Question

a particle is at rest at t=0. If acceleration of the particle is given as a=sinπt+cosπt in SI units,then maximum speed of the particle is

Answer 1

particle is rest at t = 0, means initial velocity of particle, u = 0

Acceleration of the particle is given as

$a=sin\pi t+cos\pi t$ in SI units.

we know, acceleration is the rate of change of velocity with respect to time.

so, a = dv/dt = sinπt + cosπt

or, $\int\limits^v_0{dv}=\int\limits^t_0{(sin\pi t+cos\pi t)}\,dt$

or, v = -cosπt/π + sinπt/π

or, v = 1/π(sinπt - cosπt)

= 1/π[√2 {1/√2 sinπt - 1/√2cosπt }]

= √2/π [ cosπ/4.sinπt - sinπ/4.cosπt ]

we know, sin(A - B) = sinA.cosB - cosA.sinB

= √2/π sin(πt - π/4)

as we know, maximum value of sine function is 1

so, maximum speed of particle must be √2/π m/s

Answer 2

Answer: it's correct

The answer to this ques is (√2+1)/π

I Know where u might have went wrong

Be careful in integrating a function then applying the limits correctly

Hope my answer would have helped u .

Explanation: