Find the linear speed v for each of the following.

Question

a point on the equator moving due to Earth's rotation, if the radius is 3960 mi

Accepted Answer

Welcome back. I am so glad you're here. We are told a wooden wheel that has a radius of 2 m was spun at a party game. It rotated at two pie radiance P four seconds. Calculate the linear speed V of the point on the edge of the wheel. Our answer choices are answer choice. A two pi meters per second. Answer choice B pi meters per second answer choice, C pi divided by 2 m per second and answer choice D eight pi meters per second. All right. So our linear speed V is given to us, we recall from previous lessons by taking the radius R and multiplying that by the angular speed omega. So what's our R and what is our Omega R? The radius is the distance from the center of the circle to the edge. That is 2 m and our omega our angular speed is the rate at which it travels around the circle. It's our theta divided by t our radiance over time. And here this is given to us in terms of radiance, we have two pie radiance pur four seconds. So now we can just plug in our 2 m for our radius and our two pi divided by four radians per second for our omega and solve for our linear speed V. So we have 2 m multiplied by two pi divided by four radians per second. Now, in the numerator, two multiplied by two pi is four pi in the denominator, we still have that four. And for our units in the numerator, we have meters and radiance. But linear speed is given in terms of distance per unit of time. And so we can disregard the radiance and we'll just have the meters per, in the denominator seconds. We can simplify this one more time. Four pi divided by four is just pi so our linear speed is pi meters per second. We look at our answer choices and this matches with answer choice. B well done. We'll catch you on the next one.

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Find the linear speed v for each of the following.

a point on the equator moving due to Earth's rotation, if the radius is 3960 mi