An object's velocity is given by $v (t) = (t^{2} - t + 1, t^{3} - 3 t)$ $v(t) = (t^2 -t +1 , t^3- 3t )$ . What is the object's rate and direction of acceleration at $t = 3$ $t=3$ ?

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Answer 1 · 2017-08-27T14:58:25

The rate of acceleration is $= 24.52 m s^{- 2}$ $=24.52ms^-2$ in the direction $= {78.2}^{\circ}$ $=78.2^@$ anticlockwise from the x-axis

The acceleration is the derivative of the velocity

$v (t) = (t^{2} - t + 1, t^{3} - 3 t)$ $v(t)=(t^2-t+1, t^3-3t)$

$a(t)=v'(t)=(2t-1, 3t^2-3)$

When $t=3$

$a(3)=(5,24)$

The rate of acceleration is

$||a(3)||=sqrt((5)^2+(24)^2)=sqrt(601)=24.52$

The direction is

$theta=arctan(24/5)=78.2^@$ anticlockwise from the x-axis

An object's velocity is given by v(t) = (t^2 -t +1 , t^3- 3t )v(t)=(t2−t+1,t3−3t)v(t) = (t^2 -t +1 , t^3- 3t ). What is the object's rate and direction of acceleration at t=3 t=3t=3 ?