Question

Starting from rest, a bicyclist travels around a horizontalcircular path of radius rho = 10m at a speed v = 0.09 t² + 0.1t. Determine the magnitudes of his velocity andacceleration when he has traveled a distance s =3m.

Answer 1

The velocity of a bicyclist traveling around a circular path, calculated by differentiating the distance function, is 0.64 m/s at t = 3 seconds. The acceleration, obtained by differentiating the velocity function, is 0.18 m/s². These values signify the cyclist's speed and acceleration along the circular path.

Step-by-step explanation:

The velocity of a bicyclist traveling around a circular path can be determined by finding the derivative of the distance traveled concerning time. In this case, the distance traveled, s, is given as 3m. To find the velocity, v, we can differentiate the equation v = 0.09t² + 0.1t concerning t. This gives us v = 0.18t + 0.1. Substituting t = 3 into the equation, we get v = 0.18(3) + 0.1 = 0.64 m/s. So the magnitude of the velocity is 0.64 m/s. The acceleration of the bicyclist can be determined by finding the derivative of the velocity with respect to time. Differentiating v = 0.18t + 0.1 concerning t gives us a = 0.18. So the magnitude of the acceleration is 0.18 m/s².