I am a free Particle!

This quite an easy one. By differentiating both equations we get $x=8t+2$ $y=2$ Now substitute the value $t=3$ $x=8*(3)+2=26$ $y=2$ Now to find the magnitude of the velocity do $v=\sqrt{26^2 +2^2}=\sqrt{680}=26.08$

differentiating both components of distance, we get the velocity vector V = $< 8t+2,2>$ . The magnitude of the velocity can be found by $√(8t+2)^2+2^2$ . Substituting $t=3$ into this expression, we get

$|V| = 26.08$

I did it in a slightly different way. Let $v_t$ be the velocity of the particle after time $t$ . Given that ----

$x=4t^2+2t+3 \quad \text{and} \quad y=2t+8$

$\implies \displaystyle \frac{dx}{dt}=8t+2 \quad \text{and} \quad \frac{dy}{dt}=2$

By vector resolution, we can split $v_t$ into two components, namely $v_x$ and $v_y$ which are velocity components along X-axis and Y-axis respectively. If $\theta_{inst}$ be the instantaneous angle after time $t$ between $v_t$ and the horizontal, then we have $v_x=\frac{dx}{dt}=v_t \cos (\theta_{inst}) \quad \text{and} \quad v_y=\frac{dy}{dt}=v_t \sin (\theta_{inst})$ . So,

$v_x=8t+2 \quad \text{and} \quad v_t=2$

$\implies v_t \cos (\theta_{inst})=8t+2 \quad \text{and} \quad v_t \sin (\theta_{inst})=2$

At time $t=3$ , taking $\theta_{inst}=\theta$ , we have ----

$v_3 \cos \theta=26 ....(i) \quad \text{and} \quad v_3 \sin \theta=2 .... (ii)$

From (i) and (ii), we have ---

$\tan \theta=\frac{2}{26}$

$\implies \theta= \arctan{(\frac{2}{26})} = \boxed{0.07677189126}$

Putting the value of $\theta_{inst}$ in (i), we get ---

$v_3=\frac{26}{\cos (0.07677189126)} = \frac{26}{0.9970544855}$

$\implies v_3=26.0768096208 \approx \boxed{26.08}$