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

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

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Answer 1 · 2017-07-17T00:08:10

$a = 10.0$ $a = 10.0$ ${LT}^{- 2}$ $"LT"^-2$

$θ = - {5.65}^{o}$ $theta = -5.65^"o"$

Explanation:

We're asked to find the the magnitude (rate) and direction of an object's acceleration at a given time, given its velocity equation.

To do this, we need too differentiate the velocity component equations, which are given as

$v_{x} (t) = 2 t^{2} - 2 t + 1$ $v_x(t) = 2t^2 - 2t + 1$

$v_{y} (t) = sin (t)$ $v_y(t) = sin(t)$

Finding the derivatives:

$a_{x} (t) = \frac{d}{d x} [2 t^{2} - 2 t + 1] = 4 t - 2$ $a_x(t) = d/(dx) [2t^2 -2t + 1] = 4t - 2$

$a_{y} (t) = \frac{d}{d x} [sin t] = cos t$ $a_y(t) = d/(dx)[sint] = cost$ (I'll assume this is in radians**)

Plugging in $t = 3$ $t = 3$ (no units), we have

$a_{x} = 4 (3) - 2 = 10$ $a_x = 4(3) - 2 = 10$ ${LT}^{- 2}$ $"LT"^-2$

$a_{y} = cos (3) = - 0.990$ $a_y = cos(3) = -0.990$ ${LT}^{- 2}$ $"LT"^-2$

(The ${LT}^{- 2}$ $"LT"^-2$ term is the dimensional form of the units for acceleration ( $distance \times {time}^{- 2}$ $"distance"xx"time"^-2$ ). I used this term here since no units were given.)

The magnitude of the acceleration is thus

$a = \sqrt{{(a_{x})}_{x}^{2} + {(a_{y})}_{y}^{2}} = \sqrt{10^{2} + {(- 0.990)}^{2}} = 10.0$ $a = sqrt((a_x)^2 + (a_y)^2) = sqrt(10^2 + (-0.990)^2) = color(red)(10.0$ ${LT}^{- 2}$ $color(red)("LT"^-2$

And the direction is

$θ = arctan (\frac{a_{y}}{a_{x}}) = arctan (\frac{- 0.990}{10}) = - {5.65}^{o}$ $theta = arctan((a_y)/(a_x)) = arctan((-0.990)/10) = color(blue)(-5.65^"o"$

Always make sure your arctangent calculation is in the right direction; it could be $180^{o}$ $180^"o"$ off!

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

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Explanation:

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

1 Answer

Explanation:

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