Question

A jet lands on an aircraft carrier at 30
m/s. It stops in 2.0 s. What is the
displacement of the plane when it stops?​

Answer 1

Given:-

• Initial velocity (u) = 30m/s

• Final velocity (v) = 0m/s ( as the jet stops).

• Time taken (t) = 2 s

To Find:-

• Displacement of the plane (s)

Solution:-

Firstly we calculate the acceleration of the jet.So using 1st equation of motion

→ v = u+at

Substitute the value we get

→ 0 = 30+ a×2

→ a = -30/2

→ a = -15m/s

here negative sign show retardation

∴ the acceleration of the jet is 15m/s²

And Now using 3rd equation of motion

→ v² = u² +2as

Substitute the value we get

→ 0² = 30² + 2× (-15)×s

→ 0 = 900 + (-30)×s

→ -900 = -30×s

→ s = -900/-30

→ s = 30m

∴ The displacement of the plane is 30 metres.

Answer 2

Answer = 30m

Initial velocity u, = 30 m/s

As it stop so final velocity, v = 0 m/s

Time t = 2 s

Distance covered = s

we know that ,v= u +at

0= 30 + a(2)

a=  - 15m/s²

S= ut+1/2at²

S= 30×2+1/2(-15) ×2²

S=30 m