An object's velocity is given by $v (t) = (2 t^{2} + t + 1, sin 2 t)$ $v(t) = (2t^2 +t +1 , sin2t )$ . What is the object's rate and direction of acceleration at $t = 5$ $t=5$ ?

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Answer 1 · 2017-08-21T11:56:30

The rate of acceleration is $= 21.07 m s^{- 2}$ $=21.07ms^-2$ in the direction of $= {4.6}^{\circ}$ $=4.6^@$ clockwise from the x-axis

The acceleration is the derivative of the velocity

$v (t) = (2 t^{2} + t + 1, sin 2 t)$ $v(t)=(2t^2+t+1,sin2t)$

$a(t)=v'(t)=(4t+1,2cos2t)$

When $t=5$

$a(5)=(21,2cos10)=(21,-1.68)$

The rate of acceleration is

$||a(5)||=sqrt(21^2+(-1.68)^2)=sqrt(443.8)=21.07$

The direction is

$theta=arctan(-1.68/21)=4.6^@$ clockwise from the x-axis

An object's velocity is given by v(t) = (2t^2 +t +1 , sin2t )v(t)=(2t2+t+1,sin2t)v(t) = (2t^2 +t +1 , sin2t ). What is the object's rate and direction of acceleration at t=5 t=5t=5 ?