Question

A jet plane has a takeoff speed of vto = 75 m/s and can move along the runway at an average acceleration of 1.3 m/s2. If the length of the runway is 2.5 km, will the plane be able to use this runway safely?

Answer 1

If the length of the runway is 2.5 km, then the plane will be able to use this runway safely.

Further explanation

These are the formulas that we have to remember before solving the problem.

Speed is the rate of change of distance.

$\large { \boxed {v = \frac{d}{t}}}$

v = speed ( m/s )

d = distance ( m )

t = time ( s )

Acceleration is the rate of change of velocity.

$\large { \boxed {a = \frac{\Delta v}{t}}}$

a = acceleration ( m/s² )

Δv = change in speed ( m/s )

t = time ( s )

Let us now tackle the problem!

Given:

Initial speed = u = 0 m/s

Final speed = v = 75 m/s

Average acceleration = a = 1.3 m/s²

Unknown:

Distance = d = ?

Solution:

$v ^ 2 = u ^ 2 + 2 a d$

$75 ^ 2 = 0 ^ 2 + 2 ( 1.3 ) d$

$5625 = (2.6) d$

$d = 5625 \div 2.6$

$d \approx 2200 ~ m$

$\large {\boxed {d \approx 2.2 ~ km} }$

The distance needed by the aircraft from a standstill to reach speeds of 75 m/s is around 2.2 km. The aircraft will be safe using the runway because the length of the runway ( 2.5 km ) is greater than the required distance ( 2.2 km ).

Learn more

• Velocity of A Runner : brainly.com/question/3813437
• Kinetic Energy : brainly.com/question/692781
• Acceleration : brainly.com/question/2283922

Answer details

Grade: Middle School

Subject: Physics

Chapter: Kinematics

Keywords: Speed , Distance , Acceleration , Time , Velocity , Jet , Plane , TakeOff , Runway

Answer 2

Initial velocity of the plane is Vo = 0.
acceleration a = 1.3 m/s2
total distance = 2.5 km = 2500m

time taken to reach 2.5 km with 1.3m/s^2 acceleration = t
S = Vo t + 0.5 a t^2
2500 = 0 + (0.5*1.3* t^2)
t^2 = 3846.15
t = 62 s

the maximum velocity plan can reach within 62 s is Vt
Vt = Vo + a t
Vt = 0 + (1.3*62)
Vt = 80.6 m/s

Since 80.6 m/s is greater than 75 m/s, plane can use this runway to takeoff with required speed.