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

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Answer 1 · 2017-11-21T06:51:28

The rate of acceleration is $==23ms^-2$ in the direction $=269.7^@$ anticlockwise from the x-axis.

The acceleration is the derivative of the velocity

$v(t)=(cost,-t^3+4t)$

$a(t)=v'(t)=(-sint,-3t^2+4)$

When $t=3$

$a(3)=(-sin(3), -3^3+4)=(-0.14,-23)$

The rate of acceleration is

$=||a(3)||=sqrt((-0.14)^2+(-23)^2)=23ms^-2$

The direction is

$theta=arctan(-23/-0.14)=269.7^@$ anticlockwise from the x-axis.

Answer 2 · 2017-11-21T06:57:13

$23$ units and $4.71$ radian

As two dimensional velocity of object is given by $v(t)=(cost,-t^3+4t)$

accelaration is given by $(dv)/(dt)$ i.e.

$a(t)=(-sint,-3t^2+4)$

and at $t=3$ , this is $-sin3hati+(-3*3^2+4)hatj$

or $a(t=3)=-0.14112hati-23hatj$

and its magnitude is $sqrt((0.14112)^2+23^2)=sqrt529.02=23$

and direction is $pi+tan^(-1)23/0.14112=pi+1.569=4.71$ radian

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