Barrons AP Calculus

(a)

(b)

SOLUTION:

Since   v(t)    =   6t^2    −   18t +   12  =   6(t −   1)(t    −   2), we  see that:

if  t   <   1, then v > 0;

if  1   <   t   <   2, then v   <   0;

if  2   <   t, then v > 0.

Thus,   the total   distance    covered between t   =   0   and t   =   4   is

When    we  replace v(t)    by  6t^2    −   18t +   12  in  (2) and evaluate,   we  obtain  34

units   for the total   distance    covered between t   =   0   and t   =   4.  This    can also    be

verified    on  your    calculator  by  evaluating

This    example is  the same    as  Example 26  in  Chapter 4,  in  which   the required

distance    is  computed    by  another method.

To  find    the displacement    of  the particle    from    t   =   0   to  t   =   4,  we  use the FTC,

evaluating

This    is  the net change  in  position    from    t   =   0   to  t   =   4,  sometimes   referred    to  as

“position   shift.” Here    it  indicates   the particle    ended   up  32  units   to  the right   of

its starting    point.

Example 3 __

The acceleration of an object moving on a line is given at time t by a = sin t;
when t = 0 the object is at rest. Find the distance s it travels from t = 0 to t =

SOLUTION: Since = sin t, it follows that