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

Question

1580 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-10T14:22:19

The rate of acceleration is $=3.01ms^-2$ in the direction $=266.6^@$ anticlockwise from the $"x-axis"$

The acceleration is the derivative of the velocity.

$v(t)=(tsin(pi/3t), 2cos(pi/2t)-3t)$

Therefore,

$a(t)=v'(t)=(sin(pi/3t)+pi/3tcos(pi/3t), -pisin(pi/2t)-3)$

When $t=2$

The acceleration is

$a(2)=v'(2)=(sin(pi/3*2)+pi/3*2*cos(pi/3*2), -pisin(pi/2*2)-3)$

$a(2)=(sqrt3/2-pi/3, -3)=(-0.18,-3)$

The rate of acceleration is

$=||a(2)||=sqrt((-0.18)^2+(-3)^2)=3.01ms^-2$

The direction is $theta=180+arctan (3/0.18)=266.6^@$ anticlockwise from the $"x-axis"$

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