#$&*
course Phy 121
9/17 3:30pm
Reason out the quantities v0, vf, Dv, vAve, a, Ds and Dt: If an object’s velocity changes at a uniform rate from 11 cm/s to 15 cm/s as it travels 117 cm, then what is the average acceleration of the object?
Using the equations which govern uniformly accelerated motion determine vf, v0, a, Ds and Dt for an object which accelerates through a distance of 117 cm, starting from velocity 11 cm/s and accelerating at .444 cm/s/s.
#1 We can figure out the vAve from (11cm/s + 15cm/s)/2 = (26cm/s)/2 = 13cm/s. If vAVe * `dt = `ds, then to find `dt we can divided `ds by vAve. 117cm/ 13cm/s = 9 s
Now we know all the quantities except a. This we can reason out from our change in velocity (vf-v0) divided by `dt. Our `dv is 15cm/s - 11cm/s = 4cm/s. We divide this 4cm/s by 9s to get a. (4cm/s)/9s = .44cm/s^2 (approx)
v0 = 11cm/s
vf = 15cm/s
vAVe = 13cm/s
a =.44cm/s^2 (approx)
`ds = 117cm
`dt = 9 s
#2. Start with the equation:
vf^2 = v0^2 + 2 a `ds Substitute
vf^2 = (11cm/s)^2 + 2(.444cm/s^2)117cm
vf^2 = 121cm^2/s^2 + 102.96cm^2/s^2
vf^2 = approx 223 cm^2/m^2 Take sqrt of each side
vf = +- 15cm/s
At this point we know
v0 = 11cm/s
a= .444cm/s/s
`ds = 117cm
vf = +- 15cm/s
I don't know if we can say for certain that the vf must be positive so we should evaluate it for both cases.
for vf = 15cm/s vAve = (11cm/s + 15cm/s)/2 = (26cm/s)/2 = 13cm/s
`dt = `ds/vAve = 117cm/ 13cm/s = 9s
for vf= -15cm/s, vAve = (11cm/s + -15cm/s)/2 = (-4cm/s)/2 = -2cm/s
`dt = 117cm/-2cm/s = -58.5s
We can't have a negative time interval so either vf cannot equal -15cm/s or the displacment 117 would have to be negative.
If we evaluate it that way,
`dt = -117/-2cm/s = 58.5s
???We would also need to reevaluate acceleration, which makes me think in this case we would not accept -15cm/s as a possible vf. Is that correct that we can assume it can only be positive???
"
@&
If you reverse your idea of the motion so that time runs backwards, this solution can actually make sense.
*@
&#This looks very good. Let me know if you have any questions. &#