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course Phy 231
2-5-11 about 9:00pm
Reason out the quantities v0, vf, Δv, vAve, a, Δs and Δt: If an object’s velocity changes at a uniform rate from 11 cm/s to 15 cm/s as it travels 104 cm, then what is the average acceleration of the object?Using the equations which govern uniformly accelerated motion determine vf, v0, a, Δs and Δt for an object which accelerates through a distance of 104 cm, starting from velocity 11 cm/s and accelerating at .5 cm/s/s.
v 0= 11cm/sec
vf= 15cm/sec
dv= (15cm/sec – 11cm/sec)= 4cm/sec
vAve= (15cm/sec + 11cm/sec)/2= 13cm/sec
ds= 104cm
104cm= 13cm/sec * dt
dt= 8sec
a= (15cm/sec – 11cm/sec)/8sec= .5cm/sec/sec
v 0= 11cm/sec
a= .5cm/sec/sec
ds= 104cm
vf^2= (11cm/sec)^2 + 2(.5cm/sec/sec)(104cm)
vf= square root of (225cm^2/sec^2)= 15cm/sec
15cm/sec= 11cm/sec + .5cm/sec^2(dt)
dt= (4cm/sec)/(.5cm/sec^2)= 8sec
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&#Very good responses. Let me know if you have questions. &#