Question

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

Answer 1

Answer

Explanation:

Ahh okay. I got it now. So solving for T using the first kinematics equation: Vf = 0 m/s and Vo = 100 m/s Vf = Vo + at 0 = 100m/s + (-5 m/s^2)(T) -100 m/s = (-5 m/s^2)(T) -100 m/s / -5 m/s^2 = T T= 20 s Using T then, plug it back into the second kinematics formula to find the distance required to land on the run way. X = 100 m/s (20 s) + (1/2)(-5 m/s^2)(20^2) Solving for x = 1000 m Convert the 1000 m into km and you get 1 km required to land. Thus the answer is no since they're asking for .800 km.

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

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

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