Question

A particle is moving in a straight line whose position varies with time as $x=8t^2+\frac{2}{\pi }\sin \pi t$, where x is in meter and t is in second. Then the position, velocity and acceleration of a particle at $t=4$s respectively are?

• A. $128.14$m, $66$ m/s, $14.63$ $m/s^2$
• B. $118$ m, $66$ m/s, $16$ $m/s^2$
• C. $128$ m, $66$ m/s, $16$ $m/s^2$
• D. $128$ m, $44$ m/s, $16m/s^2$

Answer 1

The correct option is C $128$ m, $66$ m/s, $16$ $m/s^2$

Given,

$x=8t^2+\frac{2}{\pi }sin\pi t$

Position,

$x{|}_{t=4sec}=8\times 4\times 4+\frac{2}{\pi }sin\left(4\pi \right)$

$x{|}_{t=4sec}=128+\frac{2}{\pi }\times 0$

$x{|}_{t=4sec}=128m$

Velocity,

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

$v=16t+\frac{2}{\pi }\pi cos\pi t$

$v=16t+2cos\pi t$

$v{|}_{t=4sec}=16\times 4+2cos\left(4\pi \right)$

$v{|}_{t=4sec}=64+2=66m/s$

Acceleration,

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

$a=16-2\pi sin\pi t$

$a{|}_{t=4sec}=16-2\pi \times 0$

$a{|}_{t=4sec}=16m/s^2$

The correct option is C.