object velocity

Question

For an object whose velocity in ft/sec is given by v(t) = sin(t), what is its distance, in feet, travelled on the interval t = 1 to t = 5? (2 points)

0.45

0.26

2.82

0.03

Answer 1

Since the question asks for distance not displacement,   the integrand in the integral over t is

the absolute value of sin(t).  

 

The easiest way to do this type of integral is to split the range integration into two parts:

  the integral of  sin(t)   from   1 to pi   ( where sine is positive )     

   and the integral of ( - sin(t)  )   from   pi to 5.  (where sine is negative)

 

  The antiderivative of sin is - cos   so   the integration   gives

 

   -[ cos(pi) - cos(1) ]  + [ cos(5) - cos(pi)  ] 

   this equals    1 +  cos(1)   +  cos(5) + 1  ~ 2.82