Question

The position of a point at time ${}^{\prime }{t}^{\prime }$ is given by $x=1+2t+3t^2$ and $y=2-3t+4t^2$. Then its acceleration at time $t$ is

Answer 1

Given $x=1+2t+3t^2$ and $y=2-3t+4t^2$
$\Rightarrow \frac{dx}{dt}=2+6t;\frac{dy}{dt}=-3+8t$
$\Rightarrow \frac{d^{2}x}{d{t}^2}=6;\frac{d^{2}y}{d{t}^2}=8$
$\therefore \text{acceleration}=\sqrt{{\left(\frac{d^{2}x}{d{t}^2}\right)}^2+{\left(\frac{d^{2}y}{d{t}^2}\right)}^2}$
$=\sqrt{6^2+8^2}=10$