Question

The velocity of a particle is $v=2t+\cos (2t)$. when $t=k$ the acceleration is $0$. show that $k=\dfrac{\pi}{4}$ ?

Answer 1

Hint: In this problem we have to show that $k=\dfrac{\pi}{4}$ by using the given value $v=2t+\cos (2t)$ and the acceleration at the given time. We know that the acceleration is the derivative of the velocity. So, we will find the derivative of the given velocity equation with respect to $t$ . Here we will use some differentiation formulas. After applying the differentiation formula, we will simplify the equation to get the equation for the acceleration of the particle. Now we will calculate the acceleration of the particle at a given time $t=k$ by substituting $t=k$ in the acceleration equation. Now we will equate it to the given value of acceleration and simplify the equation, then we will get the required solution.

Formulas Used:
1.$\dfrac{d}{dx}\left( x \right)=1$.
2. $\dfrac{d}{dx}\left( \cos x \right)=-\sin x$.
3. $\dfrac{d}{dx}\left( \cos at \right)=-a\sin \left( at \right)$.

Complete step by step solution:
Given that, $v=2t+\cos (2t)$.
Differentiating the above equation with respect to $t$ , then we will get
$\Rightarrow \dfrac{d}{dt}\left( v \right)=\dfrac{d}{dt}\left( 2t+\cos \left( 2t \right) \right)$
Applying the differentiation formulas, and simplify the obtained equation, then we will get
$\Rightarrow {{v}^{'}}=2-2\sin \left( 2t \right)$
We know that the derivative of the velocity is acceleration, so replacing the ${{v}^{'}}$ in the above equation with the $a$ , then we will get
 $\Rightarrow a=2-2\sin \left( 2t \right)$
Now the acceleration of the particle at a given time $t=k$ will be
$\Rightarrow a=2-2\sin 2k$
Given that the acceleration of the particle at $t=k$ is zero, hence
$\Rightarrow 0=2-2\sin 2k$
Simplifying the above equation, then we will get
$\begin{align}
  & \Rightarrow 2\sin 2k=2 \\
 & \Rightarrow \sin 2k=1 \\
\end{align}$
We have the value $\sin \left( \dfrac{\pi }{2} \right)=1$ , then we will have
$\begin{align}
  & \Rightarrow \sin 2k=\sin \dfrac{\pi }{2} \\
 & \Rightarrow 2k=\dfrac{\pi }{2} \\
 & \Rightarrow k=\dfrac{\pi }{4} \\
\end{align}$

We know that $0 < k < 2$ in the periodicity of the $\sin (2x)$ ; its function always is $\pi $. But in the above question they mention $a=0$, so the answer we get $k=\dfrac{\pi }{4}$.

Note:
In this problem they have given the equation of the velocity and the value of the acceleration at a given time so we have differentiated the velocity and calculated the value of acceleration. If they have given the equation of the acceleration and value of velocity. Then we will integrate the given acceleration equation and use the velocity value to get the result.