The inclination (tilt) of an amusement park ride is accelerating at a rate of 2160 degrees/min^2. What is the ride's acceleration in degrees/s^2?

[tex]\bf \begin{array}{llll} minutes&seconds\\ ------&------\\ 1m&60s\\\\ (1m)^2&(60s)^2\\ m^2&3600s^2\\\\ (1m)^3&(60s)^3\\ m^3&216000s^3 \end{array}\\\\ -------------------------------\\\\ \cfrac{2160^o}{\underline{m^2}}\cdot \cfrac{\underline{m^2}}{3600~s^2}\implies \cfrac{2160^o}{3600~s^2}\implies \cfrac{3^o}{5~s^2}\implies 0.6\frac{deg}{s^2}[/tex]

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DelcieRiveria DelcieRiveria · Answer 2 · 2018-10-09T10:08:05+02:00

Solution: The acceleration of ride in degrees per second square is 0.6 degree/[tex]s^2[/tex].

Explanation:

It is given that the inclination of the amusement park ride is accelerating at a rate of 2160 degree/[tex]s^2[/tex].

Since we know that 1 minute = 60 seconds.

[tex]1 min^2=3600sec^2[/tex]

[tex]3600^{\circ}/min^2=1^{\circ}/sec^2[/tex]

[tex]1^{\circ}/min^2=\frac{1}{3600}^{\circ}/sec^2[/tex]

[tex]2160^{\circ}/min^2=\frac{2160}{3600}^{\circ}/sec^2\\[/tex]

[tex]2160^{\circ}/min^2=0.6^{\circ}/sec^2[/tex]

Therefore, the The acceleration of ride in degrees per second square is 0.6 degree/[tex]s^2[/tex].