Kinematics question: free-fall distance traveled during a given second

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SUSUSUSUSUSUSUSU

Homework Statement



An object is released from rest in the gravitational field of the Earth. Air resistance is negligible. How far does the object move during the fourth second of its motion?

A. 15 m

B. 25 m

C. 35 m

D. 45 m

Homework Equations


v = u + at
s = ut + 1/2 at^2
v^2=u^+2as
s=(u+v)t/s

The Attempt at a Solution


u=0
a=10m/s^2
t=4

Now I am blocked... which equation should I use?

Thank you
 
on Phys.org
Hint: If you knew the total distance it had traveled by the end of each second, would that help?
 
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gneill said:
Hint: If you knew the total distance it had traveled by the end of each second, would that help?
could you give me more hint? I do not get it since that there is no total distance...
 
SUSUSUSUSUSUSUSU said:
could you give me more hint? I do not get it since that there is no total distance...
There is a total distance it has traveled from the point of release at time equals zero to any given instant of time after that. One of your equations will give you that distance for any time t.
 
gneill said:
There is a total distance it has traveled from the point of release at time equals zero to any given instant of time after that. One of your equations will give you that distance for any time t.

I have used s= (v+u)t/2
v= u+at = 0+10t=10t
therefore s= (0+10t)t/2 =10t^2/2?
 
SUSUSUSUSUSUSUSU said:
I have used s= (v+u)t/2
v= u+at = 0+10t=10t
therefore s= (0+10t)t/2 =10t^2/2?

That's what you want. I'm not sure where your "s= (v+u)t/2" came from, but your second Relevant Equation gives you the correct form directly given that the initial velocity, u, is zero :wink:
 
gneill said:
That's what you want. I'm not sure where your "s= (v+u)t/2" came from, but your second Relevant Equation gives you the correct form directly given that the initial velocity, u, is zero :wink:

ahhhh i got it!

when,
t= 0, 1, 2, 3, 4
v=0,10,20,30,40
s=0,5,20,45,80

therefore 80-45= 35m

??