An object's two dimensional velocity is given by $v (t) = (cos t, - t^{3} + 4 t)$ $v(t) = ( cost , -t^3 +4t )$ . What is the object's rate and direction of acceleration at $t = 1$ $t=1$ ?

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Answer 1 · 2018-04-13T07:28:07

The rate of acceleration is $= 1.31 m s^{- 1}$ $=1.31ms^-1$ in the direction $= 130^{\circ}$ $=130^@$ anticlockwise from the x-axis

The acceleration is the derivative of the velocity

$v (t) = (cos t, - t^{3} + 4 t)$ $v(t)=(cost, -t^3+4t)$

$a(t)=v'(t)=(-sint, -3t^2+4)$

When $t=1$

$a(1)=(-sin1, -3+4)=<-0.84, 1>$

The rate of acceleration of the object is

$||a(1)||= ||<-0.84, 1>||= sqrt((-0.84)^2+1^2)=sqrt1.71=1.31ms^-2$

The direction is

$theta=arctan(-1/0.84)=arctan(-1.19)=130^@$ anticlockwise from the x-axis

An object's two dimensional velocity is given by v(t) = ( cost , -t^3 +4t )v(t)=(cost,−t3+4t)v(t) = ( cost , -t^3 +4t ). What is the object's rate and direction of acceleration at t=1 t=1t=1 ?