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

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Answer 1 · 2017-01-13T09:39:58

The acceleration is $= ⟨ \frac{10}{9}, 6 ⟩$ $=〈10/9,6〉$ and the direction is $= 79.5$ $=79.5$ º

$v = ⟨ t - \frac{1}{t}, t^{2} ⟩$ $v=〈t-1/t,t^2〉$

Acceleration is

$a (t) = \frac{d v}{d t} = ⟨ 1 + \frac{1}{t^{2}}, 2 t ⟩$ $a(t)=(dv)/dt=〈1+1/t^2,2t〉$

Therefore,

$a (3) = ⟨ 1 + \frac{1}{9}, 6 ⟩ = ⟨ \frac{10}{9}, 6 ⟩$ $a(3)=〈1+1/9,6〉=〈10/9,6〉$

The direction is $= arctan (\frac{6}{\frac{10}{9}})$ $=arctan(6/(10/9))$

$= arctan (\frac{54}{10}) = 79.5$ $=arctan(54/10)=79.5$ º

An object's two dimensional velocity is given by v(t)=(t−1t,t2)v(t) = ( t-1/t, t^2). What is the object's rate and direction of acceleration at t=3t=3 ?