course phy 201
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Week 2 Quiz 2 Version 1Reason out the quantities v0, vf, v, vAve, a, s and t: 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
The initial velocity vo = 11cm/s
the final velocity vf=15cm/s
the change in displacement ‘ds=117cm
the change in velocity ‘dv = vf-vo
15cm/s – 11cm/s = 4cm/s
the average velocity vAve = (vf+vo)/2
(15cm/s + 11cm/s)/2 = 13cm/s
the change in time ‘dt = ‘ds/vAve
117cm / 13cm/s = 9s
The average acceleration aAve = ‘dv/’dt
4cm/s / 9s = 0.444cm/s^2
Using the equations which govern uniformly accelerated motion determine vf, v0, a, s and t for an object which accelerates through a distance of 117 cm, starting from velocity 11 cm/s and accelerating at .444 cm/s/s.
The initial velocity is 11 cm/s = vo
The average acceleration is a = 0.444cm/s^2
The change in displacement ‘ds = 117cm
To find the final velocity vf = sqt(vo^2 + 2*a*’ds)
sqt((11cm/s)^2 + 2*0.444cm/s^2 * 117cm) = 15cm/s
The change in time ‘dt = ‘ds/ vAve = ‘ds/((vf + vo)/2)
117cm / ((15cm/s + 11cm/s)/2) = 9s
vf = v0 + a `dt
`ds = (vf + v0) / 2 * `dt
`ds = vAve * `dt = (vf + v0) / 2 * `dt.
Excellent solution. Keep up the great work.