An object's two dimensional velocity is given by $v(t) = ( t^3, t-t^2sin(pi/8)t)$ . What is the object's rate and direction of acceleration at $t=12$ ?

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An object's two dimensional velocity is given by $v(t) = ( t^3, t-t^2sin(pi/8)t)$ . What is the object's rate and direction of acceleration at $t=12$ ?

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Answer 1 · 2017-07-31T16:08:06

$a = 462$ $"LT"^-2$

$theta = -20.8^"o"$

Explanation:

We're asked to find the magnitude and direction of the acceleration of an object at a certain time, given the velocity equations as functions of time.

To do this, we need to find the acceleration of the object as a suction of $t$ , by differentiating the velocity component equations.

We're given:

$v_x(t) = t^3$

$v_y(t) = t-t^2sin(pi/8)t$

So

$a_x(t) = d/(dt) [t^3] = ul(3t^2$

$a_y(t) = d/(dt) [t-t^2sin(pi/8)t] = ul(1-3t^2sin(pi/8)$

Now, we plug in $t = 12$ :

$a_x = 3(12)^2 = ul(432$

$a_y = 1-3(12)^2sin(pi/8) = ul(-164$

The magnitude of the acceleration is thus

$a = sqrt((a_x)^2 + (a_y)^2) = sqrt((432)^2 + (-164)^2)$

$= color(red)(ul(462color(white)(l)"LT"^-2$

(the $"LT"^-2$ term is the dimensional form of the units for acceleration; I used it here since no units were given)

The direction is

$theta = arctan((a_y)/(a_x)) = arctan((-164)/(432)) = color(blue)(ul(-20.8^"o"$

Always be sure to check your arctangent calculation, as it could be $180^"o"$ off!

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An object's two dimensional velocity is given by $v(t) = ( t^3, t-t^2sin(pi/8)t)$ . What is the object's rate and direction of acceleration at $t=12$ ?

1 Answer

Explanation:

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An object's two dimensional velocity is given by v(t) = ( t^3, t-t^2sin(pi/8)t)v(t) = ( t^3, t-t^2sin(pi/8)t). What is the object's rate and direction of acceleration at t=12 t=12 ?

1 Answer

Explanation:

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