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If we consider a ball moving at an acceleration of $5\ \mathrm{m\ s^{-2}}$, over a time of 4 seconds, the distance covered by the ball in the first second is $5\ \mathrm m$. In the 2nd second will $5\ \mathrm m+5\ \mathrm m=10\ \mathrm m$. In the third second it will cover a distance of $5\ \mathrm m+5\ \mathrm m+5\ \mathrm m=15\ \mathrm m$ and so on and so forth. Now, when we substitute this answer in the equations of motion derived from the area under velocity-time and distance-time graphs, we see a variation: $$\begin{align} s&=\frac12at^2\\ &=\frac12\times5\ \mathrm{m\ s^{-2}}\times(4\ \mathrm s)^2\\ &=\frac12\times5\ \mathrm{m\ s^{-2}}\times16\ \mathrm{s^2}\\ &=40\ \mathrm m \end{align}$$ is the distance covered. Now if we go back to our initial description of acceleration we that in the first second $5\ \mathrm m$, second second $10\ \mathrm m$, third second $15\ \mathrm m$, and fourth second $20\ \mathrm m$. Total distance covered in this case is $5\ \mathrm m+10\ \mathrm m+15\ \mathrm m+20\ \mathrm m=50\ \mathrm m$?

$40\ \mathrm m\neq 50\ \mathrm m$. Why this disparity between the values? Can someone please explain?

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You say:

If we consider a ball moving at an acceleration of $5\ \mathrm{m/s^2}$, over a time of 4 seconds, the distance covered by the ball in the first second is $5\ \mathrm m$. etc

But that's not true. Why do you think it would travel $5\ \mathrm m$? You already know the correct equation:

$$s = ut + \frac{1}{2}at^2$$

and if you use this to calculate the distance travelled in 1 second it comes out at $2.5\ \mathrm m$.

Look at this another way:

If the acceleration is $5\ \mathrm{m\ s^{-2}}$ then at the end of 1 second the ball is travelling at $5\ \mathrm{m\ s^{-1}}$, and that means for most of that first second the ball must have been travelling at less than $5\ \mathrm{m\ s^{-1}}$. So it can't have travelled $5\ \mathrm m$. To travel $5\ \mathrm m$ in the first second the average speed over the first second must be $5\ \mathrm{m\ s^{-1}}$, not the final speed at the end of the first second.

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    $\begingroup$ @RamSidharth If you choose this time-interval smaller (0.5s, 0.1s), you will get closer to the actual value. You are actually performing a Riemannsummation here. $\endgroup$
    – Bernhard
    Feb 10, 2012 at 15:07
  • $\begingroup$ Thanks, that is truly enlightening. However, I shall have to read up on Riemann Summation. $\endgroup$
    – Ram Sidharth
    Feb 10, 2012 at 15:13
  • $\begingroup$ Agree with @Bernhard, the value disparity can be interpreted as the difference between an integral and its upper Riemann sum, cf. en.wikipedia.org/wiki/Riemann_sum $\endgroup$
    – Qmechanic ♦
    Feb 10, 2012 at 15:45
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It seems to me you have confused velocity and acceleration.

If we consider a ball moving at an acceleration of 5m/s^2

This doesn't really make sense, acceleration is change of rate of movement. Movement is defined physically by 'velocity' in units of $ms^{-1}$. I think you have this mostly sussed out but just have your setup wrong. To correctly answer this type of question you typically need need to consider either an inital or final velocity (or in some variants distance).

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