Question

A plane has a take-off speed of $120m{s}^{-1}$ after covering a distance of $2400m$. Determine the acceleration $\left(a\right)$ of the plane (assuming it to be constant) and the time $\left(t\right)$ required to reach that speed.

• A. $a=4m{s}^{-2},t=30s$
• B. $a=3m{s}^{-2},t=30s$
• C. $a=4m{s}^{-2},t=40s$
• D. $a=3m{s}^{-2},t=40s$

Answer 1

The correct option is D $a=3m{s}^{-2},t=40s$

Given:
Initial velocity, $u=0m{s}^{-1}$
Final velocity, $v=120m{s}^{-1}$
Distance covered, $s=2400m$

Let 'a' be the acceleration
Let 't' be the time taken

By the third equation of motion:
$v^2=u^2+2as$
${120}^2=0^2+2\times a\times 2400$
$14400=4800a$
$a=3m{s}^{-2}$

By the first equation of motion:
$v=u+at$
$120=0+3t$
$t=40s$