Velocity and Acceleration

Calculus Level 4

A particle moves according to the following movement's law:

$\vec{r}(t)=(3{ t }^{ 2 }+5)\vec{{ E }_{ x }} - 2t\vec{{ E }_{ y }} (I.S.)$

For $t=1s$ , calculate the angle (to the nearest degree) between $\vec{v}$ and $\vec{a}$ .

The answer is 18.

1 solution

Joao Miguel Coelho
May 10, 2016

$\vec{v}(t)=\frac { d\vec { r } }{ dt }$ = $6t \vec{{ E }_{ x }} - 2\vec{{ E }_{ y }}$

$\vec{v}(1)$ = $6 \vec{{ E }_{ x }} - 2\vec{{ E }_{ y }}$ .

$\vec { a } (t)$ = $\frac { d\vec { v } }{ dt }$ = $6 \vec{{ E }_{ x }}$

$\vec { a } (1)$ = $6 \vec{{ E }_{ x }}$

$\vec { a }(1) \cdot \vec{v}(1)$ = $\left\| \vec { a }(1) \right\|\times \left\| \vec{v}(1) \right\| \times \ cos\theta$ $\Leftrightarrow$ $cos\theta$ = $\frac { \vec { a }(1) \cdot \vec{v}(1) }{ \left\| \vec { a }(1) \right\|\times \left\| \vec{v}(1) \right\| }$ $\Leftrightarrow$ $cos\theta$ = $\frac { 6\times 6+(-2)\times 0 }{ \sqrt { { 6 }^{ 2 } } \times \sqrt { { 6 }^{ 2 }+{ (-2) }^{ 2 } } }$ $\Leftrightarrow$ $\theta \approx 18º$

Velocity and Acceleration

The answer is 18.

1 solution

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