1. ## velocity and acceleration

An airplane takes off after it travelled adistance of 1350 m on the runway. The airplane started from rest,moved with aconstant acceleration for 3o seconds before taking off .what was the speed at the time of take off ? ( I need to solve by integral with my greatfull)

An airplane takes off after it travelled adistance of 1350 m on the runway.
The airplane started from rest, moved with a constant acceleration
for 30 seconds before taking off.
What was the speed at the time of take-off?

Let $\displaystyle \,a$ be the constant acceleration (in $\displaystyle \text{m/sec}^2$)

Then the velocity is: .$\displaystyle v(t) \;=\;\int a\,dt \:=\:at + v_o$
. . where $\displaystyle \,v_o$ is the plane's initial velocity.

The plane started from rest, so $\displaystyle v_o = 0.$
. . The velocity function is: .$\displaystyle v(t) \:=\:at$ .[1]

Then the position function is: .$\displaystyle s \;=\;\int at\,dt \:=\:\frac{1}{2}at^2 + s_o$
. . where $\displaystyle \,s_o$ is the plane's initial position.
Assume that $\displaystyle s_o = 0$, we have: .$\displaystyle s(t) \:=\:\frac{1}{2}at^2$ .[2]

We are told that $\displaystyle s = 1350$ when $\displaystyle t = 30.$

Substitute into [2]: .$\displaystyle 1350 \:=\:\frac{1}{2}a(30^2) \quad\Rightarrow\quad a \:=\:3$

Substitute $\displaystyle t = 30,\;a = 3$ into [1]:

. . $\displaystyle v(30) \:=\:(3)(30) \:=\:90\text{ m/sec.}$