Question

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 = 0.96 m and that v = 1 m/s when x = 0.

1)The velocity of the particle

2)The position where velocity is maximum

3)Determine the maximum velocity. (Round the final answer to three decimal places.)
The maximum velocity

Answer 1

1)

we know b = 0.96

We know Acceleration = a =

when x = 0 and v = 1 m/sec

1) velocity of the particle at a position of x = -1 m. ( this is random position since it is not given in the problem)

the velocity of the particle when x = -1 m

velocity v when x = -1 m= ?

1/2 V² = -0.1*-1+0.96*cos(1/0.96)-0.46

velocity = V² = 2(-0.1*-1+0.9598-0.46 )

velocity = V = 0.8942 m/sec.

2)

velocity is maximum when acceleration = a =0.

x = -5.50 m

3)

Maximum velocity at x = -5.50 m

1/2*V²_max = -0.1*(-5.50)+0.96*cos(-5.50/0.96)-0.46

maximum velocity = V = 1.44 m/sec