Is speed vector or scalar?

$a\cdot b=\left|a\times b\right|ab$ $cos \theta =ab sin \theta$

$\frac{\underset{\_}{ab}}{\underset{\_}{ab}}=\frac{sin\theta }{cos\theta }$

$tan \theta =1$

$\left|\widehat{A}+\widehat{B}\right|=\sqrt[]{A^2+B^2+2ABcos\theta }$

$=\sqrt[]{1+1+2cos\theta }$

$=\sqrt[]{2\left(1+cos\theta \right)}$

= $\sqrt[]{2\times 2co{s}^2\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$\left|\widehat{A}+\widehat{B}\right|=2cos\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$  -(1)

$\left|\widehat{A}-\widehat{B}\right|=\sqrt[]{A^2+B^2-2ABcos\theta }$

$=\sqrt[]{1+1-2cos\theta }$

$=2sin\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$  -(2)

(2) $÷$ $$ (1)

$\frac{\left|\widehat{A}-\widehat{B}\right|}{\left|\widehat{A}+\widehat{B}\right|}=\frac{2sin\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{2cos\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$\Rightarrow \left|\widehat{A}-\widehat{B}\right|=\left|\widehat{A}+\widehat{B}\right|tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Mass and temperature are scalars. We can fully describe them by magnitude alone and do not involve direction. 

Displacement, velocity, and acceleration are common vectors. These three vectors they have both magnitude and direction, and  follow vector addition rules.