Question

A stone, thrown vertically upward from the surface of the moon at a velocity of \[24m/\sec \] reaches a height of \[s = 24t - 0.8{t^2}\] metre after \[t\] second. The acceleration due to gravity \[m/{s^2}\] at the surface of the moon is
A) \[0.8\]
B) \[ - 1.6\]
C) \[2.4\]
D) \[ - 4.9\]

Answer 1

Hint: The derivation of a function representing the position of a moving particle along an axis at a time is the velocity at that time. Again the second derivation of the position function of that particle or the derivation of the velocity represents the acceleration of that moving particle at that time.

Formula used: If the position function of the object is represented as \[s\] then the velocity at a time \[t\] is represented as \[v = \dfrac{{ds}}{{dt}}\] and acceleration is represented as \[a = \dfrac{{{d^2}s}}{{d{t^2}}}\]

Complete step by step answer:
According to the given problem, height \[s = 24t - 0.8{t^2}\] metre,
As we know, velocity \[v\] is the derivative of distance \[\dfrac{{ds}}{{dt}} = 24 - 1.6t\]
And acceleration \[a\] is the derivative of velocity \[\dfrac{{{d^2}s}}{{d{t^2}}} = - 1.6\]
\[\therefore \]Acceleration of the stone is \[ - 1.6m/{\sec ^2}\]. Since it is thrown vertically upward, because of the moon's gravity, the acceleration is negative.
Therefore, the correct answer is option B).

Note: A positive velocity implies that the position of the particle increases with the increase of time while negative velocity implies the decreasing position of the particle with the increase of time. If the distance remains constant the velocity will be zero on an interval of time.