Changing speed (non-uniform circular motion)

Question

A cart is going around a circular track. The cart is located at a position. Which is defined by the angle theta, 28 degrees above horizontal.The radius of the track is 2.5 m. At this time the cart is speeding up at a rate of 2.1 m/s². What is the speed (v₄) of the cart at this time?

Answer 1

Hey Kevin, I am going to make some basic assumptions from your question for the sake of coming up with a solution to what I think your question is:

From what is given we know that we have a circular track with a radius of 2.5 meters.

We know that we have a tangential (linear) acceleration of 2.1 m/s²

And it looks like we would like to know the linear velocity at time t=4, since a V was used and not the common w for angular velocity.

With the assumptions that were made I will proceed:

a = acceleration

v= linear velocity

t = time

acceleration can be defined as the rate of change of velocity i.e., a=v/t

therefore v= at

given that we have an acceleration of 2.1 m/s² and at time t =4

we have V(4) = 2.1 m/(seconds)^{2 *} 4(seconds) = 8.4 m/second