Cracking The Ap Calculus ab Exam 2018

The intervals where the velocity and the acceleration have the same sign are 2 < t < and t > 5.

PROBLEM 3. Given that the position of a particle is found by x(t) = t^3 − 6t^2 + 1, t > 0, find the distance
that the particle travels from t = 2 to t = 5.

Answer: First, find v(t).

v(t)    =   3t^2    −   12t

Second, set v(t) = 0 and find the critical values.

3 t^2   −   12t =   0^3 t(t −   4)  =   0 t =   {0, 4}

Because the particle changes direction after four seconds, you have to figure out two time intervals
separately (from t = 2 to t = 4 and from t = 4 to t = 5) and add the absolute values of the distances.

|x(4)   −   x(2)|   +   |x(5)   −   x(4)|   =   |(−31)  −   (−15)|  +   |(−24)  −   (−31)|  =   23

PRACTICE PROBLEM SET 13

Now try these problems. The answers are in Chapter 19.

1.Find  the velocity    and acceleration    of  a   particle    whose   position    function    is  x(t)    =   t^3     −   9t^2    +   24t,    t   >

0.

2.Find  the velocity    and acceleration    of  a   particle    whose   position    function    is  x(t)    =   sin(2t) +   cos(t).

3.If    the position    function    of  a   particle    is  x(t)    =   sin ,   0   <   t   <   4π  find    when    the particle    is  changing

direction.

4.If    the position    function    of  a   particle    is  x(t)    =   3t^2    +   2t  +   4,  t   >   0,  find    the distance    that    the particle

travels from    t   =   2   to  t   =   5.

5.If    the position    function    of  a   particle    is  x(t)    =   t^2     +   8t, t   >   0,  find    the distance    that    the particle    travels

from    t   =   0   to  t   =   4.

6.If    the position    function    of  a   particle    is  x(t)    =   2sin^2 t    +   2cos^2 t,   t   >   0,  find    the velocity    and acceleration

of  the particle.

7.If    the position    function    of  a   particle    is  x(t)    =   t^3     +   8t^2    −   2t  +   4,  t   >   0,  find    when    the particle    is