# An object's velocity is given by v(t) = (t^2 +t +1 , t^3- 3t ). What is the object's rate and direction of acceleration at t=6 ?

Apr 23, 2016

$\vec{v} \left(t\right) = \left({t}^{2} + t + 1\right) \hat{i} + \left({t}^{3} - 3 t\right) \hat{j}$
Differentiating w.r.t. twe have acceleration
$\vec{a} \left(t\right) = \frac{d}{\mathrm{dt}} \left(v \left(t\right)\right) = \left(2 t + 1\right) \hat{i} + \left(3 {t}^{2} - 3\right) \hat{j}$
Acceleration at t=6 is
$\vec{a} \left(6\right) = \left(2 \times 6 + 1\right) \hat{i} + \left(3 \times {6}^{2} - 3\right) \hat{j} = 13 \hat{i} + 105 \hat{j}$
Magnitude of acceleration
$| \vec{a} \left(6\right) | = \sqrt{{13}^{2} + {105}^{2}} = 105.8$
Direction =${\tan}^{-} 1 \left(\frac{105}{13}\right) \approx {83}^{o}$ with the x-axis

Apr 23, 2016

$a = 105 , 80 \text{ } \frac{u n i t}{s} ^ 2$

#### Explanation:

$a \left(t\right) = \frac{d}{d t} v \left(t\right) \text{ derivative of v(t) give us a(t)}$

$\text{derivative of v(t) for x direction:}$
$\text{...................................................}$
${a}_{x} \left(t\right) = \frac{d}{d t} \left({t}^{2} + t + 1\right)$

${a}_{x} \left(t\right) = 2 t + 1$

$\text{fill in t=6}$

${a}_{x} \left(6\right) = 2 \cdot 6 + 1 = 13$

${a}_{x} \left(6\right) = 13$

$\text{derivative of v(t) for y direction}$
$\text{...................................................}$
${a}_{y} \left(t\right) = \frac{d}{d t} \left({t}^{3} - 3 t\right)$

${a}_{y} \left(t\right) = 3 {t}^{2} - 3$

$\text{fill in t=6}$

${a}_{y} \left(6\right) = 3 \cdot {6}^{2} - 3 = 3 \cdot 36 - 3 = 108 - 3 = 105$

$\text{acceleration is a vector quantity so that we have to add " a_x and a_y " as vector}$

$a = \sqrt{{13}^{2} + {105}^{2}}$

$a = \sqrt{169 + 11025}$

$a = \sqrt{11194}$

$a = 105 , 80 \text{ } \frac{u n i t}{s} ^ 2$