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The velocity of a particle constrained to move along the x-axis as a function of time t is given by:
v(t)=-(15/t_0) sin(t/t_0). a)If the particle is at x=4 m when t = 0, what is its position at t = 4t_0. You will not need the value of t_0 to solve any part of this problem.
b)Denote instantaneous acceleration of this particle by a(t). Evaluate the expression 6 + v(0) t + a(0) t^2/2 at t = 4t_0.

Respuesta :

shinmin

The answer is:

v = 15/T sin t/T 


then integrate 


x = -15 cos t/T + c 


if x = 4 at t = 0 then 


4 = -15 + c so c = 19 


so 


x = 19 - 15 cos t/T 


at t = 4T 


x = 19 - 15 cos 4 = 3.99 or 4

b) 


a = dv/dt = 15/T^2 cos t/T 


v(0) = 0 since sin 0/T = 0 


a(0) = 15/T^2 


so 


6 + 0 +(7.5/T^2) (16 T^2) 


= 6 + 7.5*16

 

= 126 is the answer