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

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Answer 1 · 2017-07-15T04:11:29

$a = 4.12$ $a = 4.12$ ${LT}^{- 2}$ $"LT"^-2$

$ϕ = - {14.0}^{o}$ $phi = -14.0^"o"$

Explanation:

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

In component form, the velocity is

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

$v_{y} (t) = cos (π t) - t$ $v_y(t) = cos(pit)-t$

To find the acceleration as a function of time, we must differentiate the velocity equations:

$a_{x} (t) = \frac{d}{d t} [t^{2} - 2] = 2 t$ $a_x(t) = d/(dt) [t^2-2] = 2t$

$a_{y} (t) = \frac{d}{d t} [cos (π t) - t] = - π sin (π t) - 1$ $a_y(t) = d/(dt) [cos(pit)-t] = -pisin(pit) - 1$

At $t = 2$ $t = 2$ (ambiguous units here), we have

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

$a_{y} (2) = - π sin (2 π) - 1 = - 1$ $a_y(2) = -pisin(2pi) - 1 = -1$ ${LT}^{- 2}$ $"LT"^-2$

(The term ${LT}^{- 2}$ $"LT"^-2$ is the dimensional form the units for acceleration, $length$ $"length"$ $\times$ $xx$ ${time}^{- 2}$ $"time"^-2$ . I used it here simply because no units were given.)

The magnitude of the acceleration is

$a = \sqrt{{(a_{x})}_{x}^{2} + {(a_{y})}_{y}^{2}} = \sqrt{4^{2} + {(- 1)}^{2}} = 4.12$ $a = sqrt((a_x)^2 + (a_y)^2) = sqrt(4^2 + (-1)^2) = color(red)(4.12$ ${LT}^{- 2}$ $color(red)("LT"^-2$

The direction is

$ϕ = arctan (\frac{a_{y}}{a_{x}}) = arctan (\frac{- 1}{4}) = - {14.0}^{o}$ $phi = arctan((a_y)/(a_x)) = arctan((-1)/4) = color(blue)(-14.0^"o"$

Always check the arctangent calculation to make sure your answer is not $180^{o}$ $180^"o"$ off!

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

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Explanation:

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An object's two dimensional velocity is given by v(t) = ( t^2 +2 , cospit - t )v(t)=(t2+2,cosπt−t)v(t) = ( t^2 +2 , cospit - 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 $v (t) = (t^{2} + 2, cos π t - t)$ $v(t) = ( t^2 +2 , cospit - t )$ . What is the object's rate and direction of acceleration at $t = 2$ $t=2$ ?