Question

$\huge\underline\mathrm{Question-}$


Explain a jêrky motion with diagram in the context of physics.


Thank You. ​

Answer 1

$\Huge{\sf{JolT : }}$

• Jolt refers to the rate of change of acceleration of an object with respect to time

• Jolt is a vector quantity

• SI unit of Jolt is m/s³

There is no clear distinction as of now between Acceleration and Jolt. Jolt isn't something you would come across in your daily physics researches

During most of the physical activities like walking, cycling etc,the acceleration of the system is to be balanced otherwise the equilibrium is disturbed and you tend to trip or fall down,the same can be expected of Jolt

During most of the physical activities like walking, cycling etc,the acceleration of the system is to be balanced otherwise the equilibrium is disturbed and you tend to trip or fall down,the same can be expected of Jolt

When was the last time you went on a merry go round and felt like vomiting,this is actually due to the unbalanced jolts you are experiencing

Usage of Jolt is restricted and some physical quantities like Pop,Sparkle are higher derivatives of Displacement and are termed as HyperJÊrK Systems

Jolt and Lower Derivates

Jolt is the first derivate of acceleration

• $\sf{J = \dfrac{da}{dt} }$

Jolt is the second derivate of velocity

• $\sf{J = \dfrac{ {d}^{2}v }{d {t}^{2} }}$

Jolt is the third derivate of displacement

• $\sf{J = \dfrac{ {d}^{3}s }{d {t}^{3} } }$

Every object under motion Jolts

For example,

Vigorous Jolts of Tectonic Plates causes earthquakes

Differentiation

• Differentiation is a branch of calculus which determines the derivate of a function at a given instant.

• In other words,it helps us measures infinitesimally small changes in physical quantities

• Assume derivative as a solution to a variable at a given instant

Suppose,

A function v(t) = 2t² + 3t⁴ is given to us and we have to find the acceleration of the particle at t = 4s. We are finding the acceleration at a particular instance and this is done by finding it's derivate at that instant which is differentiation.

• $\sf{ {x}^{n} = n {x}^{n - 1} }$

Here,

• n refers to the power of the variable

Here,

Velocity is given as a function of time,i.e.,it is defined with respect to time.

Velocity is given as a function of time,i.e.,it is defined with respect to time. Likewise if velocity is given as a function of displacement,you can find the derivative with respect to time. That function wouldn't exist

Consider the given relation v(t) = 2t² + 3t⁴

$\sf{a = \dfrac{dv}{dt} } \\ \\ \longrightarrow \: \sf{a = \dfrac{d(2t {}^{2} + {3t}^{4} )}{dt} } \\ \\ \longrightarrow \: \sf{a = \dfrac{d(2 {t}^{2} )}{dt} + \frac{d(3 {t}^{4} )}{dt} } \\ \\ \longrightarrow \: \boxed{ \boxed{\sf{a =(4t + 12 {t}^{3}) \: {ms}^{ - 2} }}}$

Substituting t = 4s,would give you acceleration at that instant

Derivative of a constant is zero

If you want to find the Jolt of the particle at t = 4s, differentiate it w.r.t to time

$\sf{J= \dfrac{da}{dt} } \\ \\ \longmapsto \: \sf{J = \frac{d(4t + 12 {t}^{3}) }{dt} } \\ \\ \longmapsto \: \boxed{ \boxed{ \sf{J =( 36t {}^{2} + 4) \: {ms}^{ - 3} }}}$

Again substituting t = 4s would give the desired result

# Refer to the attachment

The attachment depicts Jolty Motion

Answer 2

$\huge{\underline{\mathscr{\purple{answer :}}}}$JolT:Jolt refers to the rate of change of acceleration of an object with respect to timeJolt is a vector quantitySI unit of Jolt is m/s³