course Phy201
Reason out the quantities v0, vf, Dv, vAve, a, Ds and Dt: If an object’s velocity changes at a uniform rate from 9 cm/s to 12 cm/s as it travels 84 cm, then what is the average acceleration of the object?v0=9cm/s
vf=12cm/s
'dv=3cm/s
vAve=(vf+v0)/2=10.5cm/s
good so far
The Average Acceleration= 'dv/'dt= 3cm/s/84cm= 28cm/s^2
84 cm is not `dt. It doesn't have the right units and it's clearly specified as the displacement of the object.
Hint: What is the definition of average velocity, and how can that definition be used to find `dt?
Using the equations which govern uniformly accelerated motion determine vf, v0, a, Ds and Dt for an object which accelerates through a distance of 84 cm, starting from velocity 9 cm/s and accelerating at .375 cm/s/s.
vf=v0 + a * 'dt=9cm/s + (.375 cm/s/s * 84cm)= 12.4369????? I don't know
v0=9cm/s
a=.375 cm/s/s
ds=(vf+v0)/2 * dt
dt=84cm
I don't believe i'm doing this right? Can you send feed back, i'm not understanding this acceleration? How to figure this out without time interval changes, i'm getting mixed up with the 'dt expression.
See my note and my hint above.
&#Please see my notes and submit a copy of this document with revisions and/or questions, and mark your insertions with &&&&. &#