Question

The co-ordinates of a particle restricted to move in a plane is given by
$X=6cos\pi t$
$y=1-4cos2\pi t$
The magnitude of acceleration of particle at $t=1.5s$ is (where $x$ and $y$ are in meter and $t$ is in seconds)

• A. $Zero$
• B. $6{\pi }^{2}m/s^2$
• C. $16{\pi }^{2}m/s^2$
• D. $8{\pi }^{2}m/s^2$

Answer 1

The correct option is C $16{\pi }^{2}m/s^2$

Given,

$x=6cos\pi t$

$y=1-4cos2\pi t$

Velocity along x-axis,

$v_x=\frac{dx}{dt}=-6\pi sin\pi t$

Acceleration along x-axis,

$a_x=\frac{d{v}_x}{dt}$

$a_x=-6{\pi }^{2}cos\pi t$

$a_x{|}_{t=1.5sec}=-6{\pi }^{2}cos\left(1.5\pi \right)$

$a_x{|}_{t=1.5sec}=-6{\pi }^2\times 0=0m/s^2$

Velocity along y-axis,

$v_y=\frac{dy}{dt}=+8{\pi }^{2}sin2\pi t$

Acceleration along y-axis,

$a_y=\frac{d{v}_y}{dt}$

$a_y=16{\pi }^{2}sin2\pi t$

$a_y{|}_{t=1.5sec}=16{\pi }^{2}sin\left(3\pi \right)=16{\pi }^2\times (-1)$

$a_y{|}_{t=1.5sec}=-16{\pi }^2$

Acceleration, $a=\sqrt{a_x^2+a_y^2}$

$a=\sqrt{(0{)}^2+(-16{\pi }^2{)}^2}$

$a=16{\pi }^{2}m/s^2$

The correct option is C.