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

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Answer 1 · 2016-09-26T22:36:06

The vector is $a(t) = ((1/2)(t - 2)^(-1/2) - 1, 2t)$ Substitute 3 for t, and then compute the magnitude and direction as you would any vector.

$a(3) = ((1/2)(3 - 2)^(-1/2) - 1, 2(3))$

$a(3) = (-1/2, 6)$

The rate of acceleration is the magnitude:

$|a(3)| = sqrt((-1/2)^2 + 6^2)$

$a ~~ 6.02 (units)/s^2$

$theta = tan^-1(6/(-1/2))$

$theta ~~ -85.2°$

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