Based on experimental observations, the acceleration of a particle is defined by the relation a = –(0.1 + sin x/b), where a and x are expressed in m/s2 and meters, respectively. Know that b = 0.98 m and that v = 1 m/s when x = 0.

a) v = +/- 0.515 m/s

b) x = -0.098164 m

c)  v = +/- 1.005 m/s

Explanation:

Given:

The relationship for the acceleration is given as follows:

                            a = - (0.1 + sin(x/b))

Where,  b = 0.98

- IVP is v = 1 m/s @ x = 0

Find:

(a) the velocity of the particle when x = -1 m

(b) the position where the velocity is maximum

(c) the maximum velocity.

Solution:

- We will compute the velocity by integrating a by dt.

                          a = v*dv / dx =  - (0.1 + sin(x/0.98))

- Separate variables:

                          v*dv = - (0.1 + sin(x/0.98)) . dx

-Integrate from v = 1 m/s to v and @ x = 0 to x:

                         0.5*(v^2) = - (0.1*x - 0.98*cos(x/0.8)) - 0.98 + 0.5

                         0.5*v^2 =  0.98*cos(x/0.98) - 0.1*x - 0.48

- Evaluate at, x = -1

                         0.5*v^2 = 0.98 cos(-1/0.98) + 0.1 -0.48

                         v = sqrt (0.2651155)

                         v = +/- 0.515 m/s

- v = v_max when a = 0. Set the given expression to zero and solve for x:

                          -0.1 = sin(x/0.98)

                           x = -0.98*0.1002

                           x = -0.098164 m

- Hence now evaluate velocity through the derived expression:

                           v^2 = 1.96 cos(-0.098164/0.98) -0.96 -0.2*-0.098164

                           v = sqrt (1.0098)

                           v = +/- 1.005 m/s