Question

A particle moves along a curve whose parametric equations are: $x=t^3+2t,y=-3e^{-2t}andz=2sin\left(5t\right)$, where x, y and z show variations of the distance covered by the particle (in cm) with time t (in s). The magnitude of the acceleration of the particle $(incm/s^{2}att=0)$ is

• A. 12

Answer 1

The correct option is A 12
$x=t^3+2t$
$y=-3e^{-2t}$
$z=2sin\left(5t\right)$

$\frac{dx}{dt}=3t^2+2$

$\Rightarrow a_x=\frac{d^{2}x}{d{t}^2}=6t$

$\frac{dy}{dt}=-3e^{-2t}\times (-2)=6e^{-2t}$

$\Rightarrow a_y=\frac{d^{2}y}{d{t}^2}=-12e^{-2t}$

$\Rightarrow \frac{dz}{dt}=2\times 5cos\left(5t\right)$

$=10cos\left(5t\right)$

$\Rightarrow a_z=\frac{d^{2}z}{d{t}^2}=-50sin5t$

$\overrightarrow{a}$ $=a_x\hat{i}+a_y\hat{j}+a_z\hat{k}$

$\overrightarrow{a}$$att=0=0\hat{i}-12\hat{j}+0\hat{k}$

$\overrightarrow{a}$ $=-12\hat{j}$

$\Rightarrow$ Magnitude of acceleration at $t=0,$
$=12cm/s^2$