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

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Answer 1 · 2017-04-26T16:56:52

The acceleration at $t=2$ is $a(t) = (-1/4, 4)$ , whose absolute value is $sqrt257/4$ and the direction is $93.576^@$ w.r.t. $x$ -axis.

Acceleration is the first derivative of velocity, so if we differentiate the velocity function, we get a function for the acceleration:

$(dv)/(dt) = (-1/t^2, 2t)$

Inserting $t=2$ gives us the acceleration at $t= 2s$ , namely

$a(t) = (-1/4, 4)$ , whose absolute value is $sqrt((-1/4)^2+4^2)=sqrt(257/16)=sqrt257/4$

The direction is given by $alpha=tan^(-1)(4/(-1/4))=tan^(-1)(-16)=93.576^@$ w.r.t. $x$ -axis.

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