# An object's two dimensional velocity is given by v(t) = ( t-2 , 5t^2-3t). What is the object's rate and direction of acceleration at t=7 ?

Mar 7, 2017

$\text{acceleration} = \left(1 , 67\right) = \left(\vec{i} + 67 \vec{j}\right) m {s}^{- 2}$

direction$= {\tan}^{- 1} 67$with x-axis

#### Explanation:

If we are given velocity as a function of time, the acceleration is found by differentiating that function.

ie. $\text{ } a \left(t\right) = \dot{v} \left(t\right) = \frac{d \left(v \left(t\right)\right)}{\mathrm{dt}}$

$v \left(t\right) = \left(t - 2 , 5 {t}^{2} - 3 t\right)$

$a \left(t\right) = \dot{v} \left(t\right) = \left(1 , 10 t - 3\right)$

$a \left(7\right) = \left(1 , 70 - 3\right) = \left(1 , 67\right) = \left(\vec{i} + 67 \vec{j}\right) m {s}^{- 2}$

direction${\tan}^{- 1} \left(\frac{67}{1}\right) = {\tan}^{- 1} 67$ with x-axis