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 $t=4$ ?

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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 $t=4$ ?

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Answer 1 · 2017-07-10T13:44:49

$a = 48.0$ $"m/s"^2$

$theta = 1.19^"o"$

Explanation:

We're asked to find the object's acceleration (magnitude and direction) given its velocity equation.

We need to find the object's acceleration as a function of time, by differentiating the velocity component equations:

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

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

Note: I changed the $y$ -velocity equation SLIGHLTY by moving the final $t$ inside the sine function, because (1) otherwise it could have said $t - t^3sin(pi/8)$ and (2) this is how I've always seen these equations.

How that we know the acceleration components as functions of time, let's plug in $t = 4$ $"s"$ to find the components at that time:

$a_x (4) = 3(4)^2 = 48$ $"m/s"^2$

$a_y = 1 - 2(4)^2cos((pi/8)(4)) = 1$ $"m/s"^2$ (cosine of $pi/2$ is $0$ )

The magnitude of the acceleration is thus

$a = sqrt((48color(white)(l)"m/s"^2)^2 + (1color(white)(l)"m/s"^2)^2) = color(red)(48.0$ $color(red)("m/s"^2$

And the direction is

$theta = arctan((1cancel("m/s"^2))/(48cancel("m/s"^2))) = color(blue)(1.19^"o"$

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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 $t=4$ ?

1 Answer

Explanation:

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1 Answer

Explanation:

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