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

Looks good!:smile:In summary, to find the distance an object moves during the fourth second of its motion, you can use the equation s = ut + 1/2 at^2, given that the initial velocity, u, is zero. By plugging in the values for time (4 seconds) and acceleration (10m/s^2), you can calculate the distance traveled during the fourth second, which is 35m.
  • #1
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
 
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  • #2
Hint: If you knew the total distance it had traveled by the end of each second, would that help?
 
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  • #3
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...
 
  • #4
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.
 
  • #5
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?
 
  • #6
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:
 
  • #7
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

??
 
  • #8
Looks good!:smile:
 

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

1. What is free-fall?

Free-fall is a type of motion in which an object is only influenced by gravity. This means that the object is falling without any other force, such as air resistance, affecting its motion.

2. How is free-fall distance calculated?

The distance traveled during free-fall can be calculated using the formula d = 1/2 * g * t^2, where d is the distance traveled, g is the acceleration due to gravity (9.8 m/s^2), and t is the time in seconds.

3. Does the mass of the object affect its free-fall distance?

No, the mass of an object does not affect its free-fall distance. This is because all objects, regardless of their mass, experience the same acceleration due to gravity.

4. Can free-fall distance be negative?

Yes, free-fall distance can be negative if the object is falling in the opposite direction of the chosen reference point. In most cases, the ground is used as the reference point and therefore the distance is positive.

5. How does air resistance affect free-fall distance?

Air resistance can significantly affect free-fall distance by slowing down the object's motion. This means that the object will not reach the same distance in the same amount of time as it would in a vacuum. In real-world scenarios, air resistance must be taken into account when calculating free-fall distance.

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