An object's two dimensional velocity is given by $v (t) = (3 t^{2} - 5 t, t)$ $v(t) = ( 3t^2 - 5t , t )$ . What is the object's rate and direction of acceleration at $t = 2$ $t=2$ ?

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An object's two dimensional velocity is given by $v (t) = (3 t^{2} - 5 t, t)$ $v(t) = ( 3t^2 - 5t , t )$ . What is the object's rate and direction of acceleration at $t = 2$ $t=2$ ?

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Answer 1 · 2016-07-03T10:06:54

$\to a (t) = (7 ˆ i + ˆ j) \frac{m}{s^{2}}$ $veca(t)=(7hati+hatj)m/s^2$

Explanation:

I'm guessing they're using normal convention replaced by brackets.

So, in vectorial form, $- - \to v (t) = (3 t^{2} - 5 t) ˆ i + t ˆ j$ $vec{v(t)}=(3t^2-5t)hati+thatj$

Acceleration is the time derivative of velocity, so you go that.
$\to a = \frac{d}{d t} (\to v (t)) = \frac{d}{d t} (3 t^{2} - 5 t) ˆ i + \frac{d}{d t} (t) ˆ j$ $veca=d/dt(vecv(t))=d/dt(3t^2-5t)hati+d/dt(t)\hatj$

I'm sure you know differentiation, hence we'll get acceleration
$\to a (t) = (6 t - 5) ˆ i + 1 ˆ j$ $veca(t)=(6t-5)hati+1hatj$

We need to find acceleration at time $t = 2$ $t=2$ , substitute,
$\to a {(t)}_{t = 2} = (6 \cdot 2 - 5) ˆ i + ˆ j$ $veca(t)_{t=2}=(6*2-5)hati+hatj$

In the end, you'll get why the answer is as given in the box.

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An object's two dimensional velocity is given by $v (t) = (3 t^{2} - 5 t, t)$ $v(t) = ( 3t^2 - 5t , t )$ . What is the object's rate and direction of acceleration at $t = 2$ $t=2$ ?

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An object's two dimensional velocity is given by v(t) = ( 3t^2 - 5t , t )v(t)=(3t2−5t,t)v(t) = ( 3t^2 - 5t , t ). What is the object's rate and direction of acceleration at t=2 t=2t=2 ?

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Explanation:

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An object's two dimensional velocity is given by $v (t) = (3 t^{2} - 5 t, t)$ $v(t) = ( 3t^2 - 5t , t )$ . What is the object's rate and direction of acceleration at $t = 2$ $t=2$ ?