# College Physics help? Much appreciated if helped. The best answer will be rewarded.?

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An amusᴇмᴇɴt park ride rotates around a fixed axis such that the angular position of a point on the ride follows the equation: θ(t) = a + bt^2 – ct^3 where a = 1.1 rad, b = 0.65 rad/s^2 and c = 0.035 rad/s^3.

part a) Determine an equation for the angular speed of the ride as a function of time, ω(t).

part b) Besides at t = 0, at what time t(1) is the ride stopped? Give your answer in seconds.

part c) What is the magnitude of the angular displacement of the ride in radians between times t = 0 and t = t(1)?

part d) Determine an equation for the angular acceleration of the ride as a function of time, α(t).

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part e) What is the angular acceleration in rad/s2 when the ride is at rest at t = t(1)?

• Update: Corrected typo in the works, but it did change the final answers.

part a) Determine an equation for the angular speed of the ride as a function of time, ω(t).

ω = d θ(t)/dt = 2bt – 3ct^2

ω(t) = 2bt – 3ct^2

ω(t) = 2*0.65t – 3*0.035t^2

ω(t) = 1.3t – 0.105t^2

part b) Besides at t = 0, at what time t(1) is the ride stopped? Give your answer in seconds.

This is

ω(t) = 0

ω(t) = 2bt – 3ct^2 =0

t(2b-3ct) = 0

t*(1.3 -0.105t) = 0

1.3-0.105t = 0

0.105t = 1.3

t(1) = 1.3 / 0.105 = 12.38095238 seconds

part c) What is the magnitude of the angular displacement of the ride in radians between times t = 0 and t = t(1)?

θ(t) = a + bt^2 – ct^3

θ(0) = a +b*0 -c *0 = a

θ(t(1) ) =1.1 +0.65t^2 -0.035t^3

θ(12.38095238) = 1.1 +0.65*(12.38095238)^2- (0.035)*(12.38095238)^3

Delta Theta = θ(t(1) ) – θ(0) = 34.31239607 – 1.1 = 33.21239607

part d) Determine an equation for the angular acceleration of the ride as a function of time, α(t).

α(t)= dω(t)/dt = d ( 2bt – 3ct^2) /dt = 2b -6ct

α(t)=2b -6ct = 2*0.65 – 6*0.035t = 1.3 – 0.21t

α(t)= 1.3 – 0.21t

part e) What is the angular acceleration in rad/s2 when the ride is at rest at t = t(1)?

α(t)= 1.3 – 0.21t

α(12.38095238)= 1.3 – 0.21*12.38095238 = -1.3 