## Statement

Knowing that the magnitude of the velocity of a body in S.I. units is:

$$v=7+2\xb7t+3\xb7{t}^{2}$$

Calculate the magnitude of the tangential acceleration.

## Solution

Knowing that the tangential acceleration is obtained by the following expression:

$$\overrightarrow{{a}_{t}}=\frac{dv}{dt}\overrightarrow{{u}_{t}}\Rightarrow \phantom{\rule{0ex}{0ex}}\left|\overrightarrow{{a}_{t}}\right|=\frac{dv}{dt}\left|\overrightarrow{{u}_{t}}\right|\Rightarrow \{\left|\overrightarrow{{u}_{t}}\right|=1\}\Rightarrow \phantom{\rule{0ex}{0ex}}\left|\overrightarrow{{a}_{t}}\right|=\frac{dv}{dt}\Rightarrow \phantom{\rule{0ex}{0ex}}\left|\overrightarrow{{a}_{t}}\right|=\frac{d(7+2t+3{t}^{2})}{dt}\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{)\left|\overrightarrow{{a}_{t}}\right|=6\xb7t+2}\phantom{\rule{0ex}{0ex}}$$