Question

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✴✴Physics Genius answer it✴✴

● A Plane has a takeoff Speed of 88.3 m/s and required 1365 m to reach that speed . Determine the acceleration of the plane and the time Required to reach this speed .

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Answer 1

S (distance) = 1365 m

u (Initial Velocity) = 0 m/s {it starts from rest}

v (final velocity) = 88.3 m /s

USING 3RD EQUATION OF MOTION

v square = u square + 2as

88 .3 square = 0 + 2 * 1365 * a

7796.89 = 2730 a

a = 7796.8 / 2730

=> a = 2.856 m/sec square

For time, USING 1 ST EQUATION OF MOTION

v = u + at

88.3 = 0 + 2.856t

t = 88.3 / 2.856

=> t = 30.917 secs

These are exact values you can also find by approx values

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Answer 2

Answer:

acceleration = 2.85 m/s^2. , time = 31 seconds

Explanation:

given,

initial speed , u = 0

final speed , v = 88.3

displacement , s = 1365 m

from

${v}^{2} - {u}^{2} = 2as \\ {88.3}^{2} - {0}^{2} = 2a(1365) \\ a \: = \frac{7796.89}{2730} \\ a \: = 2.85m {s}^{ - 2}$

$a = \frac{v - u}{t} \\ a = \frac{88.3 - 0}{t} \\ 2.85 \ = \frac{88.3}{t} \\ t \: = \frac{88.3}{2.85} \\ t \: = 30.9 = 31seconds$

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