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

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(2b3ct) = 0
t*(1.3 0.105t) = 0
1.30.105t = 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(0) = 1.1 radians
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
θ(12.38095238) = 34.31239607 radians
Delta Theta = θ(t(1) ) – θ(0) = 34.31239607 – 1.1 = 33.21239607
Delta Theta =33.21239607 radians
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