An object's two dimensional velocity is given by $v (t) = (3 t^{2} - 2 t, 1 - 3 t)$ $v(t) = ( 3t^2 - 2t , 1- 3t )$ . What is the object's rate and direction of acceleration at $t = 5$ $t=5$ ?

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An object's two dimensional velocity is given by $v (t) = (3 t^{2} - 2 t, 1 - 3 t)$ $v(t) = ( 3t^2 - 2t , 1- 3t )$ . What is the object's rate and direction of acceleration at $t = 5$ $t=5$ ?

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Answer 1 · 2017-06-24T00:10:10

The acceleration is $28.2$ $28.2$ $m s^{- 2}$ $ms^-2$ at $- 84^{o}$ $-84^o$ : that is, at an angle in the third quadrant, downward and to the right.

Explanation:

The acceleration is the rate of change of the velocity, and is found as the first derivative of the velocity. We can treat the $x$ $x$ and $y$ $y$ directions independently.

$v_{x} = 3 t^{2} - 2 t$ $v_x=3t^2-2t$

$\frac{d v_{x}}{d t} = 6 t - 2$ $(dv_x)/dt=6t-2$

$v_{y} = 1 - 3 t$ $v_y=1-3t$

$\frac{d v_{y}}{d t} = - 3$ $(dv_y)/dt=-3$

We can recombine these dimensions:

$a (t) = (6 t - 2, - 3)$ $a(t)=(6t-2,-3)$

Interestingly, the acceleration in the $y$ $y$ direction is constant, while the acceleration in the $x$ $x$ direction changes with time.

At $t = 5$ $t=5$ $s$ $s$ , we substitute in 5, and get:

$a (t) = (6 (5) - 2, - 3) = (28, - 3)$ $a(t)=(6(5)-2,-3)=(28,-3)$

We can use Pythagoras' theorem to find the magnitude and the tangent to find the direction of the resultant acceleration vector.

$r = \sqrt{x^{2} + y^{2}} = \sqrt{28^{2} + {(- 3)}^{2}} = \sqrt{793} = 28.2$ $r=sqrt(x^2+y^2)=sqrt(28^2+(-3)^2)=sqrt(793)=28.2$ $m s^{- 2}$ $ms^-2$

$tan θ = \frac{28}{- 3} \to θ = - 84^{o}$ $tan \theta = 28/(-3) -> \theta = -84^o$

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An object's two dimensional velocity is given by $v (t) = (3 t^{2} - 2 t, 1 - 3 t)$ $v(t) = ( 3t^2 - 2t , 1- 3t )$ . What is the object's rate and direction of acceleration at $t = 5$ $t=5$ ?

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Explanation:

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Science

Math

Humanities

... and beyond

An object's two dimensional velocity is given by v(t)=(3t2−2t,1−3t)v(t) = ( 3t^2 - 2t , 1- 3t ). What is the object's rate and direction of acceleration at t=5t=5 ?

1 Answer

Explanation:

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An object's two dimensional velocity is given by $v (t) = (3 t^{2} - 2 t, 1 - 3 t)$ $v(t) = ( 3t^2 - 2t , 1- 3t )$ . What is the object's rate and direction of acceleration at $t = 5$ $t=5$ ?