# 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=1 ?

##### 1 Answer
Feb 18, 2017

"acceleration"=dotv(1))=(1,7)ms^(-2)

" direction tan^(-1)(7)

#### Explanation:

if velocity is given a s a function of time the acceleration is found by differentiating the velocity function.

thus.

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

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

the acceleration is then (assuming the measurements all are SI

$\dot{v} \left(1\right) = = \left(1 , 10 - 3\right) = \left(1 , 7\right) m {s}^{- 2}$

direction is ${\tan}^{- 1} \left(\frac{y}{x}\right)$

" direction tan^(-1)(7)