A jumbo jet must reach a speed of 365 km/h on the runway for takeoff. What is the least constant acceleration needed for takeoff from a 2.05 km runway?

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- Sep 11th 2008, 04:35 PMLegendsn3verdiehelp (acceleration)
A jumbo jet must reach a speed of 365 km/h on the runway for takeoff. What is the least constant acceleration needed for takeoff from a 2.05 km runway?

- Sep 11th 2008, 04:54 PMskeeter
use the kinematics equation ...

$\displaystyle v_f^2 = v_0^2 + 2a(\Delta x)$

rearranged ...

$\displaystyle a = \frac{v_f^2 - v_0^2}{2(\Delta x)}$

you might want to convert to meters and seconds unless a solution in $\displaystyle \frac{km}{hr^2}$ is o.k. - Sep 11th 2008, 04:59 PMLegendsn3verdie
- Sep 11th 2008, 05:02 PMticbol
The airplane starts from rest, so initial velocity Vo = 0

V = Vo +at

365 = 0 +at

a*t = 365 -----------(i)

We don't know a, we don't know t.

But we know the distance to be travelled to get to V = 365 km/h.

d = (average velocity)*t

ave V = (1/2)(Vo +V) = (1/2)(0 +at) = at/2

So,

d = (at/2)*t = (a*t^2)/2

We know d, and we can get t in terms of a from (i), so, substitutions,

2.05 = (a*(365/a)^2)/2

4.10 = (365^2)/a

a = (365^2) / 4.10 = 32,494 km/hr/hr

In meters/sec/sec, that would be

(32,494 km/(hr)^2)*(1000m/1km)*[(1hr)^2 / (3600sec)^2]

= 2.50725 m/sec/sec ------------------answer. - Sep 11th 2008, 05:13 PMLegendsn3verdie
- Sep 11th 2008, 05:20 PMticbol
(Happy) It is called, er, "building up".

Kidding aside, it is basic Physics of uniform linear motion. It is starting from scratcth, building up.

Oh, I lost you from the 365 = at?

Or, was it from the distance = (ave velocity)*time?

Uh, sorry, but I do not memorize too many formulas. I always rely on the basic ones only and then I build up or derive the other formulas from those basics only. I am very poor at memorization. - Sep 11th 2008, 05:22 PMskeeter
that would be almost correct ... the numerator has units km^2/hr^2, so your solution would be in units of km/hr^2.

**But**, this is your original post ...

Quote:

A jumbo jet must reach a speed of**365 km/h**on the runway for takeoff. What is the least constant acceleration needed for takeoff from a**2.05 km**runway?

- Sep 11th 2008, 05:32 PMLegendsn3verdie

because i m doing the problems in the book.. i dont want to know the answer.. i kinda want to get the answer myself, but what does it matter if i see all the steps? .. so i almost had the answeR? all i had to do was convert? i m clueless on how to convert km^2/hr^2 to km/hr^2 - Sep 11th 2008, 05:36 PMskeeter
there was no conversion involved ... when you square the velocity in units of km/hr in the numerator, you get units of km^2/hr^2.

km^2/hr^2 divided by km is km/hr^2, a valid (but not often used) unit for acceleration. - Sep 11th 2008, 05:41 PMLegendsn3verdie
- Sep 11th 2008, 05:53 PMskeeter
you forgot to square the 360.

- Sep 11th 2008, 05:59 PMLegendsn3verdie
(360km^2/hr^2 = 0^2 +2(a)1.91km

12960km^2/hr^2 = 2(a) 1.91km

(12960km^2hr^2)/2(1.91km)= a

a = 33926.70 km/hr^2

.. u sure u need to square the 360?