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polet [3.4K]
3 years ago
13

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.
Physics
1 answer:
Fiesta28 [93]3 years ago
8 0

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.

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