How displacement due to acceleration formula works? [closed]

Assume an object begins from rest, accelerates 2m every second and ran for about 4 seconds.

I see that as

In 1st second covered: 2m

In 2nd second covered: 4m

In 3rd second covered: 6m

In 4th second covered: 8m

That gives total distance covered as 20m

Now, if I go by formula s = ut + 1/2(at^2)

I get, s = (0)4 + 1/2(2*4^2) = 1/2(32) = 16m

That's doesn't match.

What I am missing here?

Thanks

closed as unclear what you're asking by David Hammen, Yashas, Jon Custer, peterh, Kyle KanosJul 16 '17 at 19:38

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• I'm fairly sure questions very much like this have been asked before on this site, but I'm having a hard time finding them. (There's a large "sea" of difficult-to-distinguish kinematics questions.) – David Z Jul 16 '17 at 7:25
• Assuming you're talking about linear motion, you are merely confusing concepts. If the body started from rest and accelerated $2 \text{ms}^{-1}$ every second, then the displacements in the 1st, 2nd, 3rd, 4th second are 1, 3, 5 and 7 m, respectively. This would give it to be 16 m. The formula does work. – vs_292 Jul 16 '17 at 7:56
• @vs_292, why 1st second displacement is 1 not 2, for the acceleration of 2m per second? – Galaxy Jul 16 '17 at 8:04
• When the mathematics and results don't agree, there are two choices: (1) you did the mathematics wrong, or (2) the mathematics is wrong. As is almost always the case, this an example of option #1, you did the mathematics wrong. With an acceleration of 2 meters/second$^2$, after one second, the distance covered is one meter, not two. – David Hammen Jul 16 '17 at 8:40
• @Galaxy : I get what you're trying to do; but you're saying that the body accelerates $2 \text{m}$ every second, but that's wrong, because the body is increasing it's *velocity* by $2 \text{m/s}$ and not displacement by $2 \text{m}$ every second. Of course, displacement's also changing, but you'd have to use the equation $S = ut + \frac{1}2 at^2$ for every second, by taking the initial velocity as the final velocity of the previous second, and the time being $1$ second, as you're finding it for every second. You will end up with 16 m. – vs_292 Jul 16 '17 at 8:47

You are saying "acceleration of 2m per second", but think over this statement for a moment. You are saying that the acceleration is 2 m/s. This is a wrong unit.

You would have moved

• $2\;\mathrm{m}$ after 1 s,
• $4\;\mathrm{m}$ after 2 s,
• $6\;\mathrm{m}$ after 3 s,
• $8\;\mathrm{m}$ after 4 s.

if the speed was $2\;\mathrm{m/s}$. But the speed is not $2\;\mathrm{m/s}$; it is changing all the time when there is acceleration. If the acceleration is $2\;\mathrm{m/s^2}$, then rather you are moving with:

• $2\;\mathrm{m/s}$ after 1 s,
• $4\;\mathrm{m/s}$ after 2 s,
• $6\;\mathrm{m/s}$ after 3 s,
• $8\;\mathrm{m/s}$ after 4 s.

Bottom-line is that there is something wrong in your sentence "acceleration of 2m per second" since it mentions acceleration but with a speed unit.

It would also be wrong to say that you accelerate 2 metres or that you move 2 seconds or that you jump 2 Joules or so. You might mean something by it, but there is definitely something wrong in such sentence.

The mistake you are making is to confuse the speed at the end of each period with the average speed in the period. Take the first second. The object starts from rest, ie zero velocity. After 1 second its speed id 2m/s, but note that after 0.5s its speed is only 1 m/s. That's why the object doesn't travel 2m in the first second.

Perhaps a velocity-time graph will show you what your error is?

The area under such a graph is equal to the displacement.

So for each interval the displacement is the area of a trapezium with parallel sides $v_{\rm final }$ and $v_{\rm initial}$ and width $\Delta t$ which is equal to $\dfrac{v_{\rm final }+v_{\rm initial}}{2}\Delta t$.
You will note that $\dfrac{v_{\rm final }+v_{\rm initial}}{2}$ is the average velocity over the time interval $\Delta t$ whereas in your calculation you have used the final velocity over that interval.