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

Answer:

Velocity,v = 0.323 m/s

Explanation:

The acceleration of a particle is given by :

$a=-(0.1+sin\dfrac{x}{b})$

b = 0.8 m when x = 0

Since, $a=v\dfrac{dv}{dx}$  

$v\dfrac{dv}{dx}=-(0.1+sin\dfrac{x}{b})$  

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

$\dfrac{v^2}{2}=-[0.1x-0.8cos\dfrac{x}{0.8}]+c$

At x = 0, v = 1 m/s

$\dfrac{1}{2}=0.8+c$

$c=-0.3$

$\dfrac{v^2}{2}=-[0.1x-0.8cos\dfrac{x}{0.8}]-0.3$

At x = -1 m

$\dfrac{v^2}{2}=-0.1(-1)+0.8cos\dfrac{(-1)}{0.8}-0.3$

${v^2}=0.1045$

v = 0.323 m/s

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