course Phy 201
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?
Since we know a,`ds and v0 we can use the equation vf^2=v0^2+2*a`ds. We have
vf^2=0^2+2*27cm/s^2*11cm,vf^2=54cm/s^2*11,vf^2=594cm/s,take the sqrt and we have vf=24.37cm/s. We can now use the equation
vf=v0+a`dt, rearrange it and we have `dt=(vf-vo)/a, `dt=24.37cm/s/27cm/s^2=.90secs
Now using the vf of our first incline as the v0 for our second incline v0=24.37cm/s. We now do the same calcualtions as the
first incline. vf^2=24.37^2+29cm/s^2*44cm,vf^2=593.9+2*396,vf^2=593.9+792,vf^2=1385.9cm/s,take the sqrt,vf=37.22cm/s. Now we
take vf=v0+a`dt, `dt=(vf-v0)/a,`dt=(37.22-24.37)/9cm/s^2,`dt=12.85cm/s/9cm/s^2,`dt=1.43sec.
So we have .90s for incline 1 and 1.43s for incline 2"
Excellent work.