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

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Answer 1 · 2017-04-20T12:15:54

The rate of acceleration is $=5.1ms^-2$ in the direction of $=11.3º$

The acceleration is the derivative of the velocity.

$a(t)=v'(t)$

$v(t) = < t-1/t , t^2>$

$a(t)= <1+1/t^2,2t>$

When $t=1/2$

$a(1/2) = <1+1/(1/2)^2,2*1/2 >$

$=<5,1 >$

The rate of acceleration is

$=||a(1/2)||$

$=||< 5,1 >||$

$=sqrt((5)^2+(1)^2)$

$=5.1ms^-2$

The direction is in the 1st quadrant

$theta=arctan(1/5)$

$=11.3º$

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