Question

a plain has a take off speed of 300km/h .what is is the acceleration in m/s2 of the plain if the plain started from rest and took 45sec to take off​

Answer 1

✿ Given :-

• Speed when flight taken off = 300km/h
• The plain started from rest
• Time taken to take off by the flight = 45s

✿ Conversion :-

We are converting all the units to the same unit ( m/s ) because we need to find the acceleration in m/s²

300 km/h → m/s

To convert we need to multiply with 5/18

→ 300 × 5/18 = 83.33 m/s

✿ To find :-

• Acceleration attained by the flight = ?

✿ Solution :-

Finding acceleration using first equation of motion

v = u + at

Where ,

• v → final velocity = 83.33 m/s
• u → Initial velocity = 0 m/s
• a → acceleration = ?
• t → time = 45s

⇒ 83.33 = 0 + a(45)

⇒ 83.33 = 45a

⇒ a = 83.33 / 45

⇒ a = 1.85 m/s²

Hence , acceleration of the flight before take off = 1.85 m/s²

Answer 2

Question:-

A plain has a take off speed of 300km/h .what is is the acceleration in m/s2 of the plain if the plain started from rest and took 45sec to take off.

To Find:-

Find the acceleration of the plane.

Formulas:-

• v = u + at
• v² - u² = 2as
• s = ut + 1/2 at²
• a = v - u/t

Answer:-

a = 1.85m/s²

Given:-

300km/h ; u = 0 ; t = 45sec ; a = ?

300 kmph = 83.333m/s

Solution:-

a = v - u

t

a = 83.333 - 0

45

a = 1.85184444m/s²

Hence Verified