Question

An aeroplane touches down at 225 km h$^{-1}$ and stops after 2 minutes. Calculate acceleration and length of runway.​

Answer 1

GivEn:

• Initial Velocity of an aeroplane, u = 225 km/hr = 62.5 m/s
• Time taken by aeroplane to stop, t = 2 min = 120 s

To find:

• Acceleration and length of runway?

Solution:

★ Now, Using 1st equation of motion,

➣ v = u + at

➯ 0 = 62.5 + a × 120

➯ 0 = 62.5 + 120a

➯ 120a = 0 - 62.5

➯ a = (0 - 62.5)/120

➯ a = - 62.5/120

➯ a = - 0.520 m/s²⠀⠀⠀⠀⠀⠀⠀ ...(1)

★ Using 3rd equation of motion,

➣ v² = u² + 2as

➯ 0² = (62.5)² + 2 × (-0.520) × s

➯ 1.041 × s = 3906.25

➯ s = 3752.4 metres

➯ s = 3.752 km

∴ Hence, Acceleration and speed or length of runway of aeroplane is - 0.520 m/s² and 3.752 km respectively.

NoTe: Here negative sign of acceleration denotes retardation.

Answer 2

Answer:

Given:-

u = 225 km h$^{-1}$ and

v = 0

t = 2 minutes = 120s.

where u means initial velocity;

v means final velocity and

t means time.

To find:-

• Acceleration
• Length of runway

Solution:-

Using Newton's first law of motion

v = u + at

Now substitute the known values

$\Longrightarrow \: 0 = 62.5 + a \times 120 \\ \\ \Longrightarrow \: a = \frac{ - 625}{120} \\ \\ = - 0.52 \: ms {}^{ - 2}$

Now length of runway

Using second equation of motion

S = ut + 1/2at$^{2}$

Substitute the known values

S = 62.5 × 120 + 1/2 × (–0.52) × (120)$^{2}$

= (7500 – 3744)m

= 3756 m.

length of runway = 3756 m.

Hence, acceleration = –0.52 ms$^{2}$ and length of runway = 3756 m.