A ball starting from rest rolls 6 cm down an incline on which its acceleration is 24 cm/s2, then onto a second incline 36 cm long on which its acceleration is constant, and on which it spends 1.513 seconds. How much time does it spend on the first incline and what is its acceleration on the For the first incline we have the following information: v0 = 0 `ds = 6 cm a = 24 cm/s^2 We need to find `dt Since v0 = 0, we can simplify vf^2 = v0^2 + 2 a `ds to be vf^2 = 2a`ds vf^2 = 2(24cm/s^2)*6cm vf^2 = 288 cm^2/s^2, we take the sqrt of each side vf = (+-)17.0 cm/s To find `dt, we can use a=`dv/`dt 24cm/^s2 = 17cm/s/`dt `dt = 17cm/s/24cm/s^2 `dt = .71 seconds For the second incline we now have the following: v0 = 17 cm/s `dt = 1.513 s `ds = 36 cm We need to find vf to find a `ds = (v0 + vf)/2*`dt Solving for vf = 2(`ds/`dt)-vo vf = 2(36cm/1.513s)-17 cm/s vf = 30.6 cm/s a = `dv/`dt a = 13.6/1.513 = 8.99 cm/s^2