course Phy 201
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?
`dv=vf-v0, 15cm/s-11cm/s=4cm/s=`dv
vAve=(v0+vf)/2, (15cm/c+11cm/s)/2=13cm/s=vAve
vAve*`dt=`ds, Flop the equation `ds/vAve=`dt, 117cm/13cm/s=9secs
a=`dv/`dt, 4cm/s/9secs=.444cm/s^2
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.
We need to use this equation to find final vf
vf^2=v0^2+2*a*`ds
vf^2=11cm/s^2+2*.444cm/s^2*117cm
vf^2=121cm/s+.888cm/s^2*117cm
vf^2=121cm/s+103.896cm/s^2
vf^2=224.896cm/s^2
Now if we taake the sqrt we will get vf
vf=15cm/s
Now if we take `dv/aAve we can find `dt
15cm/s-11cm/s=4cm/s
4cm/s/.444=9secs"
&#Very good work. Let me know if you have questions. &#