A particle is at rest at t = 0. If acceleration of the particle is given as $a=sin\pi t+cos\pi t$ in unit then the maximum speed of the particle is

Given Data:

Acceleartion of particle, $a=sin\pi t+cos\pi t$

As we know rate of change of velocity is known as acceleration so,

$\frac{dv}{dt}=sin\pi t+cos\pi t$

$dv=(sin\pi t+cos\pi t)dt$

$\int dv={\int }_0^t(sin\pi t+cos\pi t)dt$

$v={\frac{1}{\pi }[(-cos\pi t)+\left(sin\pi t\right)]}_0^t$

$v=\frac{1}{\pi }[(-cos\pi t)+\left(sin\pi t\right)]-\frac{1}{\pi }[-cos(\pi \times 0)+sin(\pi \times 0)]$

$v=\frac{sin\pi t-cos\pi t}{\pi }-\frac{1}{\pi }(-1+0)$

$v=\frac{sin\pi t-cos\pi t+1}{\pi }$

$v=\frac{1}{\pi }\{1+\sqrt{2}(sin\pi t\frac{1}{\sqrt{2}}-cos\pi t\frac{1}{\sqrt{2}})\}$

$v=\frac{1}{\pi }\{1+\sqrt{2}sin(\pi t-\frac{\pi }{4})\}$

$v_{max}=\frac{1}{\pi }(1+\sqrt{2})=\frac{\sqrt{2}+1}{\pi }unit$