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=0.80 m and that v = 1 m/s when x= 0. Determine the velocity of the particle when x = -1 m. (You must provide an answer before moving to the next part.) The velocity of the particle is m/s.

Answer 1

Velocity,v = 0.323 m/s

Step-by-step explanation:

The acceleration of a particle is given by :

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

b = 0.8 m when x = 0

Since,
`a=v(dv)/(dx)`

`v(dv)/(dx)=-(0.1+sin(x)/(b))`

`\int{v.dv}=\int{-(0.1+sin(x)/(b))}.dx`

`(v^2)/(2)=-[0.1x-0.8cos(x)/(0.8)]+c`

At x = 0, v = 1 m/s

`(1)/(2)=0.8+c`

`c=-0.3`

`(v^2)/(2)=-[0.1x-0.8cos(x)/(0.8)]-0.3`

At x = -1 m

`(v^2)/(2)=-0.1(-1)+0.8cos((-1))/(0.8)-0.3`

`{v^2}=0.1045`

v = 0.323 m/s

So, the velocity of the particle is 0.323 m/s. Hence, this is the required solution.