Solution Found!
Answer: A dog running in an open field has components of
Chapter 3, Problem 6E(choose chapter or problem)
Problem 6E
A dog running in an open field has components of velocity vx = 2.6 m/s and vy = -1.8 m/s at t1 = 10.0 s. For the time interval from t1 = 10.0 s to t2 = 20.0 s, the average acceleration of the dog has magnitude 0.45 m/s2 and direction 31.0o measured from the + x-axis toward the + y-axis. At t2 = 20.0 s, (a) what are the x- and y-components of the dog’s velocity? (b) What are the magnitude and direction of the dog’s velocity? (c) Sketch the velocity vectors at t1 and t2. How do these two vectors differ?
Questions & Answers
QUESTION:
Problem 6E
A dog running in an open field has components of velocity vx = 2.6 m/s and vy = -1.8 m/s at t1 = 10.0 s. For the time interval from t1 = 10.0 s to t2 = 20.0 s, the average acceleration of the dog has magnitude 0.45 m/s2 and direction 31.0o measured from the + x-axis toward the + y-axis. At t2 = 20.0 s, (a) what are the x- and y-components of the dog’s velocity? (b) What are the magnitude and direction of the dog’s velocity? (c) Sketch the velocity vectors at t1 and t2. How do these two vectors differ?
ANSWER:
Solution 6E
Step 1
(a) Initially at the x component of the velocity is , and y component of the velocity is .
In the interval between to the magnitude of the average acceleration is and the angle is .
So in this interval the x component is
And the y component of the acceleration is .
And in this interval of time the change in time is .
So the x component of the velocity at at is
And the y component of the velocity is