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# Finding Speed from Position Vector

APCALC-ZYE1M1

The position vector of a particle moving in the xy-plane for $t\geq0$ is $\vec{p}=\left(t^2+t+1\right)\vec{i}+\left(3t^2-8t\right)\vec{j}$.

Find the function representing the speed of the particle for $t\geq0$.

A

$\text{speed}=\sqrt{2t+1}$

B

$\text{speed}=\sqrt{6t-8}$

C

$\text{speed}=8t-7$

D

$\text{speed}=\sqrt{40t^2-92t+65}$