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 $t = 1$ $t=1$ ?

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Answer 1 · 2018-06-02T08:14:21

The rate of acceleration is $= 5.069 m s^{- 2}$ $=5.069ms^-2$ in the direction $= {9.43}^{\circ}$ $=9.43^@$ clockwise from the x-axis.

The acceleration is the derivative of the velocity.

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

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

Therefore,

When $t=1$

$a(1)=(5,2cos2)= (5, -0.83)$

The rate of acceleration is

$||a(1)||=sqrt(5^2+(-0.83)^2)=5.069ms^-2$

The direction is

$theta=arctan(-0.83/5)=9.43^@$ 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=1 t=1t=1 ?