An object's two dimensional velocity is given by $v (t) = (5 t - t^{3}, 3 t^{2} - 2 t)$ $v(t) = ( 5t-t^3 , 3t^2-2t)$ . What is the object's rate and direction of acceleration at $t = 7$ $t=7$ ?

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Answer 1 · 2017-05-14T07:15:16

The rate of acceleration is $= 147.53 m s^{- 1}$ $=147.53ms^-1$ in the direction of $=164.27º$ anticlockwise from the x-axis.

The acceleration is the derivative of the velocity.

$v(t)=(5t-t^3, 3t^2-2t)$

$a(t)=v'(t)=(5-3t^2, 6t-2)$

Therefore,

$a(7)=(5-3*49, 6*7-2)=(-142,40)$

The rate of acceleration is

$||a(4)||=sqrt(142^2+40^2)$

$=sqrt21764$

$=147.53ms^-2$

The direction is

$theta=arctan(-40/142)$

$theta$ lies in the 2nd quadrant

$theta=180-15.73=164.27$

An object's two dimensional velocity is given by v(t) = ( 5t-t^3 , 3t^2-2t)v(t)=(5t−t3,3t2−2t)v(t) = ( 5t-t^3 , 3t^2-2t). What is the object's rate and direction of acceleration at t=7 t=7t=7 ?