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.

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 

the force represented by vector a is normal force. if the hockey puck is moving to the right, when it is hit with a significant force by the hockey stick the puck will move off in a different direction. correct on edge.