Distance, velocity & Acceleration

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otomanb
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taking derivation of distance equation is easy as every body knows.
differentiating distance we get velocity and differentiating velocity we get acceleration
but if a bird fly from one tree to another of distance 50m in 3 sec. we can get it's velocity by
v=s/t but how can we get it's acceleration ?
 
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hi otomanb! :smile:
otomanb said:
if a bird fly from one tree to another of distance 50m in 3 sec. we can get it's velocity by v=s/t but how can we get it's acceleration ?

if v = s/t, then the acceleration is zero :wink:
 
Only when the function x=f(t) is given, can we use differentiation
 
Try thinking this way: there are many ways that the bird could have flown the 50 meters in 3 seconds. It could have rapidly accelerated to say 17 m/sec and flown the whole way there at that speed, and then quickly slowed down at the end. Or it could have gradually accelerated through the first 25 meters and then gradually slowed during the final 25 meters. Or it could have done a crazy flight speeding up and slowing down repeatedly. So as the others have said, the 50/3 = 16.7 m/sec is just the average, and to find the acceleration you need to know the actual location at each point along the way. (That's azureth's function; f(t)).
 
Because v=s/t gives us the constant value and derivation of constant value gives us ZERO answer.
 
but if i use equation of motion
2as=vf^2-vi^2
it give me some answer like 2.78m/s^2 :confused:
 
otomanb said:
but if i use equation of motion
2as=vf^2-vi^2
it give me some answer like 2.78m/s^2 :confused:

vi = 0, but you don't know vf

you'll have to use another constant acceleration equation, s = vit + 1/2 at2 :wink:

btw, is this a rocket-powered bird?

what makes you think it flies with constant acceleration? :biggrin:
 
not constant acceleration with constant "Velocity" and if velocity is constant acceleration is always ZERO
 
2as=vf^2-vi^2
sorry don't know that it's constant acceleration equation