A plane has a take-off speed of 120 $\frac{m}{s}$ and it reaches that speed after covering 2400 m. Determine the acceleration of the plane (assuming it to be constant) and the time required to reach this speed.

a = 4 m/s², t = 30 s

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a = 3 m/s², t = 30 s

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a = 4 m/s², t = 40 s

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a = 3 m/s², t = 40 s

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Solution

The correct option is D
a = 3 m/s², t = 40 s

Initial velocity, u = 0

Final velocity, v = 120 m/s

By the third equation of motion, $v^{2} = u^{2} + 2 a s$

$120^{2}$ = 0 + $2 \times a \times 2400$

Solving we get, a = 3 $m / s^{2}$

Now by first equation of motion, v = u + at

120 = 0 + 3 t

t = 40 s

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A plane has a take-off speed of 120 ms and it reaches that speed after covering 2400 m. Determine the acceleration of the plane (assuming it to be constant) and the time required to reach this speed.

The correct option is D a = 3 m/s2, t = 40 s Initial velocity, u = 0 Final velocity, v = 120 m/s By the third equation of motion, v2=u2+2as 1202= 0 + 2×a×2400 Solving we get, a = 3 m/s2 Now by first equation of motion, v = u + at 120 = 0 + 3 t t = 40 s

A plane has a take-off speed of 120 $\frac{m}{s}$ and it reaches that speed after covering 2400 m. Determine the acceleration of the plane (assuming it to be constant) and the time required to reach this speed.