Question

A car travels three quarters of the way around a circle of radius $20.0 \mathrm{m}\( in a time of \)3.0 \mathrm{s}$ at a constant speed. The initial velocity is west and the final velocity is south. (a) Find its average velocity for this trip. (b) What is the car's average acceleration during these 3.0 s? (c) Explain how a car moving at constant speed has a nonzero average acceleration.

Short answer

Question: Explain how a car moving at a constant speed can have a nonzero average acceleration and determine the average velocity and average acceleration of the car given a radius of 20.0 m and a time of 3 seconds to travel three-quarters of the way around the circle. Answer: A car moving at a constant speed can have a nonzero average acceleration when there is a change in its direction, such as in a circular path. In this case, there is a centripetal acceleration acting towards the center of the circle, which is nonzero. The average velocity of the car is approximately 5.71 m/s, and the average acceleration is approximately 4.0 m/s².

Step-by-step solution

(a) Finding average velocity

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: $$\vec{v}_{avg} = \frac{|\vec{d}|}{3\,{s}}$$

(a) Conclusion

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

(b) Finding average acceleration

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: $$\vec{a}_{avg} = \frac{|\Delta \vec{v}|}{3\,{s}}$$

(b) Conclusion

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

(c) Explaining nonzero average acceleration

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.