Question

An airplane starts from rest at the end of a runway and accelerates at a constant rate. In the first second, the airplane travels 1.11 m. What is the speed of the airplane at the end of the second second?

Answer 1

Answer:

$v=4.44\frac{m}{s}$

Explanation:

Given that the airplane starts from the rest (this is initial velocity equals to zero)  and accelerates at a constant rate, position can be described like this: $x=v_{0}t +\frac{1}{2} at^{2}$ where x is the position, t is the time a is the acceleration and $v_{0}$ is initial velocity. In this way acceleration can be found. $a=\frac{2(x-v_{0}t) }{t^{2} } =\frac{2(1.11m-0)}{1s^{2} } =2.22\frac{m}{s^{2} }$.

Now we are able to found velocity at any time with the formula: $v=v_{0} +at = 0\frac{m}{s} +(2.22\frac{m}{s^{2}}.2s)=4.44\frac{m}{s}$