An object's two dimensional velocity is given by $v(t) = ( t^2 - 2t , 1- 3t )$ . What is the object's rate and direction of acceleration at $t=7$ ?

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Answer 1 · 2018-03-13T07:20:18

The rate of acceleration is $=12.37ms^-2$ in the direction $-14.04^@$ clockwise from the x-axis.

The acceleration is the derivative of the velocity

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

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

When $t=7$

$a(7)=(2*7-2,-3)=(12,-3)$

The rate of acceleration is

$||a(7)||=||(12,-3)||=sqrt(12^3+(-3)^2)=sqrt(144+9)=sqrt153=12.37ms^-2$

The angle is

$theta=arctan(-3/12)=arctan(-1/4)=-14.04^@$ clockwise from the x-axis.

An object's two dimensional velocity is given by v(t) = ( t^2 - 2t , 1- 3t )v(t) = ( t^2 - 2t , 1- 3t ). What is the object's rate and direction of acceleration at t=7 t=7 ?