# SIMPLE QUESTION, NEED HELP!

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1. Aug 13, 2015

### Infinty

• Warning: Thread titles should be descriptive of the thread content, not a plea for help.
1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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)2

1720 M

..that is not what I got.

2. Aug 13, 2015

### 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.

3. Aug 13, 2015

### billy_joule

4. Aug 13, 2015

### Infinty

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.

5. Aug 13, 2015

### Infinty

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?

6. Aug 13, 2015

### e.bar.goum

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

7. Aug 13, 2015

### Infinty

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

8. Aug 13, 2015

### e.bar.goum

Sure. What do you get?

9. Aug 13, 2015

### Infinty

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. :)

10. Aug 13, 2015

### 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 !!

11. Aug 13, 2015

### e.bar.goum

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