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course Phy 201
9/25 8
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=11cm/s, vf=15cm/s', 'ds=117cm
vAve= (15cm/s+11cm/s)/2= 13cm/s
'dv= 15cm/s-11cm/s= 4cm/s
vf^2=v0^2 +2a*'ds
(15cm/s)^2= (11cm/s)^2 + 2a(117cm)
225cm^2/s^2= 121cm^2/s^2+a(234cm)
225cm^2/s^2-121cm^2/s^2=a(234cm)
104cm^2/s^2=a(234cm)
(104cm^2/s^2)/(234cm)=a
.444cm/s^2=a
a='dv/'dt
a*'dt='dv
'dt='dv/a
'dt=4cm/s/.44cm/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.
v0=11cm/s, a= .444cm/s^2, 'ds=117cm
vf^2=v0^2 +2a*'ds
vf^2= (11cm/s)^2 + 2(.444cm/s^2)(117cm)
vf^2=121cm^2/s^2+ 104cm^2/s^2
sqrt(vf^2)= sqrt(225cm^2/s^2)
vf=15cm/s
vf-v0='dv
15cm/s-11cm/s='dv
4cm/s='dv
a='dv/'dt
.444cm/s^2=(4cm/s)/'dt
'dt=9sec
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