
A stone is falling freely and describes a distance $s$ in $t$ seconds given by equation $s=\dfrac{g{{t}^{2}}}{2}$. The acceleration of the stone is
A. uniform
B. 0
C. non-uniform
D. indeterminate
Answer
497.1k+ views
Hint: We start the differentiation process for velocity and acceleration keeping in mind that the given function is of distance. We assume the instantaneous changes as the velocity and acceleration represented by $\dfrac{ds}{dt}=v,\dfrac{dv}{dt}=a$ respectively. We differentiate the function to find the solution.
Complete step by step answer:
We have to find the velocity and acceleration as functions of t where t represents time.
We know that velocity is considered as the rate of change of displacement and acceleration is considered as the rate of change of velocity.
These rates of change will be considered for instantaneous.
If we consider displacement, velocity and acceleration as $s,v,a$ respectively, then the instantaneous changes will be considered with respect to time.
Therefore, instantaneous change of velocity is acceleration which gives $\dfrac{dv}{dt}=a$ and instantaneous change of displacement is velocity which gives $\dfrac{ds}{dt}=v$.
It is given $s=\dfrac{g{{t}^{2}}}{2}$. We now differentiate $s=\dfrac{g{{t}^{2}}}{2}$ to get $v,a$. Here $g$ is constant.
The differentiation gives $v=\dfrac{ds}{dt}=\dfrac{g}{2}\times 2t=gt$ and $a=\dfrac{dv}{dt}=g$.
Therefore, the acceleration of the stone is uniform.
So, the correct answer is “Option A”.
Note: We completed the formulas assuming that no external forces were applied. To find velocity, we take the derivative of the original position equation. To find acceleration, we take the derivative of the velocity function.
Complete step by step answer:
We have to find the velocity and acceleration as functions of t where t represents time.
We know that velocity is considered as the rate of change of displacement and acceleration is considered as the rate of change of velocity.
These rates of change will be considered for instantaneous.
If we consider displacement, velocity and acceleration as $s,v,a$ respectively, then the instantaneous changes will be considered with respect to time.
Therefore, instantaneous change of velocity is acceleration which gives $\dfrac{dv}{dt}=a$ and instantaneous change of displacement is velocity which gives $\dfrac{ds}{dt}=v$.
It is given $s=\dfrac{g{{t}^{2}}}{2}$. We now differentiate $s=\dfrac{g{{t}^{2}}}{2}$ to get $v,a$. Here $g$ is constant.
The differentiation gives $v=\dfrac{ds}{dt}=\dfrac{g}{2}\times 2t=gt$ and $a=\dfrac{dv}{dt}=g$.
Therefore, the acceleration of the stone is uniform.
So, the correct answer is “Option A”.
Note: We completed the formulas assuming that no external forces were applied. To find velocity, we take the derivative of the original position equation. To find acceleration, we take the derivative of the velocity function.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

The pH of the pancreatic juice is A 64 B 86 C 120 D class 12 biology CBSE

Give 10 examples of unisexual and bisexual flowers

