#$&*
course PHY 121
9/30/11
Determine the acceleration of an object whose velocity is initially 24 cm/s and which accelerates uniformly through a distance of 66 cm in 3.8 seconds.Since we know v0, `ds, and `dt. We can determine vf.
I can use the equation:
`ds = 1/2(v0+vf)*`dt
So if I substitute the numbers in,
66cm = 1/2(24cm/s + vf)* 3.8s
First you would divide both sides by 3.8s.
66cm/3.8s = 17.4cm/s = 1/2(24cm/s+vf)
Now we can divide both sides by .5
17.4cm/s / .5 = 34.7cm/s = 24cm/s + vf.
Now subtract 24cm/s to 34cm/s which is 10.73cm/s = vf.
So now that we have vf and v0, we can get the acceleration. So 10.7cm/s - 24cm/s = -13.3cm/s. This is the change in velocity, so -13.3cm/s over 3.8s will give us acceleration which is -3.5cm/s^2. "
@& Very good.*@