1
a)  1.8/1.2
b)  (1.8/1.2)*(180/p)
c)  0.620*2.4
5
Q(t) = 4t-3t²+t³
a)   w(t) = dQ/dt = 4 - 6t +3t²        w(2)   w(4)
b)  [w(4) - w(2) ]/2
c)  a(t) = dw/dt = -6 + 6t             a(2)     a(4)
9
h = 1/2 gt²      so t = sqrt(2h/g)
(Q_final - 0) /t = 2.5*2p/t
11
a)   a = [w_final - w_initial]/t   assuming constant acceleration   and w = 2p*rev/min*60
b)   Q = w_initial*t + 1/2 at²    and divide by 2p  to have revolutions
33
a)   w = v/r
b)   a_r = v²/r
c)   0 (zero)
35
a)   a = [w_final - w_initial]/t  = (0 - 150)/(2.2*60)
b)   Q = w_initial*t + 1/2 at² = 150*2.2*60 - 1/2 a (2.2*60)²   where a from a)
c)   acceleration is constant so it is the same at any moment between 0 and 2.2 hours.
      linear acceleration  a = a*r where a  is from a) and r = 0.5 m
d)   net acceleration is the vector sum of tangential and radial acceleration (they are always at right angle)
      the magnitude of the sum is   sqrt(a_r² + a_t²)  where a_t is from c)  and a_r = 75²/0.5
39
a)   linear velocity = v = at = 0.5*15    so  radial acceleration   a_r = v²/r
and  magnitude of the net acceleration is    sqrt(a_r² + a_t²)  where a_t = 0.5
b)  angle can be calculated from  Q = tan-1 (a_r /a_t)