An object's two dimensional velocity is given by $v (t) = (t - 2, 2 cos (\frac{π}{2} t) - t)$ $v(t) = ( t-2 , 2cos(pi/2t )- t )$ . What is the object's rate and direction of acceleration at $t = 3$ $t=3$ ?

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Answer 1 · 2017-05-21T07:12:59

The rate of acceleration is $= 2.36 m s^{- 2}$ $=2.36ms^-2$ in the direction
$theta=65º$ anticlockwise from the x-axis

The acceleration is the derivative of the velocity.

$v(t)=(t-2, 2cos(pi/2t)-t)$

$a(t)=v'(t)=(1, -pisin(pi/2t)-1)$

Therefore,

$a(3)=(1, pi-1)=(1,2.14)$

The rate of acceleration is

$||a(4)||=sqrt(1+2.14^2)$

$=sqrt5.59$

$=2.36ms^-2$

The direction is

$theta=arctan((pi-1))$

$theta$ lies in the 1st quadrant

$theta=65º$ anticlockwise from the x-axis

An object's two dimensional velocity is given by v(t) = ( t-2 , 2cos(pi/2t )- t )v(t)=(t−2,2cos(π2t)−t)v(t) = ( t-2 , 2cos(pi/2t )- t ). What is the object's rate and direction of acceleration at t=3 t=3t=3 ?