man and wall, pushing the man away from the wall. At the end of the 0.24 seconds how many miles per hour would the man be moving towards the other end of the room? Second part to the same question: suppose that the main direction of earthquake acceleration was vertical, and that the man was already seated on the ground. At what speed (mph) would the man be propelled upward? And how high would he be propelled before reaching the apogee or highest point, in feet? THANKS!

Answer 1

"12.9 G's in 0.24 seconds". I am going to assume that you mean to say "12.9 G's for 0.24 seconds" here.

If that is the case, then you can treat this as mostly a kinematics question.

For the first part of the question, the man has an initial velocity of 0 m/s and an acceleration of 12.9g's = 12.9 * 9.8 m/s/s = 126.42 m/s/s over a duration of 0.24 seconds. You are looking for the final velocity after 0.24 seconds. You can use the kinematics equation

v_f = v_i+at = 0 + (126.42)(0.24) = 30.34 m/s.

To convert from m/s to miles/hour, remember that there are about 1600 m in a mile (four laps around a track at a typical American high school), there are 60 minutes in an hour, and 60 seconds in a minute. Your final answer will be about 65 miles/hour, but I won't do that part for you.

Second part of the question: The man is initially propelled upwards at 30.34 m/s, but once he leaves the ground, he accelerates from gravity at -9.8 m/s/s. (Because the ground is no longer touching him, it can no longer push him upwards, leaving just gravity to affect his motion.) Given a vertical motion as stated in the problem, the velocity at the top of the trajectory is 0 m/s. You can figure out the height by using the equation

v_f² = v_i²+2ah → 0 = 30.34²+2(-9.8)h

You can solve for h in meters from this. To convert it to feet, use the fact that there are 100 cm in 1.0 meters and 30.5 cm in 1.0 feet. Your final answer will be near 150 feet!