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

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

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Answer 1 · 2017-07-16T17:40:56

$a = 8.06$ $a = 8.06$ ${LT}^{- 2}$ $"LT"^-2$

$θ = - {7.13}^{o}$ $theta = -7.13^"o"$

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

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

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

Finding the derivatives:

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

$a_{y} (t) = \frac{d}{d x} [cos (π t) - t] = - π sin (π t) - 1$ $a_y(t) = d/(dx)[cos(pit)-t] = -pisin(pit) - 1$ (I'll assume this is in radians**)

Plugging in $t = 5$ $t = 5$ (no units), we have

$a_{x} = 2 (5) - 2 = 8$ $a_x = 2(5)-2= 8$ ${LT}^{- 2}$ $"LT"^-2$

$a_{y} = - π sin (π (5)) - 1 = - 1$ $a_y = -pisin(pi(5))-1 = -1$ ${LT}^{- 2}$ $"LT"^-2$

(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

$a = \sqrt{{(a_{x})}_{x}^{2} + {(a_{y})}_{y}^{2}} = \sqrt{8^{2} + {(- 1)}^{2}} = 8.06$ $a = sqrt((a_x)^2 + (a_y)^2) = sqrt(8^2 + (-1)^2) = color(red)(8.06$ ${LT}^{- 2}$ $color(red)("LT"^-2$

And the direction is

$θ = arctan (\frac{a_{y}}{a_{x}}) = arctan (\frac{- 1}{8}) = - {7.13}^{o}$ $theta = arctan((a_y)/(a_x)) = arctan((-1)/8) = color(blue)(-7.13^"o"$

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

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Explanation:

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

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Explanation:

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