Starting from rest, a boulder rolls down a hill with constant acceleration and travels 3.00 during the first second.
How far does it travel during the second second?
How fast is it moving at the end of the first second?
How fast is it moving at the end of the second second?
If we use s=(1/2)a*t^2, t=1 and s=3 then a=6m/s^2
To answer this
How far does it travel during the second second?
Find the distance it would travel in 2s and subtract the distance it traveled in the first second.
For this
How fast is it moving at the end of the first second?
Use v=a*t Since you know a and t, just evaluate them in the formula.
...and finally
How fast is it moving at the end of the second second?
Use v=a*t with the a from above and t=2.
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yeah annonymous shut up
I'm glad you think bunnies are nice, but let's stay focused on the original questions.
For the first question, "How far does it travel during the second second?" we can use the formula s = (1/2)at^2, where s is the distance, a is the acceleration, and t is the time.
In this case, we are given that during the first second, the distance traveled is 3.00. So we can plug in the values and solve for the distance traveled during the second second.
s = (1/2)(6)(2)^2 - (1/2)(6)(1)^2
s = (1/2)(6)(4) - (1/2)(6)(1)
s = 12 - 3
s = 9
Therefore, the boulder travels 9.00 during the second second.
For the second question, "How fast is it moving at the end of the first second?", we can use the formula v = at, where v is the velocity and t is the time.
Again, we know the acceleration is 6 and the time is 1, so we can calculate the velocity.
v = (6)(1)
v = 6
Therefore, the boulder is moving at a velocity of 6 m/s at the end of the first second.
For the final question, "How fast is it moving at the end of the second second?", we can use the same formula v = at, but with a time of 2.
v = (6)(2)
v = 12
Therefore, the boulder is moving at a velocity of 12 m/s at the end of the second second.