Question

A particle's position in the x-y plane varies as $x\left(t\right)=3t^3+4t^2,y\left(t\right)=16t^4+2$

Find the acceleration vector of the particle after 1 sec of start.

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Particles position vector in x is given as $x\left(t\right)=3t^3+4t^2$
$v_x\left(t\right)=\frac{dx\left(t\right)}{dt}=9t^2+8t$
$a_x\left(t\right)=\frac{d{v}_x\left(t\right)}{dt}=18t+8$
At t = 1
$a_x\left(t\right)=18+8=26m/s^2$

Particles position vector y is given as $y\left(t\right)=16t^4+2$
$v_y=\frac{dy\left(t\right)}{dt}=64t^3$
$a_y=\frac{d{v}_y\left(t\right)}{dt}=192t^2$
At t = 1 sec
$a_y=192m/s^2$
$\Rightarrow \overrightarrow{a}=a_x\hat{i}+a_y\hat{j}$
$\overrightarrow{a}=26m/s^2\hat{i}+192m/s^2\hat{j}$