Launching aircraft by steam

By the Equation of motion, $v^2 = u^2 + 2as$

where $v$ is the final velocity , $u$ is the initial velocity , $a$ is the acceleration and $s$ is the displacement .To get the minimum acceleration we are considering the greatest displacement that is $95$ m . As the jet starts from rest $u = 0$ .

$\displaystyle a = \frac{v^2 - u^2}{2s}$ ,by the equation

$\displaystyle =\frac{80^2 - 0^2}{2 \times 95}$

$=\boxed{33.684}$

$d=ut+\frac{1}{2}at^{2}$

$v=u+at$

Plugging in numbers we get

$95=\frac{1}{2}at^{2}$

$190=at^{2}$

$80=at$

Solving the system we get $t=2.375$

Plugging it back into the equations gives us $80=a(2.375)$

$a=\frac{80}{2.375}=\boxed{33.68}$

By third law of motion we know - v²-u²=2 a s.

So the final velocity of the aircraft to be attained is 80m/s. And it starts from rest so initial velocity is zero.The distance to be travelled is 95m.Onapplying this equation we get that the acceleration required must be33.68m/s²

Acceleration is uniform, so, we can use the equation of uniform accelerated bodies, i.e $2aS = (v_{f})^{2} - (v_{i})^{2}$ ,
Where,
$a$ is the uniform acceleration (which is required in this problem)
$v_{f}$ is the final velocity which is attained by the body (in this case the necessary takeoff velocity)
$v_{i}$ is the initial velocity of the body (in this case is zero because the jet is initially at rest), and
$S$ is the distance covered during the motion (in this case is the length of the stream catapult)
rearranging the formula we get:
$a = \frac{(v_{f})^{2}}{2S}$
substituting values in the formula we have:
$a = \frac{80^{2}}{2*95}$
Therefore, $a = \boxed{33.68}$

v = 80m/s, s = 95m, a =? using relation: a = v^2 / 2 *S putting the values, we get a = 33.68

We know that the distance is equal to the initial velocity times t, plus the integration on acceleration in terms of t which gives the velocity at each point t... We know this distance to be 95 m and the initial velocity to be 0 so...

95 = 1/2a(t^2)

We also know that a = (vf - vi)/t. The final velocity of the jet must be at least 80m/s so a = 80/t

Substituting we get...

95 = 1/2(80/t)(t^2)

190 = 80t^2/t

190 = 80t

t = 2.375

Since a = 80/t

a = 80/2.375 = 33.68m/s^2

use the v^2=u^2+2as (80)^2=u^2+2a95 a=33.68approx=34

v^2 - u^2 = 2as

v=final velocity = 80 m/s

u=initial velocity = 0 m/s

s=distance=95m

Therefore, a=acceleration=(6400/(2*95))=3200/95

as we know

$v^{2}$ = $u^{2}$ + 2 a s

here v=80m/s ; $v^{2}$ =1600 $(m/s)^{2}$

u=0 [time of taking off u of aircraft is fixed ]

s=95m

we get 1600\ $(m/s)^{2}$ = 2 a 95m

a= 33.68

using the third equation of motion ie is squareof final velocity is equal to intial velocity square plus 2as where a is the accelaration and s is the distance to be covered ...now substitute and find the answer

Given that-> initial velocity u=0, final velocity=80, distance x=95 using v^{2} = u^{2} + 2ax a= \frac{v^{2} - u^{2}}{2x} on put the values a=33.6842105263158m/s^{2}