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?
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.
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.
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.
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?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?
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