course PHY 231
v_0 = 10cm/s
'ds = 36cm
'dt = 3s
v_avg = (36/3)cm/s = 12cm/s
v_avg = (v_0 + v_f) / 2
v_f = 2v_avg - v_0 = 2*12cm/s - 10cm/s = 14cm/s
'dv = v_f - v_0 = 4cm/s
a = 'dv/'dt = 1.3cm/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 36 cm, ending with velocity 10 cm/s and accelerating at 1.333 cm/s/s.
v_0 = ?
v_f = 10cm/s
v_avg = ?
'ds = 36cm
'dt = ?
'dv = ?
a = 1.333cm/s^2
(v_f - v_0) = 'dv
(v_f + v_0)/2 = v_avg = 'ds*'dt
a = 'dv/'dt
'dt = 'dv/a
v_avg = 'ds*'dv/a = ('ds/a)*(v_f - v_0) = (v_f + v_0)/2
('ds/a)*(v_f - v_0) = (v_f + v_0)/2
(36cm/s)(s^2/1.333cm)*(10cm/s - v_0) = (10cm/s + v_0)/2
(36cm/1.333cm)(s^2/s)*(10cm/s - v_0)*2 = 10cm/s + v_0
27*2s(10cm/s - v_0) = 10cm/s + v_0
54s(10cm/s - v_0) = 10cm/s + v_0
540*cm*(s/s) - 54*v_0cm(s/s) = 10cm/s + v_0cm/s
????I cannot seem to work this problem out. Is the object starting from rest???
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After doing the next quiz I came back to this one because I had a better Idea. My algebra was off a little. Here's my complete answer,
v_0 = ?
v_f = 10cm/s
v_avg = ?
'ds = 36cm
'dt = ?
'dv = ?
a = 1.333cm/s^2
v_avg = (v_f + v_0)/2
'dv = v_f + v_0
'ds = v_avg * 'dt
'dt = 'dv/a
TF 'ds = (v_avg * 'dv)/a
'ds = (v_f + v_0)(v_f - v_0)/2a
'ds*2a = (v_f^2 - v_0^2)
v_0 = sqrt(v_f^2 - 2a'ds)
v_0 = sqrt(100cm^2/s^2 - 2*36*1.333cm^2/s^2)
v_0 = sqrt[ (100-96)cm^2/s^2 ]
v_0 = 2cm/s
From here it is easy. It follows that 'dv = v_f - v_0 which works out to 8cm/s
v_avg = (10cm/s + 2cm/s)/2 = 6cm/s
and finally 'dt = 'dv/a = (8cm/s)(s^2/1.333cm) = 6s
Excellent work. You have used reasoning to derive the fourth equation of motion from the definitions.
There's also a calculus-based derivation, which in some ways is simpler, but you'll see that later.
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