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