# Homework Help: Distance and question

1. Sep 10, 2011

### emaviator

Hi,

I am struggling to find the answer to calculate what should be a simple question. Maybe someone on here can shed some light for me.

Q. What distance would be needed for an aeroplane with a mass of 50 000kg to accelerate from stationary to 77m/s, prior to leaving the runway surface? Assuming the aeroplane has a constant acceleration of 1.2m/s.

a.2470m
b.779m
c.4941m
d.1540m

Could you please show how to calculate this?

Thanks

2. Sep 10, 2011

### kuruman

Hi emaviator and welcome to PF. Please follow the rules of this forum and use the template when you seek help with homework. Show us the relevant equations and tell us what you tried and what you think about the problem. We just don't give answers away.

3. Sep 11, 2011

### emaviator

Hi Kuruman thanks. Its not a homework question its more revision. I already have the answer of a but I just dont understand how to get there and have looked through my notes and on the net for answers but I just can't find how to calculate this.

I tried f=m*a (in this question 50000*1.2 giving a net force of 60000N)

Then dividing the net force by the acceleration required (60000/77=779.2) rounded back to give answer b but i know this isn't the correct answer and can't see where im going wrong.

I just wanted to know how to do this calculation not the answer..

thanks =)

4. Sep 11, 2011

### cepheid

Staff Emeritus
The mass is a bit of a red herring here, because you have already been given the acceleration. A discussion of of forces (dynamics) is not needed. You can solve the problem using basic kinematics. Have a look at the kinematics equations for constant acceleration. Also have a look at the quantities you've been given: initial speed, final speed, and acceleration. Can you see an equation that will let you solve for distance given those things?

5. Sep 11, 2011

### emaviator

Hi cepheid, thanks for your reply it was just what i was looking for! I am now alot closer to that answer than before but it doesn't seem perfect. I have looked at the kinematic equations and found the equation for displacement: d = vi*t + 1/2*a*t2 (squared). Also given the constant acceleration of 1.2m/s accelerating to 77m/s woulkd take 64.2 seconds based on 77/1.2.

So the answer for my question becomes 0(accelerating from stationary)*64.2=0 + 1/2*1.2*64.2squared = 2472.984 giving the closest answer as a.

Are these calculations correct as even though I have got to the correct answer it may just be luck :)

Thanks again

6. Sep 11, 2011

### cepheid

Staff Emeritus

Yeah, those calcs look fine. If you keep all digits (e.g. 64.166666667 s rather than 64.2 s) in intermediate calculation steps and only round the final answer, you will find that you will avoid rounding error, and your final answer will be closer to the one given in the multiple choice question.

HOWEVER, those two equations would not have been my first choice, since both involve the time variable, and you haven't been given any information about the time interval in this problem (which means you needed to use one equation to solve for time and then a second one to solve for distance). You could have arrived at the answer more directly by using a third kinematic equation. This equation has been produced by using the first two equations together to eliminate the time variable already, resulting in an equation that expresses the relation between only the initial and final velocities, acceleration, and distance.

Last edited: Sep 11, 2011
7. Sep 11, 2011

### emaviator

Thanks for your help with that, was one that i was really struggling to get my head around. I have looked at the kinematic equations and can't see how to use the third equation without using the time. Both of the equations I have looked at for finding the displacement involve a time variable :s

8. Sep 11, 2011

### cepheid

Staff Emeritus
I was referring to this equation:

vf² = vi² + 2ad​

vf - final velocity

vi - initial velocity

a - acceleration

d - distance travelled

This equation can be derived from the other two that you posted. Just use the first one (v = v0 + at) to solve for t and then plug that expression for t into the second equation (d = v0t + ½at²). The result will be the equation above. In this way, time has been eliminated as a variable. EDIT: I'm not saying you have to derive it. Obviously you can just go ahead and use it. I was just trying to explain where it comes from.

Last edited: Sep 11, 2011
9. Sep 12, 2011

### emaviator

Ahhh I see how to use that equation to get the answer for displacement without the need for knowing the time now.
Thanks again for your help with that, it helped alot with my understanding of the caculations necessary.

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