Calculus Application Problem2





In this page calculus application problem2 we are going to see solution of the second problem of the topic application problems in calculus.calculus application problem1

Question 2:

A particle of unit mass moves so that displacement after t seconds is given by x = 3 cos (2t - 4). Find the acceleration and kinetic energy at the end of 2 seconds. (K.E = (1/2) m v²)

Solution:

Here the variable "x" represents the displacement of the object and "t" displacement of a particle in seconds

x (t) = 3 cos (2t - 4)

mass = 1

to find acceleration we have to different the given equation two times

velocity dx/dt = 3 [- sin (2t - 4)] (2(1) - 0)

             (v)  = - 6 sin (2t - 4)

velocity at t = 2

               v = -6 sin (2(2) - 4)

                  = -6 sin (4-4)

                  = -6 sin (0)

                  = 0

  acceleration d²x/dt² = -6 cos (2 t - 4) (2(1)- 0)

                               = -6 cos (2 t - 4) 2

                               = -12 cos (2 t - 4)

now we have to put t = 2

                               = -12 cos (2(2) - 4)

                               = -12 cos (4 - 4)

                               = -12 cos (0)

                               = -12 (1)

                               = -12

Kinetic energy K.E = (1/2) m v²               

                         = (1/2) (1) (0)²

                         = (1/2) (0)

                         = 0