Solution Found!
As a function of time, the velocity of the football
Chapter 3, Problem 69GP(choose chapter or problem)
As a function of time, the velocity of the football described in Problem 68 can be written as \(\mathrm {\vec{v}=(16.6\ m/s)\hat{x}-[(9.81\ m/s^2)}t]\mathrm {\hat{y}}\). Calculate the average acceleration vector of the football for the time periods (a) \(t=0\) to \(t=1.00\ \mathrm s\), (b) \(t=0\) to \(t=2.50\ \mathrm s\), and (c) \(t=0\) to \(t=5.00\ \mathrm s\). (If the acceleration of an object is constant, its average acceleration is the same for all time periods.)
Equation Transcription:
Text Transcription:
vec{v}=(16.6 m/s)hat{x}-[(9.81 m/s^2)t]hat{y}
t=0
t=1.00 s
t=0
t=2.50 s
t=0
t=5.00 s
Questions & Answers
QUESTION:
As a function of time, the velocity of the football described in Problem 68 can be written as \(\mathrm {\vec{v}=(16.6\ m/s)\hat{x}-[(9.81\ m/s^2)}t]\mathrm {\hat{y}}\). Calculate the average acceleration vector of the football for the time periods (a) \(t=0\) to \(t=1.00\ \mathrm s\), (b) \(t=0\) to \(t=2.50\ \mathrm s\), and (c) \(t=0\) to \(t=5.00\ \mathrm s\). (If the acceleration of an object is constant, its average acceleration is the same for all time periods.)
Equation Transcription:
Text Transcription:
vec{v}=(16.6 m/s)hat{x}-[(9.81 m/s^2)t]hat{y}
t=0
t=1.00 s
t=0
t=2.50 s
t=0
t=5.00 s
ANSWER:
Step 1 of 4
Calculate the average acceleration of the football for given time range.