v = Vo + a t
v = 0 + 9.81 (8)
d = Vo t + (1/2)(9.81)(64)
= 0 +4.9(64)
so yes, you have the right idea.
a.Calculate the stone's velocity after 8.0 s.
b. What is the stone's displacement during this time?
Given:
t = 8.0 s
a = 9.81 m/s^2
For a, in order to calculate the velocity, should i use d = vit + 1/2 at^2 and only use the 1/2 at^2.
For b, should i use the d = vit + 1/2at^2 to calculate the distance.
Thanks I appreciate this a lot
v = 0 + 9.81 (8)
d = Vo t + (1/2)(9.81)(64)
= 0 +4.9(64)
so yes, you have the right idea.
a. To calculate the stone's velocity after 8.0 s, you can indeed use the equation d = vit + 1/2 at^2. However, if we assume the stone starts from rest, the initial velocity (vi) would be zero. So the equation simplifies to d = 1/2 at^2. Just plug in the values, and voila! You'll get the velocity.
b. For the stone's displacement during this time, you can indeed use the same equation d = vit + 1/2 at^2. Since the initial velocity is zero, the equation becomes d = 1/2 at^2. Just substitute the given values and calculate the distance traveled.
So grab your calculator, and let the numbers fly! Hope that helps, and remember, don't let gravity get you down!
Using the equation d = vit + (1/2)at^2, the initial velocity term (vit) becomes zero, so the equation simplifies to d = (1/2)at^2.
Now, substituting the known values:
a = 9.81 m/s^2 (acceleration due to gravity)
t = 8.0 s (time)
We can calculate the displacement (d) during this time using d = (1/2)at^2:
d = (1/2) * 9.81 m/s^2 * (8.0 s)^2
= 39.24 m/s^2 * 64 s^2
= 2511.36 m
Therefore, the displacement of the stone after 8.0 s is 2511.36 meters.
For part b, to calculate the distance traveled by the stone, you can again use the same equation: d = (1/2)at^2. The difference between displacement and distance is that displacement considers the starting and ending points, while distance only considers the total length traveled.
Since the stone starts at rest, its displacement and distance will be the same in this case.
So, the answer to part b is also 2511.36 meters.