Question

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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Answer 1

Kinematics Formulae

v = u + at
s = ut + 0.5at²

using second equation substitute s = 3meters , u =initial velocity = at rest = 0 , t = 1 second

3 = 0 + 0.5*a*1*1

a = 6 meters/second^{2 = acceleration}

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1) distance travelled in two seconds substitute a =6 , u =0 , t = 2

s = 0 + 0.5*6*2*2 = 12 meters

distance travelled during 2nd second = 9 meters

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2) velocity at end of first second v = u + at

Substitute u = 0 , a = 6 ,t = 1second

v = 0 + 6 = 6 meters/second

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3) velociity at end of 2nd second v = u + at = 0 + 6*2 = 12 meters/second

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