1. ## help (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?

2. 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.

3. Originally Posted by skeeter
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.

how do you that you would use that equation??

so it would go like this?

360km^2 = 0^2 +2(a)1.91km

360km^2 = 2(a) 1.91km

(360km^2)/2(1.91km)= a

that right?

4. Originally Posted by Legendsn3verdie
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?
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]

5. i lost you from here:

Originally Posted by ticbol
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]

looks like your using another equation i m not aware of

6. Originally Posted by Legendsn3verdie
i lost you from here:

looks like your using another equation i m not aware of
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.

7. Originally Posted by Legendsn3verdie
how do you that you would use that equation??

so it would go like this?

360km^2 = 0^2 +2(a)1.91km

360km^2 = 2(a) 1.91km

(360km^2)/2(1.91km)= a

that right?
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 ...

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?
why the change in values?

8. Originally Posted by skeeter
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 ...

why the change in values?

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

9. 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.

10. Originally Posted by skeeter
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.
ok so did i do this right?

(360km^2/hr^2 = 0^2 +2(a)1.91km

360km^2/hr^2 = 2(a) 1.91km

(360km^2hr^2)/2(1.91km)= a

a = 94.24km/hr^2

11. you forgot to square the 360.

12. (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?