Question

(20%) Problem 4: The shadow of a pendulum cast on a flat board moves on a straight line. By placing the x-axis on the straight line with the origin at the middle of the total path, the x-coordinate of the shadow is given by the following function: x(t) = 49cos(u), where t is in seconds and x is in centimeters. 425% Part (a) Find the speed, in centimeters per second, of the shadow at t = 1/4 s. VE Grade Summary Deductions 0% Potential 100% 9 HOME JT 7 8 E 14 5 6 sino cotan atan cosh cos tan asino acos acotan sinh tanh cotanh Degrees Radians Submissions Attempts remaining: 999 (% per attempt) detailed view 1 2 3 0 VO BACKSPACE DER + END CLEAR Submit Hint Feedback I give up! Hints: 0% deduction per hint. Hints remaining: 1 Feedback: 1% deduction per feedback. A 25% Part (b) Find the speed, in centimeters per second, of the shadow at t = 1/2 s. A 25% Part (c) Find the magnitude of the acceleration, in meters per second squared, of the shadow's motion at t = 1/3 s. A 25% Part (d) How much distance, in centimeters, does the shadow travel in 20 sec?

Answer 1

v = dx/ dt

v = - 49 $\pi$ sin ($\pi$t)

so,

(a)

at t = 1/4 s

v = - 108.8 cm/s

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(b)

at t = 1/2 s

v = - 154 cm/s

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(c)

a = dv / dt = - 49 $\pi$ ² cos ($\pi$t)

so,

at t = 1/3

a = - 242 cm² / s

or

a = 0.0242 m/s²

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(d)

time period = 2 seconds ( one oscillation)

so,

distance = 10 * 4 * 49

distance = 1960 meters