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 $t=1$ ?

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Answer 1 · 2017-02-18T09:51:53

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

$" direction tan^(-1)(7)$

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

thus.

$v(t)=(t-2,5t^2-3t)$

$a=(dv(t))/(dt)=dotv(t)=(1,10t-3)$

the acceleration is then (assuming the measurements all are SI

$dotv(1)==(1,10-3)=(1,7)ms^(-2)$

direction is $tan^(-1)(y/x)$

$" direction tan^(-1)(7)$

An object's two dimensional velocity is given by v(t) = ( t-2 , 5t^2-3t)v(t) = ( t-2 , 5t^2-3t). What is the object's rate and direction of acceleration at t=1 t=1 ?