An object's two dimensional velocity is given by $v (t) = (t^{3}, t - t^{2} sin (\frac{π}{8}) t)$ $v(t) = ( t^3, t-t^2sin(pi/8)t)$ . What is the object's rate and direction of acceleration at $t = 4$ $t=4$ ?

Question

An object's two dimensional velocity is given by $v (t) = (t^{3}, t - t^{2} sin (\frac{π}{8}) t)$ $v(t) = ( t^3, t-t^2sin(pi/8)t)$ . What is the object's rate and direction of acceleration at $t = 4$ $t=4$ ?

1 Answer

Impact of this question

1553 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-10T13:44:49

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

$θ = {1.19}^{o}$ $theta = 1.19^"o"$

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:

$a_{x} = \frac{d}{d t} [t^{3}] = 3 t^{2}$ $a_x = d/(dt) [t^3] = 3t^2$

$a_{y} = \frac{d}{d t} [t - t^{2} sin ((\frac{π}{8}) t)] = 1 - 2 t^{2} cos ((\frac{π}{8}) t)$ $a_y = d/(dt) [t - t^2 sin((pi/8)t)] = 1 - 2t^2cos((pi/8)t)$

Note: I changed the $y$ $y$ -velocity equation SLIGHLTY by moving the final $t$ $t$ inside the sine function, because (1) otherwise it could have said $t - t^{3} sin (\frac{π}{8})$ $t - t^3sin(pi/8)$ 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 $t = 4$ $t = 4$ $s$ $"s"$ to find the components at that time:

$a_{x} (4) = 3 {(4)}^{2} = 48$ $a_x (4) = 3(4)^2 = 48$ ${m/s}^{2}$ $"m/s"^2$

$a_{y} = 1 - 2 {(4)}^{2} cos ((\frac{π}{8}) (4)) = 1$ $a_y = 1 - 2(4)^2cos((pi/8)(4)) = 1$ ${m/s}^{2}$ $"m/s"^2$ (cosine of $\frac{π}{2}$ $pi/2$ is $0$ $0$ )

The magnitude of the acceleration is thus

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

And the direction is

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

Science

Math

Humanities

... and beyond

An object's two dimensional velocity is given by $v (t) = (t^{3}, t - t^{2} sin (\frac{π}{8}) t)$ $v(t) = ( t^3, t-t^2sin(pi/8)t)$ . What is the object's rate and direction of acceleration at $t = 4$ $t=4$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

An object's two dimensional velocity is given by v(t) = ( t^3, t-t^2sin(pi/8)t)v(t)=(t3,t−t2sin(π8)t)v(t) = ( t^3, t-t^2sin(pi/8)t). What is the object's rate and direction of acceleration at t=4 t=4t=4 ?

1 Answer

Explanation:

Related questions

Impact of this question

An object's two dimensional velocity is given by $v (t) = (t^{3}, t - t^{2} sin (\frac{π}{8}) t)$ $v(t) = ( t^3, t-t^2sin(pi/8)t)$ . What is the object's rate and direction of acceleration at $t = 4$ $t=4$ ?