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course Phy 201
6/26 4:00 pm
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? 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.
Average Velocity : (11 cm/s + 15 cm/s) / 2
= 13 cm/s
Change in time:
117 cm / 13 cm/s
= 9 s
Average Accleration = ‘dv / ‘dt
(4 cm/s )/ 9 s
= 4/9 cm/s^2
vf = 15 cm / sec (given)
v0 = 11 cm / sec (given)
a = .444 cm/s^2 (given)
‘ds = 117 cm (given)
‘dt = 9 sec (found)
2nd Question:
V0= 11 cm/s (given)
‘ds= 117 cm (given)
a = .444 cm/s/s (given)
dv= vf-v0
dv= 15 cm/s - 11 cm/s
dv= 4 cm/s
vAve= (v0+vf)/2
vAve= ( 11 cm/s + 15 cm/s)/2
vAve= 13 cm/s
‘dt= ds/vAve
Dt= 117 cm/ 13 cm/s
Dt= 9 s
Vf=?
vf^2 = v0^2 + 2 a `ds
vf^2 =( 11cm/s)^2 + 2 (.444cm/s/s)(117 cm)
vf^2= 122 cm/s + 103.9 cm/s
squart(vf)= squart [ 225.9 cm/s]
vf= 15 cm/s
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