Respuesta :
For the answer to the question above asking what is the time tTO needed to take off?
s = ut + 0.5at^2
s = distance = 1800m, u = initial velocity = 0, a = acceleration = 5m/s^2, t = time
1800 = 0 + 0.5*5*t^2
t^2 = 1800/2.5 = 720
t = 26.83 s (to 4 sig figs)
I hope this helps
s = ut + 0.5at^2
s = distance = 1800m, u = initial velocity = 0, a = acceleration = 5m/s^2, t = time
1800 = 0 + 0.5*5*t^2
t^2 = 1800/2.5 = 720
t = 26.83 s (to 4 sig figs)
I hope this helps
The time required to take off the plane at the end of the runway is [tex]\boxed{26.83{\text{ s}}}[/tex].
Further explanation:
Here, we have to find the time needed to take off the plane.
Given:
The constant acceleration of the plane is [tex]5{\text{ }}{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^2}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^2}}}[/tex]
Length of the runway is [tex]1800{\text{ m}}[/tex].
Formula and concept used:
The time required to take off the plane can be calculated by using the kinematic equation of the motion.
By using second kinematic equation of the motion,
[tex]\boxed{s = ut + \frac{1}{2}a{t^2}}[/tex] ….. (1)
Here, [tex]s[/tex] is the length of the runway.
[tex]u[/tex] is the initial velocity.
[tex]a[/tex] is the initial velocity.
[tex]t[/tex] is the time required to take off the plane.
Initially plane was in the rest position. So, initial velocity of the plane will be zero.
Here, plane takes take off at the end of the runway. So, we will consider the distance travelled by the plane is equal to the length of the runway.
Calculation:
Substitute [tex]0{\text{ }}{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}[/tex] for [tex]u[/tex] , [tex]1800{\text{ m}}[/tex] for [tex]s[/tex] and [tex]5{{{\text{ m}}} \mathord{\left/{\vphantom {{{\text{ m}}} {{{\text{s}}^2}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^2}}}[/tex] for [tex]a[/tex] in equation (1).
[tex]\begin{gathered}1800 = 0 \times t + \frac{1}{2} \times 5{\left( t \right)^2} \hfill \\1800 = \frac{{5{t^2}}}{2} \hfill \\ \end{gathered}[/tex]
Multiply both the side by [tex]\dfrac{2}{5}[/tex] in the above equation.
[tex]\begin{aligned}\frac{2}{5}\times1800&=\frac{2}{5}\times\frac{{5{t^2}}}{2}\\\frac{{3600}}{5}&={t^2}\\\end{aligned}[/tex]
Simplify the above equation,
[tex]\begin{aligned}t&=\sqrt {\frac{{3600}}{5}} \\&=\frac{{60}}{{\sqrt 5}}\\&=26.83{\text{s}}\\\end{gathered}[/tex]
Thus, the time required to take off the plane is [tex]\boxed{26.83{\text{ s}}}[/tex].
Learn more:
1. Water is a compound: https://brainly.com/question/4636675.
2. Acceleration of the box against the friction: https://brainly.com/question/7031524.
3. Conservation of momentum https://brainly.com/question/9484203.
Answer details:
Grade: Senior school
Subject: Physics
Chapter: Kinematics
Keyword:
Plane accelerates, constant acceleration, runway, takeoff velocity, end of the runway, time needed to take off, acceleration, deceleration, equation of motion, Newton’s laws and inertial law.
