Experiencing Jerk

Credits goes to Razvan Barbu. I just did the LaTeX job

We already know the law of linear movement at constant acceleration rate to be ( displacement ) $x = x_0+ v_0t +\frac{1}{2}at^2$ If the acceleration is not constant, but increases at a constant pace, we simply introduce a new concept of Accelerated acceleration , the Jerk, j . The formula becomes: $X = x_o + v_o t +\frac{1}{2}at^2 +\frac{1}{2}\times\frac{1}{3}jt^3$ Where, Time, $t = 6 sec$ Initial Position, $x_0= 0$ Initial Velocity, $v_0=0$ Initial Acceleration, $a_0 = 0$ Accelerated Acceleration, Jerk, $j= 2ms^{3}$ Solution: $X = 0 + 0 + 0 +\frac{1}{6}2\times 6^3= 2*36 =72m$

We already know the law of linear movement at constant acceleration rate to be ( displacement ) X = Xo + Vo t + ( A t^2 ) / 2 If the acceleration is not constant, but increases at a constant pace, we simply introduce a new concept of "accelerated acceleration" - Aa. The formula becomes: X = Xo + Vo t + ( Ao t^2 ) / 2 +( Aa t^3 ) / 6
where t - time = 6 seconds, Xo - initial position = 0, Vo - initial velocity = 0, Ao - initial acceleration = 0, Aa - "accelerated acceleration" = 2. Solution: X = 0 + 0 + 0 + ( 2
6^3 ) / 6 = 2*36 = 72m

Guys, there is a mistake in this question.its actually unsolvable without an initial condition for acceleration."initially at rest " simply implies v(t=0)=0 not a(t=0)=0.Therefor upon first integration we gets a=2t+C where C is a constant and finally S=(t^3)/3+C(t^2)/2.I solved it by assuming C=0.actually one more initial condition was needed.

da/dt =2. hence a = 2t (BY INTEGRATION). NOW a = dv/dt, INTEGRATING AGAI WE GET v = (t)(t). NOW V = ds/dt, INTEGRATING AGAIN WITH LIMITS WE GET S=72M

first integration of 2 m/s^3 gives you acceleration, 2t. Second integration gives you speed, 2t^2/2 or t^2. Third integration gives you distance, t^3/3. In 6 seconds, the car would have moved a distance of 6^3/3= 72m