Context: The function v(t) represents the velocity of a particle moving along a horizontal line at any time, t, greater than or equal to zero. If the velocity is positive, the particle moves to the right. If the velocity is negative, the particle is moving to the left.

The question: A particle has a velocity function v(t)=Asin(B(t-C))+D, with the minimum velocity at the point (1, -4) and the maximum velocity at the point (4, 2). Evaluate the constants A,B, C, and D and write the complete function.

3 answers

v = A sin(B(t-C))+D
max v = D+A = 2
min v = D-A = -4
So, D = -1 and A = 3

v = 3sin(B(t-C))-1

max of sinθ occurs at θ = π/2
max of sinθ occurs at θ = 3π/2

so, see what you can do with that, the way I git A and D above.

Recall that the period of sin(kt) is 2π/k
What I ended up getting was
v(t)=3 sin( pi/3 (t- ((4-pi/2)))) -1

so B= pi/3 and C=4–pi/2
I'll have to see where you went wrong. Too bad you didn't show your work, like me. wolframalpha says you are off:

http://www.wolframalpha.com/input/?i=3+sin(+pi%2F3+(t-+((4-pi%2F2))))+-1