Calculating Distance Traveled: Airplane Acceleration Equation Explained

[Infinty]

Homework Statement

An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.

Homework Equations

The Attempt at a Solution

Okay, so i get how to write it, and I understand what the majority of the equation means. But when I do the actual math, the answer is totally effed up. Haha. How am I calculating wrong? Could someone please break down the calculations part for me so I can better understand this kind of question?

d = vi*t + 0.5*a*t2

d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s2)*(32.8 s)2The answer in the answer key says it is:

1720 M

..that is not what I got.
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[e.bar.goum]

What did you get?

Are you aware that "2" at the end of your equation should be squared?

that is,

$d = v_i t + 0.5 a t^2$

I get the same answer as the key using that equation.

 

[billy_joule]

It's t squared.

d = vi*t + 0.5*a*t²
not
d = vi*t + 0.5*a*t2

Pasting your work into wolfram and adding the missing caret gives the correct answer:

http://www.wolframalpha.com/input/?i=d+=+(0+m/s)*(32.8+s)++0.5*(3.20+m/s2)*(32.8+s)^2

 

[Infinty]

What did you get?

Are you aware that "2" at the end of your equation should be squared?

that is,

$d = v_i t + 0.5 a t^2$

I get the same answer as the key using that equation.

Ahh yes. I am aware is should be squared. I'm completely new to physics and am struggling a bit. Is the first part of the equation: d = (0 m/s)*(32.8 s)
not equal to zero since you are multiplying 32.8 by zero? From there I'm just..sort of lost. I apologize if this seem ignorant and basic.

 

[Infinty]

It's t squared.

d = vi*t + 0.5*a*t²
not
d = vi*t + 0.5*a*t2

Pasting your work into wolfram and adding the missing caret gives the correct answer:

http://www.wolframalpha.com/input/?i=d+=+(0+m/s)*(32.8+s)++0.5*(3.20+m/s2)*(32.8+s)^2

Haha my ignorance is killing me. I guess I'm just not sure how to go about the math. I'm assuming the first part d = (0 m/s)*(32.8 s) is equal to zero since you are multiplying it...and then what?

 

[e.bar.goum]

Ahh yes. I am aware is should be squared. I'm completely new to physics and am struggling a bit. Is the first part of the equation: d = (0 m/s)*(32.8 s)
not equal to zero since you are multiplying 32.8 by zero? From there I'm just..sort of lost. I apologize if this seem ignorant and basic.

Yep! That part is absolutely equal to zero. What about 0.5*(3.20 m/s2)*(32.8 s)^2 ?

 

[Infinty]

Yep! That part is absolutely equal to zero. What about 0.5*(3.20 m/s2)*(32.8 s)^2 ?

So then I assume I take half of 3.20 and multiply it by 32.8 squared?

 

[e.bar.goum]

So then I assume I take half of 3.20 and multiply it by 32.8 squared?

Sure. What do you get?

 

[Infinty]

Sure. What do you get?

By George, I think I've got it!

I'm not even sure what I was doing wrong now before! Hahah. 1721.344

Ahh thank you for humoring me on this...I think perhaps I wasn't even squaring at the end. Couldn't even tell you now. :)

 

[Noctisdark]

Since there is no initial velocity, $x = \frac{at^2}{2}$, you know what the acceleration is, the time is also given and $\frac{1}{2}$ is $\frac{1}{2}$, work it out !

 

[e.bar.goum]

By George, I think I've got it!

I'm not even sure what I was doing wrong now before! Hahah. 1721.344

Ahh thank you for humoring me on this...I think perhaps I wasn't even squaring at the end. Couldn't even tell you now. :)

Great! And no worries, we've all been there before!