Question

the mtion of a particle is defined by the relation x(t)= 3sin2t+3cos2t+20 in metre m at rime t. find time at what time patricle attains maximum acceleration ​

Answer 1

Answer:

Given:

Displacement time relationship has been given as follows :

$x = 3 \sin(2t) + 3 \cos(2t) + 20$

To find:

Time at which max acceleration os obtained

Calculation:

Acceleration is obtained by the 2nd order derivative of Displacement function.

$v = \dfrac{dx}{dt}$

$= > v = \dfrac{d(3 \sin(2t) + 3 \cos(2t) + 20)}{dt}$

$= > v = 6 \cos(2t) - 6 \sin(2t)$

Again differentiation of velocity function :

$a = \dfrac{dv}{dt}$

$= > a = \dfrac{d \{6 \cos(2t) - 6 \sin(2t) \}}{dt}$

$= > a = - 12 \sin(2t) - 12 \cos(2t)$

$= > a = - 12\{ \sin(2t) + \cos(2t) \}$

Now for max value of sin and cos functions, we follow this rule :

$= > a = - 12 \sqrt{2} \{ \frac{1}{ \sqrt{2} } \sin(2t) + \frac{1}{ \sqrt{2} } \cos(2t) \}$

$= > a = - 12 \sqrt{2} \{ \cos(45 \degree) \sin(2t) + \sin(45 \degree) \cos(2t) \}$

Following formula of sin (A + B) :

$= > a = - 12 \sqrt{2} \{ \sin(2t + 45 \degree) \}$

Max value of sin(2t + 45°) is 1

$= > a = - 12 \sqrt{2} \times \{ 1 \}$

$= > a = - 12 \sqrt{2} \: m {s}^{ - 2}$

So max Acceleration is :

$\boxed{ \red{ \huge{ \bold{ a_{max} = - 12 \sqrt{2} \: m {s}^{ - 2} }}}}$

Time for max acceleration can be found out by this way :

$\dfrac{da}{dt} = 0$

$=> \dfrac{d(-12\sqrt{2}\{\sin(2t+45\degree)\}}{dt} = 0$

Simplifying , we get :

$\cos(2t+45\degree)=0$

We need to take general solution in this case :

$=>2t+45\degree= (2n+1)\dfrac{\pi}{2}$

$=>2t+\dfrac{\pi}{4}= (2n+1)\dfrac{\pi}{2}$

$=>2t=\pi n + \dfrac{\pi}{4}$

$=> t = \dfrac{\pi n + \dfrac{\pi}{4}}{2}$

So final answer :

$\boxed{\huge{\blue{\bold{=> t = \dfrac{\pi n + \dfrac{\pi}{4}}{2}}}}}$

Answer 2

AnswEr :

The position of the particle is defined as :

$\sf \: x(t) = 3sin2t + 3cos2t + 20$

Differentiating x w.r.t t,we get velocity :

$\sf \: v = \dfrac{dx}{dt} \\ \\ \longrightarrow \: \sf \: v = \dfrac{d(3sin2t + 3cos2t + 20)}{dt} \\ \\ \longrightarrow \: \sf \: v = 3cos2t \times \dfrac{d(2t)}{dt } - 3sin2t \times \dfrac{d(2t)}{dt} + 0 \\ \\ \longrightarrow \: \sf \: v = 6(cos2t - sin2t) \: \: {ms}^{ - 1}$

Differentiating v w.r.t to t,we get acceleration :

$\sf \: a \: = \dfrac{dv}{dt} \\ \\ \longrightarrow \: \sf \: a = \dfrac{d( - 6sin2t +6cos2t )}{dt} \\ \\ \longrightarrow \: \sf \: a \: = - 6cos2t \times \dfrac{d(2t)}{dt } - 6sin2t \times \dfrac{d(2t)}{dt} \\ \\ \longrightarrow \: \sf \: a = - 12(cos2t + sin2t) \: \: {ms}^{ - 2}$

We have to calculate the max acceleration :

$\sf \: \dfrac{da}{dt} = 0 \\ \\ \implies \: \sf \: \dfrac{d \bigg( - 12( sin2t + cos2t) \bigg)}{dt} = 0 \\ \\ \implies \: \sf \: -24cos2t + 24sin2t = 0 \\ \\ \implies \: \sf \: 24sin2t = 24cos2t \\ \\ \implies \: \sf sin2t = sin(90 - 2t) \: \\ \\ \implies \: \sf \: 2t = 90 - 2t \\ \\ \implies \: \sf \: 4t = 90 \\ \\ \implies \: \sf \: t = \dfrac{\pi}{2} \times \dfrac{1}{4} \\ \\ {\implies \: \boxed{ \boxed{ \sf \: t = \dfrac{\pi}{8} }}}$

At t = π/8,the particle achieves maximum acceleration