Question

A particle moves along the curve given by $x=t,y=$ $\frac{t^2}{2}andz=\frac{t^3}{3},$ where $x,y,z,t$ are in SI units. The magnitude of the tangential acceleration, when $t=1\text{s}$ is

Answer 1

We know that differentiating position wrt time gives velocity,
$x=t\Rightarrow v_x=\frac{dx}{dt}=1$
$y=\frac{t^2}{2}\Rightarrow v_y=\frac{dy}{dt}=t$
$z=\frac{t^3}{3}\Rightarrow v_z=\frac{dz}{dt}=t^2$
velocity vector can be written as
$\overrightarrow{v}=\hat{i}+t\hat{j}+t^2\hat{k}$
Magnitude of velocity is
$\therefore v=\sqrt{1+t^2+t^4}$
We know that, tangential acceleration is rate of change of magnitude of velocity. Hence,
$\therefore a_T=\frac{dv}{dt}=\frac{1}{2\sqrt{1+t^2+t^4}}(2t+4t^3)$
At $t=1\text{s}$,
$a_T=\frac{1}{2\sqrt{1+1+1}}(2+4)=\frac{6}{2\sqrt{3}}=\sqrt{3}{\text{m/s}}^2$