Question

For the one-dimensional motion, described by x = t–sint
(a) x (t) > 0 for all t > 0.
(b) v (t) > 0 for all t > 0.
(c) a (t) > 0 for all t > 0.
(d) v (t) lies between 0 and 2.

Answer 1

Initially: at  t= 0,  x = 0 - sin 0 = 0..
     Differentiate:   x' = 1 - Cos t 
                 as cos t <= 1,  x' >= 0...  Hence,  x raises with time..
  So  x(t) > 0 for all t > 0..
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  v(t) = x ' (t) =  1 - cos t       =    0  for t = 2 n π
       so  v (t) > 0 is not always true
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  a (t) =  Sin t    = 0 for  t = 2n π
           so  a(t)  is not always  > 0..
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     v(t)  = 1 - cos t          as  -1 <= cos t <= 1
           so    0 <=   v (t)   <=  2

Answer 2

Answer:

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