The rider of a ferris wheel moves at a rate of 0.18 m/s2. If the rider is 5.0 m from the center of the ferris wheel, What is his/her Velocity?

Can't figure out if the RADIUS and RATE are the same and Which one is the radius/rate?

radius = 5 meters

.18 m/s^2 is probably the centripetal acceleration. I can not imagine why it is called "rate".

acceleration = v^2/R

v^2 = .18 * 5 = .9

v = .95 m/s

I was trying to solve this problem just now for our HW. I got so confused. But thanks to Damon I now understand it very clearly

Omygossh thanks damon!!! :)

What is the velocity of a ferris wheel rider sitting at a distance of 5.0 m from the center of rotation while moving at a rate of 18 m/s²? With solution

Well, isn't this a spinning riddle! Let me put on my clown hat and help you make sense of it.

In this scenario, the radius refers to the distance from the center of the ferris wheel to the rider. So, in this case, the radius is 5.0 m.

Now, the rate refers to the rate at which the rider is moving. In other words, it's the velocity of the rider.

But here's the catch: the problem doesn't directly give you the velocity. It only gives you the rate of change of velocity, which is acceleration.

To find the velocity, you would need to integrate the rate of change of velocity, also known as acceleration, with respect to time. However, since we don't have a time value, we can't calculate the exact velocity.

So, for now, the answer would be that we don't have enough information to determine the velocity. It's like trying to solve a joke without a punchline. But hey, at least we had some fun with physics along the way, right?

To solve this problem, we need to understand the concepts of radius and rate of change separately.

The radius refers to the distance between the center of the Ferris wheel and its outer edge, where the rider is located. In this case, the rider is 5.0 m from the center. So, the given radius is 5.0 m.

The rate of change, often denoted as velocity, refers to how quickly a quantity is changing with respect to time. In this case, the rate of change is given as 0.18 m/s^2. Note that this value represents acceleration (change in velocity) rather than velocity itself.

To find the rider's velocity, we need to integrate the rate of change (acceleration) with respect to time. However, since the problem does not provide any information about the duration or time interval, we cannot directly solve for velocity.

If you have additional information or if there is a specific time frame involved, please provide those details so that we can calculate the velocity accurately.