Part A Learning Goal: To be able to calculate position, velocity, and acceleration of an object in curvilinear motion using a rectangular coordinate system. An object's motion can be described along a path represented by a fixed x, y, z coordinate system. In such a system, the position vector, r, is described as r=zi+j+zk The object's velocity, V, can be found by taking the first time derivative of the position vector,: v = dr = vzi+vyj + vzk The object's acceleration, a, can be found by taking the time derivative of the velocity, V: = azi+ayj+a₂k A car drives on a curved road that goes down a hill. The car's position is defined by the position vector, r = { [30.0 cos(t)] + [30.0 sin(t)] j-(At)k} ftwhere A== 25.0ft/s. The image below shows the system projected onto the x-y plane. What are the car's velocity and acceleration vectors at this position? Draw your vectors starting at the black dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded and the trigonometric function arguments are in radians. ▸ View Available Hint(s) Part B No elements selected Submit Previous Answers ✓ Correct What is the magnitude, 1, of the car's velocity. v, at t = 1.00 s? Express your answer numerically in feet per second to three significant figures. ▸ View Available Hint(s) v = 26.7 ft/s Submit Previous Answers ✓ Correct Part C What is the magnitude, a, of the car's acceleration, a, at t = 1.00 s? Express your answer numerically in feet per second squared to three significant figures. ▸ View Available Hint(s) a= Submit ΜΕ ΑΣΦ Η vec 0 12 ? Provide Feedback ft/s² P -30 ft Next >
Part A Learning Goal: To be able to calculate position, velocity, and acceleration of an object in curvilinear motion using a rectangular coordinate system. An object's motion can be described along a path represented by a fixed x, y, z coordinate system. In such a system, the position vector, r, is described as r=zi+j+zk The object's velocity, V, can be found by taking the first time derivative of the position vector,: v = dr = vzi+vyj + vzk The object's acceleration, a, can be found by taking the time derivative of the velocity, V: = azi+ayj+a₂k A car drives on a curved road that goes down a hill. The car's position is defined by the position vector, r = { [30.0 cos(t)] + [30.0 sin(t)] j-(At)k} ftwhere A== 25.0ft/s. The image below shows the system projected onto the x-y plane. What are the car's velocity and acceleration vectors at this position? Draw your vectors starting at the black dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded and the trigonometric function arguments are in radians. ▸ View Available Hint(s) Part B No elements selected Submit Previous Answers ✓ Correct What is the magnitude, 1, of the car's velocity. v, at t = 1.00 s? Express your answer numerically in feet per second to three significant figures. ▸ View Available Hint(s) v = 26.7 ft/s Submit Previous Answers ✓ Correct Part C What is the magnitude, a, of the car's acceleration, a, at t = 1.00 s? Express your answer numerically in feet per second squared to three significant figures. ▸ View Available Hint(s) a= Submit ΜΕ ΑΣΦ Η vec 0 12 ? Provide Feedback ft/s² P -30 ft Next >
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter2: Motion In One Dimension
Section: Chapter Questions
Problem 30P: A speedboat increases its speed uniformly from vi = 20.0 m/s to Vf = 30.0 m/s in a distance of 2.00 ...
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