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

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Answer 1 · 2018-01-09T08:37:39

The rate of acceleration is $=sqrt(5)ms^-2$ in the direction $=153.4^@$ anticlockwise from the $"x-axis"$

The acceleration is the derivative of the velocity

$v(t)=(1/t, t^2)$

Therefore,

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

When $t=1$ , the acceleration is

$a(1)=(-1,2)$

The rate of acceleration is

$||a(1)||=sqrt((-1)^2+(2)^2)=sqrt(5)$

The direction of acceleration is

$theta=arctan(-1/2)=153.4^@$ anticlockwise from the $"x-axis"$

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