Question

The position vector of a particle changes with time according to the relation $r(t)= 15\,t^2 \hat{i} + (4 -20\, t^2)\hat{j}$. What is the magnitude of the acceleration (in $m/s^2)$ at $t = 1$?

Answer 1

$\vec{ r} = 15t^2 \hat{i} + (4 - 20\, t^2) \hat{j}$
$\vec{v} =\frac{d\vec{r}}{dt} = 30 t\hat{i} - 40 t\hat{j}$
Acceleration, $\vec{a} = \frac{d\vec{v}}{dt} = 30 \hat{i} - 40 \hat{j}$
$\therefore a = \sqrt{30^2 + 40^2} = 50\,m/s^2$