asked 21.8k views
3 votes
The inclination (tilt) of an amusement park ride is accelerating at a rate of 2160\,\dfrac{\text{degrees}}{\text{min}^2}2160 min 2 degrees ​ 2160, start fraction, start text, d, e, g, r, e, e, s, end text, divided by, start text, m, i, n, end text, squared, end fraction. What is the ride's acceleration rate in \dfrac{\text{degrees}}{\text{s}^2} s 2 degrees ​ start fraction, start text, d, e, g, r, e, e, s, end text, divided by, start text, s, end text, squared, end fraction?

2 Answers

4 votes

Answer:

0.6

Explanation:

answered
User Reed Sandberg
by
8.0k points
4 votes

Answer:

0.6 deg/s²

Explanation:

Acceleration is the time rate of change of velocity, it is the ratio of velocity to time. The formula for acceleration is given by:

Acceleration = Change in velocity / time taken = (Final velocity - Initial velocity) / time

Given that the acceleration is 2160 deg/min², we have to convert it to deg/s²

1 min = 60 seconds

1 min² = 60² s² = 3600 s²


2160\ deg/min^2=(2160\ deg)/(1\ min^2*3600\ (s^2)/(min^2) )=0.6\ deg/s^2\\ \\2160\ deg/min^2=0.6\ deg/s^2

answered
User PAC
by
8.1k points
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