# Help double-checking my work on acceleration

1. Oct 13, 2013

### AFSteph

Hello! I'm new here and I would just like to make sure my work is correct.
1. The problem statement, all variables and given/known data
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

2. Relevant 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$^{2}$

3. 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.

2. Oct 14, 2013

### NihalSh

You know the acceleration!!!!.....What are you trying to solve for???....You haven't mentioned anything in your question about finding anything!!!.....The question is just :

If the race is for a short period of time, the initial velocity plays more important role because vaguely speaking acceleration get much time to set in. If the race is for a longer period of time, the acceleration of the object would create a disparity in the distance travelled (ever increasing with time).

You solved for overtaking point (11 s), which clearly gives you the above explanation.

Short Period<11 s
Long Period>11 s

Edit: you can easily calculate borderline length of short track and long track from overtaking point, which you have. It is 88 m.

Last edited: Oct 14, 2013
3. Oct 14, 2013

### AFSteph

@NihalSh
I just wanted to make sure that my calculations were correct. Like I said, I haven't done algebra in a long time and was kind of unsure of myself. So I guess what I wrote is right?
Thank you for talking the time to comment! I really appreciate a second opinion :)

4. Oct 15, 2013

### NihalSh

yes, calculations were right.

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