The equation models the height h in centimeters after t seconds of a weight atta
ID: 3112691 • Letter: T
Question
The equation models the height h in centimeters after t seconds of a weight attached to the end of a spring that has been stretched and then released. H = 7 cos(pi/3 t) a. Solve the equation for t. b. Find the times at which the weight is first at a height of 1 cm, of 3 cm, and of 5 cm above the rest position. Round your answers to the nearest hundredth. c. Find the times at which the weight is at a height of 1 cm, of 3 cm, and of 5 cm below the rest position for the second time. Round your answers to the nearest hundredth.Explanation / Answer
a. The height ( h) (in cm) of a weight attached to the end of a spring , from the rest position, t seconds after the spring is stretched and then released, is given by h = 7cos(t/3). Then cos(t/3) = h/7 so that t/3 = cos- 1(h/7) or, t = (3/) cos-1(h/7) or, t = (3/) arccos(h/7).
b. When h = -1, we have t = (3/)arccos(-1/7) = (3/3.14159265)*(98.21321065)=93.78670782 =93.787 seconds( on rounding off to the nearest hundredth).
When h = -3, we have t = (3/)arccos(-3/7) = (3/3.14159265)*(115.37693349)= 110.1768557 = 110.177 seconds( on rounding off to the nearest hundredth).
When h = -5, we have t = (3/)arccos(-5/7) = (3/3.14159265)*(135.58469138)=129.4738431 = 129.474 seconds( on rounding off to the nearest hundredth).
c. When h = 1, we have t = (3/)arccos(1/7) = (3/3.14159265)*(81.78678935)=78.10063092 = 78.101 seconds( on rounding off to the nearest hundredth).
When h = 3, we have t = (3/)arccos(3/7) = (3/3.14159265)*(64.62306651)=61.71048291 = 61.710 seconds( on rounding off to the nearest hundredth).
When h = 5, we have t = (3/)arccos(5/7) = (3/3.14159265)*(44.41530862)= 42.41349554= 42.413 seconds( on rounding off to the nearest hundredth).
Note: Since the spring is stretched in a direction below the weight, the height h will be positive in a direction below and negative in a direction above the position of rest of the weight.
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