An object's two dimensional velocity is given by $v(t) = ( t-2 , 2cos(pi/2t )- t )$ . What is the object's rate and direction of acceleration at $t=5$ ?

Question

1341 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-19T06:57:46

The rate of acceleration is $=4.26ms^-2$ in the direction of $=76.4º$ clockwise 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,

$a(5)=(1, -pi-1)=(1,-4.14)$

The rate of acceleration is

$||a(4)||=sqrt(1+4.14^2)$

$=sqrt18.15$

$=4.26ms^-2$

The direction is

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

$theta$ lies in the 4th quadrant

$theta=-76.4º$

An object's two dimensional velocity is given by v(t) = ( t-2 , 2cos(pi/2t )- t )v(t) = ( t-2 , 2cos(pi/2t )- t ). What is the object's rate and direction of acceleration at t=5 t=5 ?