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QUESTION: Hi,

I am finding it very difficult to understand why is SUVAT differing from this real life example?

Let's assume Usain Bolt runs a 100m in 10sec(for calculation ease)

That means his acceleration would be

(v-u)/t = (10-0)/10 = 1m/s/s

Now if I calculate displacement, then

s=ut + (0.5*a*t*t)

=0*10 + (0.5*1*10*10)

=50

I have also made per/second break up calculating s at each 1second stage..

And that too totals to 50!?!?

I would really like to know.. What am I doing wrong??? It should 100??

Will await ur reply

ANSWER: Well, to start off with, no. His final velocity is not 10 m/s. You made that assumption, that's your problem with this derivation. If he only got up to that at the end, he couldn't actually make 100 m in 10 seconds, because he'd be traveling an average velocity of 5 m/s. See the inconsistency? His actual final velocity if he was moving at constant acceleration (your assumption here) would be 20 m/s in order to cover 100 m in 10 s. That's an acceleration of 2 m/s^2, so plug that in and your problem is solved. Be careful with your assumptions! :)

In reality this is unrealistic, he reaches a maximum velocity quickly and maintains it against resistive forces (both in his legs and air drag).

---------- FOLLOW-UP ----------

QUESTION: Yes!!

You are right .. i get it now, so weird!!!

So given the constant acceleration assumption if started from inertia(u=0), and covered a particular distance(s=100) in t=10 .. how do i get v correctly?

because my starting point is to find a =(v-u)/t, that would immediately send me down the wrong path

SO, how do i start my calculations when tackling such a problem

basically to arrive at v=20m/s... or a=2m/s^2...??

thanks,

appreciated :-)

Since you didn't know v, you can't solve a single equation with two unknowns (you also don't know a). So your second equation gives you a, since you know s, u, and t. Generally you need to find an equation (and that's really the only equation, everything else is derivation) that you have only one unknown for. If you can find two equations and two unknowns, you can use one to substitute into the other to get rid of it, I call that the fundamental trick of algebra... It helps a lot of people to draw up their knowns and unknowns in two lists, and find equations or sets of equations that they can solve.

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