Question

What is the smallest distance, in meters, needed for an airplane touching the runway with a velocity of 360 km/h and an acceleration of -10 m/s2 to come

Answer 1

Answer:

The Distance is 500 m or, 0.5 km.

Explanation:

Acceleration (a) = -10 m/s²

Initial Velocity (u) = 360 km/hr or,

= (360*5/18) m/s

= 100 m/s

Final Velocity (v) = 0 m/s   [ as the airplane is coming to rest ]

So,

v² = u²+2aS  

Here, S means distance,

= 0² = 100²-2*10*S

= 0² = 10000-20S

= 20S = 10000

= S = 10000/20

= S = 500 m or 0.5 km

Hope this helps.

Answer 2

Aɴꜱᴡᴇʀ

$\huge \sf d = 500m$

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Gɪᴠᴇɴ

Initial Velocity (u) = 360 km/h and this can also be written as 100 m/s

Final velocity (v) = 0 m/s

Acceleration (a) = -10 m/s²

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ᴛᴏ ꜰɪɴᴅ

The smallest distance in the unit of meter for the aeroplane to touch the runway with a v of 360 km/h and a of -10 m/s²

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Sᴛᴇᴘꜱ

With the use of 3rd law of motion

$\begin{lgathered}\dashrightarrow \boxed{ \fbox{\sf{v^2 \: - \: u^2 \: = \: 2as}}} \\ \\ \leadsto\tt{0 \: - \: 100^2 \: = \: 2 \: \times \: -10 \: \times \: s} \\ \\ \leadsto\tt{-10000 \: = \: -20s} \\ \\ \leadsto \tt{s \: = \: \dfrac{10000}{20}} \\ \\ \leadsto \tt{s \: = \: 500} \\ \\{\sf \pink{\leadsto\: Distance \: travelled \: is \: 500 \: m}}\end{lgathered}$

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$\huge{\mathfrak{\purple{hope\; it \;helps}}}$