Phy 201
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A ball starting from rest rolls 11 cm down an incline on which its acceleration is 27 cm/s2, then onto a second incline 44 cm long on which its acceleration is 9 cm/s2. How much time does it spend on each incline?
11=0+1/2*27*t square
27t square=22
Tsquare=22/27=.9sec
Tsquare = 9 sec is simply not a correct statement, but .9 sec itself is about right and you are on the right track. Done more carefully you get
If t^2 = 22 / 27, then t = .9, approximately.
If you use units you have
t^2 = 22 cm / (27 cm/s^2) = .8 s^2, approximately, so that
t = sqrt(.8 s^2) = .9 s, approximately.
Velocity=.9*27+0=24.3cm/sec
good
44=24.3dt+1/2*9 t square
Good; you have correctly formulated this problem in terms of the third equation, though your equation would be improved if it read
44 cm =24.3 cm/s * ` dt + 1/2 *9 cm/s^2 * `dt^2.
You would then need to solve the equation for `dt.
However this would require the use of the quadratic formula, which most students find confusing in this context. Most students will find the equation easier to solve if they use the fourth equation.
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Solution
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