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

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Answer 1 · 2016-11-18T19:38:29

$| ¯ a | \approx 12.04, θ \approx 6.2 radians$ $|bara| ~~ 12.04, theta ~~ 6.2" radians"$

$\frac{d v_{x}}{d t} = 2 t + 2$ $(dv_x)/dt = 2t + 2$

$\frac{d v_{y}}{d t} = - 2 π sin (π t) - 1$ $(dv_y)/dt = -2pisin(pit) - 1$

Evaluate both at t = 5:

$\frac{d v_{x}}{d t} ∣_{t = 5} = 12$ $(dv_x)/dt|_(t=5) = 12$

$\frac{d v_{y}}{d t} ∣_{t = 5} = - 1$ $(dv_y)/dt|_(t=5) = -1$

The acceleration vector is:

$¯ a = 12 ˆ i - ˆ j$ $bara = 12hati - hatj$

The magnitude of the acceleration is:

$| ¯ a | = \sqrt{12^{2} + {(- 1)}^{2}} = \sqrt{145} \approx 12.04$ $|bara| = sqrt(12^2 + (-1)^2) = sqrt(145) ~~ 12.04$

The direction is:

$θ = 2 π + {tan}^{- 1} (- \frac{1}{12})$ $theta = 2pi + tan^-1(-1/12)$

Note: we add $2 π$ $2pi$ , because the angle is in the 4th quadrant.

$θ = 2 π + {tan}^{- 1} (- \frac{1}{12})$ $theta = 2pi + tan^-1(-1/12)$

$θ \approx 6.2 radians$ $theta ~~ 6.2 " radians"$

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