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

Question

1580 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-23T06:12:04

The rate of acceleration is $=24.52ms^-2$ in the dorection $=78.2^@$ anticlockwise from the x-axis

The acceleration is the derivative of the velocity

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

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

At $t=3$ , the acceleration is

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

The rate of acceleration is

$=||a(3)||=sqrt(5^2+24^2)=sqrt601=24.52ms^-2$

The direction is

$theta=arctan(24/5)=78.2^@$

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