the height, d, in feet of a ball suspended from a spring as a function of time, t, in seconds can be modeled…

the height, d, in feet of a ball suspended from a spring as a function of time, t, in seconds can be modeled by the equation (d = - 2sinleft(pileft(t+\frac{1}{2}\right)\right)+5). the ball is released from its lowest - point at (t = 0) seconds. using your knowledge of the general form of sine and cosine functions, which of the following equations can also model this situation?\n(d=-2cos(pi t)+5)\n(d = - 2cosleft(pileft(t+\frac{1}{2}\right)\right)+5)\n(d = 2cos(pi t)+5)\n(d = 2cosleft(pileft(t+\frac{1}{2}\right)\right)+5)

the height, d, in feet of a ball suspended from a spring as a function of time, t, in seconds can be modeled by the equation (d = - 2sinleft(pileft(t+\frac{1}{2}\right)\right)+5). the ball is released from its lowest - point at (t = 0) seconds. using your knowledge of the general form of sine and cosine functions, which of the following equations can also model this situation?\n(d=-2cos(pi t)+5)\n(d = - 2cosleft(pileft(t+\frac{1}{2}\right)\right)+5)\n(d = 2cos(pi t)+5)\n(d = 2cosleft(pileft(t+\frac{1}{2}\right)\right)+5)

Answer

Explanation:

Step1: Recall the phase - shift relationship between sine and cosine

The general relationship is $\sin(A)=\cos(A - \frac{\pi}{2})$. Given the equation $d=- 2\sin(\pi(t+\frac{1}{2})) + 5$.

Step2: Apply the phase - shift formula

We know that $\sin(\pi(t+\frac{1}{2}))=\cos(\pi(t+\frac{1}{2})-\frac{\pi}{2})$. Simplify $\pi(t+\frac{1}{2})-\frac{\pi}{2}=\pi t+\frac{\pi}{2}-\frac{\pi}{2}=\pi t$. So, $-2\sin(\pi(t + \frac{1}{2}))+5=-2\cos(\pi t)+5$.

Answer:

$d = - 2\cos(\pi t)+5$