Question

An Aeroplane starts from rest with an acceleration of 3 metre per second square and takes a run for 35 second before taking off.What is the minimum length of runway and with what velocity the plane took off?​

Answer 1

Given :-

Initial velocity of Aeroplane (u)= 0

Acceleration (a) = 3m/s^2

Time (t)= 35 seconds

To find :-

◆ Final velocity of Plane and length of runway ( distance )

• First find the velocity

$\boxed{\bigstar{\sf{ v= u+at}}}$

$\implies\sf v = 0+ 3\times 35\\ \\ \implies\sf v= 105m/s$

$\underline{\textsf{\dag\ Final \ velocity = {\textbf{ 105m/s}}}}$

$\boxed{\sf{\dag\ \ s= ut+\dfrac{1}{2}at^2}}$

• where ,

u= Initial velocity

t= time

a= acceleration

s= distance

$\implies\sf s= 0\times (35) +\dfrac{1}{2}\times (3)\times (35)^2\\ \\ \implies\sf s= 0+ \dfrac{3\times 1225}{2}\\ \\ \implies\sf s= \cancel{\dfrac{3675}{2}}\\ \\ \implies\sf s = 1837.5m$

$\underline{\textsf{\dag\ Distance \ covered \ by \ plane = {\textbf{ 1837.5\ m}}}}$

Answer 2

Answer:

Explanation:

Given :-

Initial velocity, u = 0 (As it starts from rest)

Acceleration, a = 3 m/s²

Time taken, t = 35 seconds.

To Find :-

Minimum length of runway i.e distance traveled

Formula to be used :-

1st equation of motion, i.e, v = u + at

2nd equation of motion, i.e s = ut + 1/2 × at²

Solution :-

Putting all the values, we get

v = u + at

⇒ v = 0 + 3 × 35

⇒ v = 105 m/s

Now distance covered, s

s = ut + 1/2 × at²

⇒ s = 0 × 35 + 1/2 × 3 × 35 × 35

⇒ s = 1/2 × 3 × 1225

⇒ s = 1/2 × 3675

⇒ s = 3675/2

⇒ s = 1837.5 m

Hence, the minimum length of runway is 1837.5 m and velocity the plane took off 105 m/s.