# Francesca, who likes physics experiments, dangles her watch from a thin piece of string, while...

## Question:

Francesca, who likes physics experiments, dangles her watch from a thin piece of string, while the jetliner she is in takes off from JFK Airport. She notices that the string makes angle of 25 degrees, with respect to the vertical, as the aircraft accelerates for takeoff, which takes about 18s. Estimate the takeoff speed of the aircraft.

## The accelerated frame of reference

Newton's laws give the correct dynamics in any inertial frame of reference. However, if the frame of reference is accelerating or non- inertial then even in the absence of applied forces objects are seen to accelerate. For instance, if you are inside a train moving with a constant velocity and then if it so happens that the brakes are applied then you will find yourself hurtling forward even though no visible force has been applied on you. The Newtonian scheme may be resurrected in a non-inertial frame by invoking a fictitious force. If the acceleration of the frame is {eq}\displaystyle { a} {/eq} simply assign a force {eq}\displaystyle {- ma} {/eq} to a particle of mass {eq}\displaystyle {m} {/eq}. Now take into account all the real forces and apply Newton's law.

## Answer and Explanation:

Since the angle made by the watch on the string with the vertical is given to be 25 degrees, it follows that the watch is under the equilibrium of three forces at this angle. The gravitational force {eq}\displaystyle {mg} {/eq} vertically down, the fictitious force {eq}\displaystyle {ma} {/eq} due to the acceleration acting horizontally opposite to the direction of the jetliner's acceleration, and finally, the tension {eq}\displaystyle {T} {/eq} of the string. The ratio of the horizontal force to the vertical force must be the tangent of the angle with the vertical. That is,

{eq}\displaystyle { \tan 25=\frac{ma}{mg}=\frac{a}{g}} {/eq}.

Therefore the acceleration of the plane is,

{eq}\displaystyle { a=g \tan25=9.8\times 0.4663=4.57\ m/s^2} {/eq}.

Since the plane accelerates for 18 seconds starting from rest, the velocity it acquires must be,

{eq}\displaystyle { v=u+at=0+4.57\times 18=82.25\ m/s=296 \ km/hr} {/eq}.