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

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Answer 1 · 2016-04-23T09:43:51

$\to a = - 0, 85 ˆ i + 4 ˆ j$ $vec a=-0,85 hati+4 hatj$
$θ = - 78^{o}$ $theta=-78^o$

$\to a (t) = \to a ˆ i + \to a ˆ j$ $vec a(t)=vec a hat i+ vec a hat j$

$\to a (t) = \frac{d}{d t} (\sqrt{t^{2} - 1} - 2 t) ˆ i + \frac{d}{d t} (t^{2}) ˆ j$ $vec a(t)=d/(d t)(sqrt(t^2-1)-2t)hat i+d/(d t)(t^2)hat j$

$\to a (t) = (\frac{2 t}{2 \cdot \sqrt{t^{2} - 1}} - 2) ˆ i + (2 t) ˆ j$ $vec a(t)=((2t)/(2*sqrt(t^2-1))-2)hat i+(2t)hat j$

$\to a (2) = (\frac{2 \cdot 2}{2 \cdot \sqrt{2^{2} - 1}} - 2) ˆ i + (2 \cdot 2) ˆ j$ $vec a(2)=((2*2)/(2*sqrt(2^2-1))-2)hat i+(2*2)hat j$

$\to a (2) = (\frac{2}{\sqrt{3}} - 2) ˆ i + 4 ˆ j$ $vec a(2)=(2/sqrt3-2)hat i+4hatj$

$\to a (2) = (\frac{2 - 2 \sqrt{3}}{\sqrt{3}}) ˆ i + 4 ˆ j$ $vec a(2)=((2-2sqrt3)/sqrt3)hat i+4 hatj$

$\to a = (\frac{2 \sqrt{3} - 6}{3}) ˆ i + 4 ˆ j$ $vec a=((2sqrt3-6)/3)hati+4 hatj$

$\to a = - 0, 85 ˆ i + 4 ˆ j$ $vec a=-0,85 hati+4 hatj$

$∣ ∣ ∣ ∣ \to a ∣ ∣ ∣ ∣ = \sqrt{{(- 0, 85)}^{2} + 4^{2}} magnitude of a$ $||vec a ||=sqrt((-0,85)^2+4^2)" magnitude of a"$

$a = 4, 09 \frac{unit}{s^{2}}$ $a=4,09 " "("unit")/s^2$

$tan θ = \frac{4}{- 0, 85}$ $tan theta=4/(-0,85)$

$θ = - 78^{o}$ $theta=-78^o$

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