Question

A person is standing on a weighing machine placed on the floor of an elevator. The elevator starts going up with the same acceleration, moves with uniform velocity for a while and finally decelerates to stop. The maximum and the minimum weights recorded are \[72kg{\text{ }}and{\text{ }}60kg\]. assuming the magnitudes of the acceleration and the deceleration are the same, find (a) the true weight of the person (b) the magnitude of the acceleration takes $g = 9.9m/{s^2}$

Answer 1

Hint: Motion is a change in position of an object with time. The magnitude of displacement may be equal or not to the path length traveled by an object.
The maximum weight will be shown when the elevator accelerates upwards.

Complete step by step solution:
Acceleration is a vector quantity. We can find the magnitude of this vector the same way you find the magnitude of any vector. If the components of the vector are known, then we have to square each component and add up all the squares of components, and then have to take the positive square root of that number.
Consider the acceleration of the elevator first upward and then downward with the same magnitude.
And
$ \Rightarrow mg - ma = 60g$
Now solve the equation,
$ \Rightarrow m = \dfrac{{132g}}{{2g}}$
After simplification,
$ \Rightarrow m = 66kg$ (a).
And,
$2ma = 12g$
Now simplify the equation,
$ \Rightarrow a = \dfrac{{6g}}{m} = 6 \times \dfrac{{9.9}}{{6.6}} = 0.9m{s^{ - 2}}$
So, the answer is $0.9m{s^{ - 2}}$.

Note:
The acceleration is known as the vector quantity,
Acceleration has magnitude and direction because it is a vector quantity.
Although the velocity of a uniform circular motion may be constant, its direction keeps changing. Therefore the acceleration is not constant.