$y=u(t-k)+\frac{1}{2}a(t-k)^2$ and $y=ut+\frac{1}{2}at^2$ are two equations depicting motions of a body.

Q.How the motions depicted by these two equations are different?

enter image description hereThe graph of $y=u(x-k)+\frac{1}{2}a(x-k)^2$ for $k=2s$ and $a=10ms^2$ and $u=5.8 ms^{-1}$

enter image description hereThe graph of equation:$y=ux+\frac{1}{2}ax^2$ for $k=2s$ and $a=10ms^2$ and $u=5.8 ms^{-1}$

These both graphs look pretty parallel and same,except the first one is shifted version of 2nd one.


(From 1st equation)$$y=ut-uk+\frac{1}{2}at^2+ak^2-2atk$$ $$\frac{\text{d}y}{\text{d}t}=u+2at-2ak$$


Here the acceleration and velocity of the 1st equation are different from the 2nd equation.

For my part,I think that the in the $1$st equation the motion started when $t=2$ and in the $2$nd equation motion started from $t=0$.That is the only difference.But the mathematics shows that the have different accelerations,which is independent of time variable.(provided I assume the accelerations to be constant).


2 Answers 2


Itachi, perhaps your calculations should be corrected as follows:

(From the 1st equation)$$y=ut-uk+\frac{1}{2}at^2+\frac{1}{2}ak^2-atk$$ $$\frac{\text{d}y}{\text{d}t}=u+at-ak$$


(From the 2nd equation)$$y=ut+\frac{1}{2}at^2$$ $$\frac{\text{d}y}{\text{d}t}=u+at$$


  • 1
    $\begingroup$ @Itachí Uchiha Is it OK? $\endgroup$ Oct 13, 2016 at 15:17

Your interpretation is correct. Only difference in the two cases is one is starting two seconds later.

One of the ways to write the equation is $$y= u(t-t_0)+\frac{1}{2} a (t-t_0)^2$$ In first case, say for $y1(t)$, $t_0=2$, while for second case, for $y2(t)$, $t_0=0$. Since $u$ and $a$ are same for both the cases, their initial velocities and acceleration are same. That is what you are getting, $y''= a$.

However, any quantity that depends on time will be shifted by $t_0$ for case 1. We can see that by writing it explicitly. For case 1, $$\frac{dy1}{dt}= u+ 2a(t-t_0)$$ while for case 2, it is $$\frac{dy2}{dt}=u+2a(t-0)=u+2at$$

  • 2
    $\begingroup$ um..Is not Accelaration of case 2 is supposed to be a? $\endgroup$
    – user132865
    Oct 13, 2016 at 10:38
  • $\begingroup$ Yes. Its a in both cases. $\endgroup$ Oct 14, 2016 at 9:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.