Every one knows the three famous equations for motions with constant acceleration . But what if the motion were having a jerk? How should then be the equations for motions? How can I find them?


It's no different than any other case with a general time dependence of acceleration. You just need to know that acceleration is the rate of velocity change: $a=\dfrac{dv}{dt}$ and velocity is the rate of position change $v= \dfrac{dx}{dt}$. Integrate twice and you get

$$v=v_0+\int_0^{t} a(\tau)d\tau$$ $$x=x_0+\int_0^t v(\tau)d\tau$$

For constant $a$, this just gives $v=v_0+at$ and $x=x_0+v_0 t + \frac12 at^2$.

Jerk is defined analogously to the previous two kinematic relations, $j=\dfrac{da}{dt}$. If it's constant, just integrate once to get $a=a_0+jt$ and then twice more to get $v=v_0+a_0t+ \frac12 jt^2$ and $x=x_0+v_0t +\frac12 a_0 t^2+\frac16 j t^3$.

  • $\begingroup$ Sir, you are right in your points. But can you tell why you used definite integral? My book used indefinite integral. $\endgroup$ – user36790 Nov 20 '14 at 7:49
  • $\begingroup$ If you write indefinite integral, you have to manually apply the initial conditions to find the integrating constant. In physics, it's really easy to write integral bounds, because they directly relate to initial conditions. In this case, the velocity goes from $v_0$ to the current velocity $v$, when time goes from $0$ to current time $t$ (it's common to set the origin of time to $0$). In physics, we sometimes don't write the limits and we still know from the context that we mean a definite integral. Also, the book probably just wanted to hint to the procedure, not apply it to an example. $\endgroup$ – orion Nov 20 '14 at 7:58
  • $\begingroup$ Then sir, you ought to answer this . I will be very grateful. $\endgroup$ – user36790 Nov 20 '14 at 8:23
  • $\begingroup$ And sir, what you have written is nothing but indefinite integral because it has variable upper limit. You can see Richard Courant's statement regarding this and also Fundamentamental theorem of Calculus 1 . Thanks. $\endgroup$ – user36790 Nov 20 '14 at 8:31
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    $\begingroup$ Yes it is. It's the same thing but knowing the lower limit tells you exactly what the integrating constant is. It's the same math, but it's easier to understand and work with if you explicitly put in the initial condition. I responded in your other question. $\endgroup$ – orion Nov 20 '14 at 9:02

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