- #1
AFSteph
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Hello! I'm new here and I would just like to make sure my work is correct.
Question as listed in my textbook:
"Go to http://www.fearofphysics.com/Xva/xva.html
Start your first object at 1 m, give it a speed of 2.5 m/s, and an acceleration of 1 m/s squared.
Start your second object at 1 m, give it a speed of 8 m/s, and an acceleration of 0 m/s squared.
Hit "Go!" to see which object wins the race. Which won, the first or the second object?
What factor ended up being most important in the race, speed or acceleration? If the race track was longer, which would have been more important, speed or acceleration? Explain your answer."
Note: You guys don't have to go to the site. It's just a simple flash game. At those settings the second object wins because it's a very short race. (You can just click here to see http://www.fearofphysics.com/cgi-bi...1=1&object2=redtruck&x2=1&v2=8&a2=0&mode=wrap
Formula given by my textbook for finding velocity and distance of an accelerating object in which v_{o} is the initial velocity.
v = v_{o}t + at
d = v_{o}t + \frac{1}{2}at^{2}
In this case speed is more important, as seen by the second object winning the short race presented by the game. However, if the race were longer, let's say it lasted for 100 seconds, the first object would have caught up to the second object 11 seconds in and by the end of the race the first object would have traveled 5250 m while the second would have gone a mere 800 m, losing by 4450 m. The longer the race is, the acceleration causes a greater and greater disparity between the distances traveled by the two objects.
My calculations
Finding the time when the two objects had traveled the same distance. (Solving for t)
d = v_{o}t + \frac{1}{2}at^{2} = r x t
d = 2.5 m/s x t + \frac{1}{2}1 m/s^{2} x t^{2} = 8 m/s x t
d = 2.5t + \frac{t^{2} }{2} = 8t
d = 5t + t^{2} = 16t
t^{2} = 11t
t = 11 seconds
Check: (First object) 2.5m/s x 11 seconds + \frac{1}{2}1 m/s^{2} x 11^{2} = 88 meters traveled
(Second object) 8m/s x 11 seconds = 88 meters traveled
Finding the distances traveled by the two objects in 100 seconds.
First object: 2.5m/s x 100 seconds + \frac{1}{2}1 m/s^{2} x 100^{2} = 250 + 5,000 = 5,250 meters traveled
Second object: 8m/s x 100s = 800 meters traveled
Sorry for the length! It's been a long time since I've done any algebra and I just want to make sure I did this right. I hope it's easy to follow my calculations.
Homework Statement
Question as listed in my textbook:
"Go to http://www.fearofphysics.com/Xva/xva.html
Start your first object at 1 m, give it a speed of 2.5 m/s, and an acceleration of 1 m/s squared.
Start your second object at 1 m, give it a speed of 8 m/s, and an acceleration of 0 m/s squared.
Hit "Go!" to see which object wins the race. Which won, the first or the second object?
What factor ended up being most important in the race, speed or acceleration? If the race track was longer, which would have been more important, speed or acceleration? Explain your answer."
Note: You guys don't have to go to the site. It's just a simple flash game. At those settings the second object wins because it's a very short race. (You can just click here to see http://www.fearofphysics.com/cgi-bi...1=1&object2=redtruck&x2=1&v2=8&a2=0&mode=wrap
Homework Equations
Formula given by my textbook for finding velocity and distance of an accelerating object in which v_{o} is the initial velocity.
v = v_{o}t + at
d = v_{o}t + \frac{1}{2}at^{2}
The Attempt at a Solution
In this case speed is more important, as seen by the second object winning the short race presented by the game. However, if the race were longer, let's say it lasted for 100 seconds, the first object would have caught up to the second object 11 seconds in and by the end of the race the first object would have traveled 5250 m while the second would have gone a mere 800 m, losing by 4450 m. The longer the race is, the acceleration causes a greater and greater disparity between the distances traveled by the two objects.
My calculations
Finding the time when the two objects had traveled the same distance. (Solving for t)
d = v_{o}t + \frac{1}{2}at^{2} = r x t
d = 2.5 m/s x t + \frac{1}{2}1 m/s^{2} x t^{2} = 8 m/s x t
d = 2.5t + \frac{t^{2} }{2} = 8t
d = 5t + t^{2} = 16t
t^{2} = 11t
t = 11 seconds
Check: (First object) 2.5m/s x 11 seconds + \frac{1}{2}1 m/s^{2} x 11^{2} = 88 meters traveled
(Second object) 8m/s x 11 seconds = 88 meters traveled
Finding the distances traveled by the two objects in 100 seconds.
First object: 2.5m/s x 100 seconds + \frac{1}{2}1 m/s^{2} x 100^{2} = 250 + 5,000 = 5,250 meters traveled
Second object: 8m/s x 100s = 800 meters traveled
Sorry for the length! It's been a long time since I've done any algebra and I just want to make sure I did this right. I hope it's easy to follow my calculations.
