Respuesta :
Answer:
a) v = +/- 0.515 m/s
b) x = -0.098164 m
c) v = +/- 1.005 m/s
Explanation:
Given:
The relationship for the acceleration is given as follows:
a = - (0.1 + sin(x/b))
Where, b = 0.98
- IVP is 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/0.98))
- Separate variables:
v*dv = - (0.1 + sin(x/0.98)) . dx
-Integrate from v = 1 m/s to v and @ x = 0 to x:
0.5*(v^2) = - (0.1*x - 0.98*cos(x/0.8)) - 0.98 + 0.5
0.5*v^2 = 0.98*cos(x/0.98) - 0.1*x - 0.48
- Evaluate at, x = -1
0.5*v^2 = 0.98 cos(-1/0.98) + 0.1 -0.48
v = sqrt (0.2651155)
v = +/- 0.515 m/s
- v = v_max when a = 0. Set the given expression to zero and solve for x:
-0.1 = sin(x/0.98)
x = -0.98*0.1002
x = -0.098164 m
- Hence now evaluate velocity through the derived expression:
v^2 = 1.96 cos(-0.098164/0.98) -0.96 -0.2*-0.098164
v = sqrt (1.0098)
v = +/- 1.005 m/s
