course phy 121
2/10 at 1:25
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?
v0=11 cm/s
vf=15 cm/s
'ds=117 cm
117cm = (11cm/s+15 cm/s)/2 *'dt
117cm= 13 cm/s * 'dt
9 s = 'dt
'ds= vAve * 'dt
117 cm= vAve* 9sec
117 cm/9 sec = vAve
13 cm/s= vAve
vf^2=v0^2+ 2a'ds
(15 cm/s)^2=(11cm/s)^2 + 2a(117 cm)
225 cm^2/s^2= 121cm^2/s^2 + 234cm(a)
104 cm^2/s2 = 234 cm(a)
.444 cm/s^2 = a
'dv=a * 'dt
'dv= .44cm/s^2 * 9s
'dv=3.96 cm/s
Very good reasoning.
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.
vf=15 cm/s
v0=11 cm/s
a=.444 cm/s/s
Ds=117 cm
Dt=9 sec
This is all based on the above solutions.
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You didn't use the equations to answer this question. It's important that you know how to do so.
To answer this specific question you can only use the information given in the question, which consists of the values of `ds, v0 and a. The previous solution does match the given quantities, so having solved the previous problem you do know what the solutions will be. However that doesn't constitute a solution.
How do you find these quantities based on only the information given in the question?
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