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# A plane has a take off speed of $88.3 m/s$ and requires 1365 m to reach that speed. Determine the acceleration of the plane.(A) $3.86m/{s^2}$ (B) $2.86m/{s^2}$(C) $2.8m/{s^2}$(D) $2.6m/{s^2}$

Last updated date: 11th Sep 2024
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Hint:Given problem is straight forward formula based problem. If initial velocity, final velocity and distance or displacement is given and we have to find acceleration then use Newton’s III equation of motion which is
${v^2} = {u^2} + 2as$
Where
u $=$ Initial velocity
v $=$ Final velocity
a $=$ Acceleration
s $=$ Displacement

Given that initial velocity of plane is $u = 0 m/s$ final velocity of plane $v = 88.3 m/s$
Distance travelled by plane
$s = 1365 m$
Let a be the acceleration of the plane.
So, using Newton’s III equation of motion
$\Rightarrow {v^2} = {u^2} + 2as$
$\Rightarrow {(88.3)^2} = {(0)^2} + 2a(1365)$
$\Rightarrow 7796.89 = 0 + (2730)a$
$\Rightarrow (2730)a = 7796.89$
$\Rightarrow a = \dfrac{{7796.89}}{{2730}}$
$\Rightarrow a = 2.865$
$\therefore a \simeq 2.86m/{s^2}$
Hence, the acceleration of plane is $2.86m/{s^2}$

So, option B is the correct answer.

Note: In many problems, students may get confused about where and which equation of Newton’s we should apply. So, If acceleration, time and initial velocity is given and we have to find final velocity of the body then we should apply Newton’s I equation.If displacement is asked then we should apply Newton’s II equation.If acceleration is asked and time is not given then we should apply Newton’s III equation.