An object's two dimensional velocity is given by $v(t) = ( t-2t^3 , 4t^2-t)$ . What is the object's rate and direction of acceleration at $t=1$ ?

Question

1463 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-18T06:26:41

The rate of acceleration is $=8.6ms^-2$ in the direction $144.5^@$ anticlockwise from the x-axis

The acceleration is the derivative of the velocity

$a(t)=(dv)/dt$

Here,

$v(t)=(t-2t^3,4t^2-t)$

$a(t)=v'(t)=(1-6t^2,8t-1)$

Therefore,

When $t=1$

$a(1)=(1-6,8-1)=(-5,7)$

The rate of acceleration is

$||a(1)||=sqrt((-5)^2+(7)^2)=sqrt(25+49)=sqrt74=8.6ms^-2$

The direction of acceleration is

$theta=arctan(-5/7)=144.5^@$ anticlockwise from the x-axis

An object's two dimensional velocity is given by v(t) = ( t-2t^3 , 4t^2-t)v(t) = ( t-2t^3 , 4t^2-t). What is the object's rate and direction of acceleration at t=1 t=1 ?