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 $t=2$ ?

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Answer 1 · 2016-05-26T15:49:43

$a=17.03" "m/s^2$

$tan alpha=17$

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

$(d v(t))/(d t)=a(t)$

$a_x(t)=d/(d t) (t-2)=1" "a_x=1" constant"$

$a_y(t)=d/(d t)(5t^2-3t)=10t-3$

$a_y(2)=10*2-3$

$a_y(2)=17$

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$a=sqrt(a_x^2+a_y^2)$

$a=sqrt(1^2+17^2)$

$a=17.03" "m/s^2$

$tan alpha=17$

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=2 t=2 ?