Need help with velocity, acceleration, time and distance

1. Jan 15, 2016

Kenzo333

1. The problem statement, all variables and given/known data

You need to calculate the acceleration (m/s 2 ) of a jet fighter which is launched from
an aircraft-carrier by catapult, given the jet’s required takeoff speed (km/hr) and the
distance available on deck to reach that speed (metres) as the catapult accelerates the
jet from rest to takeoff. Assume constant acceleration. You are also to calculate the
time (secs) for the jet to be accelerated to takeoff speed. You are also required to
report the acceleration in terms of its g-force where g = 9.80665m/s 2 .

The analysis part of this lab is based around 2 simultaneous equations, with unknowns
for acceleration and time given velocity and distance. Solve these equations so that
you can calculate both acceleration and time in terms of distance and velocity. The
easiest way to do this is to solve the first equation for a (acceleration) and then to
substitute that expression into the second equation and then solve it for t (time).

2. Relevant equations

Relevant formulae are (simple Newtonian motion with initial velocity of 0):
v = at
s = ½at 2
where v = velocity, a = acceleration, t = time, s = distance.

3. The attempt at a solution

Hello
I dont understand the question, could someone please explain it to me?
I know the input or given data is velocity and distance
but how to solve everything? Thank you!

2. Jan 15, 2016

Suraj M

Which part didn't you understand?
You're given distance and final velocity, is there a way you can calculate the acceleration? Any equation you know?

3. Jan 15, 2016

Kenzo333

Is this the correct equation? but then how to find the time and where do we get the vf and vi?
Thanks
http://media.wiley.com/Lux/18/329818.image1.png [Broken]

Last edited by a moderator: May 7, 2017
4. Jan 15, 2016

haruspex

The first part of that, $a=\frac{\Delta v}{\Delta t}$, is correct since you are told the acceleration is constant and the initial velocity is zero. But you cannot use velocity=distance/time since that only gives an average velocity. Instead, the second of your two Relevant Equations is appropriate. That is valid for constant acceleration and an initial velocity of zero.
I recommend that you remember the more general form of it, $s=v_it+\frac 12at^2$. That will cover the cases where the initial velocity is not zero.
For the present problem, you know the initial speed, the final speed and the distance; not time. So this equation is not sufficient either.
However, you can combine it with your v=at equation. That gives you a pair of equations each containing unknowns a and t, so you can solve.

Last edited by a moderator: May 7, 2017
5. Jan 16, 2016

Kenzo333

Hi, how to combine it? thank you.
Can you give me step by step equation which one I should do it first? thank you

6. Jan 16, 2016

haruspex

Have you never solved simultaneous equations?

7. Jan 16, 2016

Kenzo333

Yes,
Please help me ><

8. Jan 16, 2016

haruspex

Take the two equations you quoted in the very first post on this thread. Use one of them to express t in terms of other variables, then use that to substitute for t in the other equation. This is how we solve simultaneous equations, isn't it? Please show more effort.

9. Jan 16, 2016

Kenzo333

thank you but "then use that to substitute for t in the other equation"
i dont understand that part
also how do I solve both equation since I dont know the t?
T = distance/velocity right?

10. Jan 16, 2016

haruspex

No, that is not one of your two equations in your original post. I already told you you cannot use velocity=distance/time here because it is not constant velocity.
The equation you do have is t=v/a. This means that wherever you see t in the other equation you can replace it by v/a. That is what is meant by substitution. It is the standard way of solving simultaneous equations.

11. Jan 16, 2016

Kenzo333

Thanks, last question. How do you know that t=v/a ?

12. Jan 16, 2016

haruspex

It's a simple rearrangement of the v=at equation you quoted in post #1.
As I mentioned in post #4, it should really be $\Delta v= at$, i.e. the increase in velocity is acceleration times time. But in this case we know the initial velocity is zero.

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