Based on experimental observations, the acceleration of a particle is defined by the relationa = - (0.1 + sin (x/b)), where a and x are expressed in m/s^2 and meters, respectively. Knowing that b = 0.8 m and that v = 1 m/s when x = 0, determine (a) the velocity of the particle when x = - 1 m, (b) the position where the velocity is maximum, (c) the maximum velocity.

Question

Engineering

Topaz

13 October, 03:03

Based on experimental observations, the acceleration of a particle is defined by the relationa = - (0.1 + sin (x/b)), where a and x are expressed in m/s^2 and meters, respectively. Knowing that b = 0.8 m and that v = 1 m/s when x = 0, determine (a) the velocity of the particle when x = - 1 m, (b) the position where the velocity is maximum, (c) the maximum velocity.

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Answers (1)

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Angeline · Answer 1

a) v = + / - 0.323 m/s

b) x = - 0.080134 m

c) v = + / - 1.004 m/s

Explanation:

Given:

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

b = 0.8

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/0.8))

- Separate variables:

v*dv = - (0.1 + sin (x/0.8)). dx

-Integrate from v = 1 m/s @ x = 0:

0.5 (v^2) = - (0.1x - 0.8cos (x/0.8)) - 0.8 + 0.5

0.5v^2 = 0.8cos (x/0.8) - 0.1x - 0.3

- Evaluate @ x = - 1

0.5v^2 = 0.8 cos (-1/0.8) + 0.1 - 0.3

v = sqrt (0.104516)

v = + / - 0.323 m/s

- v = v_max when a = 0:

-0.1 = sin (x/0.8)

x = - 0.8*0.1002

x = - 0.080134 m

- Hence,

v^2 = 1.6 cos (-0.080134/0.8) - 0.6 - 0.2*-0.080134

v = sqrt (0.504)

v = + / - 1.004 m/s