course Phy231
2/23/10 12:30pm
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?
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.
12cm/s = vf 9cm/s = v0 84cm='ds 'dv=12cm/s-9cm/s=3cm/s
12cm/s + 9cm/s= 21cm/s /2= 10.5cm/s is vAve
12cm/s + 9cm/s= 21cm/s /2= 10.5cm/s implies that
12cm/s + 9cm/s = 10.5cm/s,
which is not so.
Equal signs indicate equality, not train of thought.
12cm/s + 9cm/s= 21cm/s
and
21cm/s /2= 10.5cm/s.
This is the average of the initial and final velocities, which since acceleration is uniform is equal to the average velocity.
(vf-v0)/'ds= (12cm/s -9cm/s)/84cm= .036cm/s^2 is the a
This is a valid solution to the first problem.
'ds/'dt=vAve 84cm/'dt=10.5cm/s 84cm=10.5cm/s *'dt 8s='dt
It's not clear where your answer to the first question stops and your answer to the second starts.
aAve=.375cm/s/s
v0=9cm/s
'ds=84cm
(vf-v0)/'ds=aAve (vf-9cm/s)/84cm=.375cm/s^2
change in velocity divided by change in position does not give you acceleration
(vf-9cm/s)=31.5cm/s vf=31.5cm/s + 8cm/s= 40.5cm/s These units don't seem right.
'ds/'dt=vAve vAve=(vf+v0)/2 (40.5cm/s+9cm/s)/2=24.75cm/s 84cm/'dt=24.75cm/s 84cm=24.75cm/s*'dt 'dt=3.39s
You need to redo your solution to the second question. You are given `ds, v0 and a and need to use the equations of motion to find your solution for the other variables, based on the given quantities.
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