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.

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

To evaluate the constants A, B, C, and D and write the complete function, we can use the given information about the minimum and maximum velocity.

1. Minimum velocity: At the point (1, -4), the velocity is -4.
- Using the equation v(t) = Asin(B(t - C)) + D, we substitute t = 1 and set v(t) = -4:
-4 = Asin(B(1 - C)) + D

2. Maximum velocity: At the point (4, 2), the velocity is 2.
- Using the equation v(t) = Asin(B(t - C)) + D, we substitute t = 4 and set v(t) = 2:
2 = Asin(B(4 - C)) + D

We now have a system of two equations with four unknowns: A, B, C, and D. To solve for these constants, we need an additional equation. We can consider the phase shift C.

3. Phase shift: The phase shift is the value of t when the sine function reaches its maximum or minimum value.

- The maximum velocity occurs at t = 4, and the minimum velocity occurs at t = 1.
- The phase shift C can be obtained by finding the difference between the values of t at the maximum and minimum points:
- C = 4 - 1 = 3

Now, we have three equations and four unknowns:

1. -4 = Asin(B(1 - C)) + D
2. 2 = Asin(B(4 - C)) + D
3. C = 3

To solve this system, we can substitute the value of C from eq. 3 into eq. 1 and solve for A, B, and D:

-4 = Asin(B(1 - 3)) + D
-4 = Asin(-2B) + D [eq. 4]

Using eq. 4 and eq. 2, we can then find A, B, and D. However, without any specific numeric value for A, B, or D, we cannot determine their exact values. The given information is not sufficient to solve for the values of A, B, C, and D.