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course Phy 241
7/21/2013 3:35 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.
v_0 = 11cm/s
v_f = 15cm/s
`d v = 4cm/s
vAve = 13cm/s
a =
`d s = 117cm
`d t = 9s
`ds = (v_0+v_f)/2*`dt
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At this point you are using an equation as opposed to reasoning.
I assume you got the 9 s by dividing 117 cm by 13 cm/s.
All you need to do at this point is divide `dv by `dt.
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`ds/(v_0+v_f)/2 = `dt
2*`ds/(v_0+v_f) = `dt
2*117cm/(11cm/s+15cm/s) = `dt
2*117cm/(11cm/s+15cm/s) = `dt
234cm/26cm/s = `dt
9s = `dt
`ds = v_0*`dt+.5*a*`dt^2
`ds-(v_0*`dt) = .5*a*`dt^2
[`ds-(v_0*`dt)]/`dt^2 = .5*a
([`ds-(v_0*`dt)]/`dt^2)*2 = a
([117cm-(11cm/s*9s)]/9s^2)*2 = a
([117cm-(99cm)]/9s^2)*2 = a
([18cm]/9s^2)*2 = a
(2cm/s^2)*2 = a
4cm/s^2 = a
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I don't see where you have solved the second problem, which does require the use of the equations.
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.
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