Can a "Jerk" be calculated?

Question

Let me make up a question:

A prototype is being tested. The research team would like to know how big of a jerk they can achieve. The pilot takes off and accelerates uniformly from the starting position at a rate that will cause his velocity to be 300 feet per second in exactly 10 seconds. When his velocity reaches 150 feet per second, he flips the test switch and his acceleration uniformly increases at a rate which will cause his velocity to reach 600 feet per second at the original 10 second mark. What was the magnitude of the jerk?

I really hope I didn't screw that up.

I knew I was going to screw it up.

Answer 1

You do have enough to calculate jerk. Jerk is the time rate change of acceleration, and acceleration is the time rate change of velocity. One important thing to note (which the answerer before me overlooked but was otherwise on the money) is that the jerk continues uniformly during the final five-second period, so the change in velocity is not linear during this time and we can't simply use average acceleration.

The original acceleration is a_0 = (300 ft/s) / (10 s) = 30 ft/s^2. If he flips the switch while traveling at 150 ft/s (which I will now call v_0 for "initial" velocity), he has only been accelerating for 5 s. The next part of the problem is a little more complicated. His velocity now increases from 150 ft/s to 600 ft/s in just 5 seconds, while jerking uniformly. The formula for this is v = v_0 + (a_0)*t + 0.5*j*t^2. This should look familiar, because it is analagous to the formula for displacement under constant acceleration, x = x_0 + (v_0)*t + 0.5*a*t^2. Anyway, we know the initial velocity (that is, the velocity at the beginning of the time period of interest), the final velocity, the initial acceleration, and the elapsed time.

So we can write 600 ft/s = 150 ft/s + (30 ft/s^2)(5 s) + 0.5*j*(5 s)^2, and j is the only unknown.

600 ft/s = 150 ft/s + (30 ft/s^2)(5 s) + 0.5*j*(5 s)^2
450 ft/s = 150 ft/s + 0.5*j*(25 s^2)
300 ft/s = 0.5*j*(25 s^2)
(600 ft/s) / (25 s^2) = j
j = 24 ft/s^3

Note that the units of j are ft/s^3, which is acceleration per second.

Answer 2

[Edit] The guy below me's right. Oops! Good catch though.


[Original Post]

So his original acceleration is dv / dt, or (300 ft/s -0 ft/s)/(10 s -0 s) = 30 ft/s^2

He hits 150 ft/s at (150 ft/s / 30 ft/s^2) = 5 sec, so he has 5 sec remaining.

Then his new acceleration is (600 ft/s -150 ft/s)/(5 s) = 90 ft/s^2

So his jerk is da/dt = (90 ft/s^2 - 30 ft/s^2 )/(5 s) = 12 ft/s^3

Answer 3

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Answer 4

Had to see the answers, Wow that math made me think duh I am stupid lol. But looks about right .