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

Question

1550 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-20T10:44:20

$v_x(t) =3t - sqrt t$

$=>a_x(t) = dotv_x(t) = 3-1/(2sqrtt)$

$:. a_x(3) = 3-1/(2sqrt3)$

$v_y(t) =t^3 - 3 t$

$=>a_y(t) = dotv_y(t) = 3t^2-3$

$:. a_y(3) = 24$

hence, the object's acceleration is given as $sqrt(a_x^2+a_y^2)$

and its direction is given as $tantheta = v_y/v_x$

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