Question

A plane of the Philippine Air Lines starts from rest and accelerates down the runway and at 29 s attains a velocity of 260 km/hr. Determine the acceleration of the plane.

Answer 1

Given:-

• Initial velocity , u = 0m/s

• Time taken , t = 29s

• Final velocity = 260km/h

To Find:-

• Acceleration of the plane , a.

Solution:-

Firstly we convert the unit here

→ 260km/h = 260×5/18 = 72.22m/s

Now we have to calculate the acceleration of the plane. So by using 1st equation of motion

→ v = u+at

Substitute the value we get

→ 72.22 = 0+ a×29

→ a = 72.22/29

→ a = 2.49 ≈ 2.50m/s²

∴ The acceleration of the plane is 2.50m/s²

Extra Information !!!

•The rate of change of velocity at per unit time is called acceleration .

• SI unit of acceleration is m/s²

•It is vector Quantity ( both magnitude and direction)

Answer 2

Aɴsᴡᴇʀ:

• The required acceleration of the plane = 2.50m/s²

Gɪᴠᴇɴ:

• A plane of the Philippine Air Lines starts from rest and accelerates down the runway and at 29 seconds attains a velocity of 260 km/hr.

Nᴇᴇᴅ ᴛᴏ ғɪɴᴅ:

• The required cceleration of the plane = ?

Solution:

Cᴏɴᴠᴇʀᴛ Uɴɪᴛ:

➜ 260 km/h

➜260 × 5/18

➜ 72.22m/s

Now, Aᴄᴄᴏʀᴅɪɴɢ ᴛᴏ ᴛʜᴇ ғɪʀsᴛ ᴇǫᴜᴀᴛɪᴏɴ ᴏғ ᴍᴏᴛɪᴏɴ:

• v = u + at

Pᴜᴛᴛɪɴɢ ᴛʜᴇ ᴠᴀʟᴜᴇs

➥ 72.22 = 0 + acceleration × 29

➥ Acceleration = 72.22/29

➥ Acceleration = 2.49 ≈ 2.50m/s²

Tʜᴇʀᴇғᴏʀᴇ:

• The required acceleration of the plane = 2.50m/s²

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