Concept explainers
Videos
(a)
An example of motion in which the magnitude of the instantaneous velocity is always larger than the average velocity.
(a)
Answer to Problem 1Q
Circular motion is an example of motion in which the magnitude of the instantaneous velocity is always larger than the average velocity.
Explanation of Solution
Instantaneous velocity is the rate of change of position with respect to time. Whereas the average velocity is the displacement divided by the corresponding time. When an object performs a circular motion, the instantaneous velocity is always greater than the average velocity. This is because the average velocity over any complete cycle is zero since the displacement is zero. But instantaneous velocity assumes some finite value at any given time.
It is obeyed for any object which is accelerating continuously in the direction of velocity.
Conclusion:
Therefore, Circular motion is an example of motion in which the magnitude of the instantaneous velocity is always larger than the average velocity.
(b)
Example of motion having instantaneous velocity is never parallel to the instantaneous acceleration.
(b)
Answer to Problem 1Q
In uniform circular motion, the instantaneous velocity is always perpendicular to the instantaneous acceleration.
Explanation of Solution
During uniform circular motion, the object is moving with constant speed, as the object moves in a circle the change in direction is also conatnt.it is an example of motion having instantaneous velocity is never parallel to the instantaneous acceleration.
Conclusion:
Therefore, in uniform circular motion, the instantaneous velocity is always perpendicular to the instantaneous acceleration.
Want to see more full solutions like this?
Chapter 5 Solutions
College Physics, Volume 1
- How can i solve this if n1 (refractive index of gas) and n2 (refractive index of plastic) is not known. And the brewsters angle isn't knownarrow_forward2. Consider the situation described in problem 1 where light emerges horizontally from ground level. Take k = 0.0020 m' and no = 1.0001 and find at which horizontal distance, x, the ray reaches a height of y = 1.5 m.arrow_forward2-3. Consider the situation of the reflection of a pulse at the interface of two string described in the previous problem. In addition to the net disturbances being equal at the junction, the slope of the net disturbances must also be equal at the junction at all times. Given that p1 = 4.0 g/m, H2 = 9.0 g/m and Aj = 0.50 cm find 2. A, (Answer: -0.10 cm) and 3. Ay. (Answer: 0.40 cm)please I need to show all work step by step problems 2 and 3arrow_forward
- look at answer show all work step by steparrow_forwardLook at the answer and please show all work step by steparrow_forward3. As a woman, who's eyes are h = 1.5 m above the ground, looks down the road sees a tree with height H = 9.0 m. Below the tree is what appears to be a reflection of the tree. The observation of this apparent reflection gives the illusion of water on the roadway. This effect is commonly called a mirage. Use the results of questions 1 and 2 and the principle of ray reversibility to analyze the diagram below. Assume that light leaving the top of the tree bends toward the horizontal until it just grazes ground level. After that, the ray bends upward eventually reaching the woman's eyes. The woman interprets this incoming light as if it came from an image of the tree. Determine the size, H', of the image. (Answer 8.8 m) please show all work step by steparrow_forward
Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning