Question

the drive mechanism of an oil pump converts circular motion into vertical motion. the vertical position of the top of the oil pump can be found using the equation below, where d is the distance from the top of the oil pump to the ground, r is the radius of the drive mechanism, ω is the velocity of the drive mechanism in degrees per second, and t is time in seconds.
$d = r\sin(\omega t) + 26$ ft
if the drive mechanism has a radius of 3 feet and is operating at a velocity of $60^\circ$ per second, what is the distance from the top of the oil pump to the ground after 4.5 seconds?
a. 28.6 feet
b. 26 feet
c. 29 feet
d. 23 feet

Explanation:

Step1: Substitute the given values into the formula

Given \(r = 3\), \(\omega=60\), and \(t = 4.5\). Substitute into \(d=r\sin(\omega t)+26\). First, calculate \(\omega t\): \(\omega t=60\times4.5 = 270^{\circ}\).

Step2: Calculate the sine value

We know that \(\sin(270^{\circ})=- 1\).

Step3: Calculate the value of \(d\)

Substitute \(\sin(270^{\circ})=-1\) and \(r = 3\) into the formula \(d=r\sin(\omega t)+26\). Then \(d=3\times(-1)+26\).

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Answer:

D. 23 feet