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

##### 1 Answer
May 6, 2017

$\text{answer :} a \left(1\right) = \left(0 , - 1\right)$

#### Explanation:

$\frac{d}{d t} v \left(t\right) = \left(\textcolor{red}{\frac{d}{d t} \left({t}^{2} - 2 t\right)} , \textcolor{g r e e n}{\frac{d}{d t} \left(\cos \pi t - t\right)}\right)$

$\frac{d}{d t} v \left(t\right) = a \left(t\right)$

$a \left(t\right) = \left(\textcolor{red}{2 t - 2} , \textcolor{g r e e n}{- \pi \sin \pi t - 1}\right)$

$\text{solve t=1}$

$a \left(1\right) = \left(2 \cdot 1 - 2 , - \pi \sin \pi \cdot 1 - 1\right)$

$a \left(1\right) = \left(0 , - \pi \sin \pi - 1\right)$

$\sin \pi = 0$

$a \left(1\right) = \left(0 , - 1\right)$
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