An object's velocity is given by $v(t) = (2t^2 +t +1 , sin2t )$ . What is the object's rate and direction of acceleration at $t=3$ ?

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Answer 1 · 2017-04-07T10:07:06

The rate of acceleration is $=13.04ms^-2$ in the direction of $=4.4º$

The acceleration is the derivative of the velocity.

$v(t)=(2t^2+t+1,sin2t)$

$a(t)=v'(t)=(4t+1,2cos2t)$

Therefore,

$a(3)=(13,2cos6)=(13,0.96)$

The rate of acceleration is

$||a(3)||=sqrt(13^2+0.96^2)=13.04ms^-2$

The direction is

$theta=arctan(0.96/13)=4.4º$

An object's velocity is given by v(t) = (2t^2 +t +1 , sin2t )v(t) = (2t^2 +t +1 , sin2t ). What is the object's rate and direction of acceleration at t=3 t=3 ?