91Ó°ÊÓ

Since the car is traveling in a circle, we need to determine the east-west and north-south change in position to find the displacement. In this problem, the car has traveled three-quarters of the way around the circle. The east-west displacement can be found from the change in position: The car starts facing west and moves south, hence the east-west displacement is \(20\,\text{m} \times (1 - \cos(\frac{3}{4}\cdot 2\pi))\). Similarly, the north-south displacement can be found from the change in position of the other half of the circle: From north to south, hence the north-south displacement is \(20\,\text{m}\times\sin(\frac{3}{4}\cdot 2\pi)\). The displacement is a vector, so we need to combine the east-west and north-south displacements: $$\vec{d} = (20\,\text{m} \times (1 - \cos(\frac{3}{4}\cdot 2\pi)), 20\,\text{m}\times\sin(\frac{3}{4}\cdot 2\pi))$$ Now we can find the magnitude of the net displacement: $$|\vec{d}| = \sqrt{(20\,\text{m} \times (1 - \cos(\frac{3}{4}\cdot 2\pi)))^2+ (20\,\text{m}\times\sin(\frac{3}{4}\cdot 2\pi))^2}$$ Finally, we can find the average velocity by dividing the displacement by the time elapsed: $$\vec{v}_{avg} = \frac{|\vec{d}|}{3\,{s}}$$

After computing the values, we find that the average velocity is \(\approx$$5.71\,\text{m/s}\).

To find the average acceleration, we first need to find the initial and final velocities. From the given data we can have: 1. Initial velocity: \(v_i\) = (speed, 0) 2. Final velocity: \(v_f\) = (0, -speed) As we calculated in part (a), the average velocity magnitude is \(\approx$$5.71\,\text{m/s}\). Now we find the change in velocity: $$\Delta \vec{v} = \vec{v}_f - \vec{v}_i = (0, -speed) - (speed, 0)$$ The magnitude of the change in velocity can be calculated as: $$|\Delta \vec{v}| = \sqrt{(0 - speed)^2 + (-speed - 0)^2}$$ Now we can find the average acceleration by dividing the change in velocity by the time elapsed: $$\vec{a}_{avg} = \frac{|\Delta \vec{v}|}{3\,{s}}$$

After computing the values, we find that the average acceleration is \(\approx$$4.0\,\text{m/s²}\).

A car moving at a constant speed can have a nonzero average acceleration when there is a change in its direction. In this case, the car is moving around a circle, so the direction of its velocity changes at each point in the circular path. When the direction of velocity changes, there is a centripetal acceleration acting towards the center of the circle, given by: $$a_c = \frac{v^2}{r}$$ Where \(v\) is the speed, and \(r\) is the radius of the circle. Thus, although the car maintains a constant speed, there is a nonzero acceleration due to the change in direction.