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

Question

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

1 Answer

Impact of this question

1669 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-06T21:22:39

$a = 6.52$ $a = 6.52$ ${m/s}^{2}$ $"m/s"^2$

$ϕ = 259^{o}$ $phi = 259^"o"$

Explanation:

We're asked to find the object's acceleration at time $t = 1$ $t = 1$ $s$ $"s"$ from a given velocity equation.

To do this, we need to find the object's acceleration components as a function of $t$ $t$ , via differentiation of the velocity equations:

Finding the derivative of velocity component equations, we have

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

$v_{y} = t - t e^{2}$ $v_y = t - te^2$

$a_{x} = \frac{d}{d t} [e^{t} - 2 t^{2}] = e^{t} - 4 t$ $a_x = d/(dt) [e^t - 2t^2] = e^t - 4t$

$a_{y} = \frac{d}{d t} [t - t e^{2}] = 1 - e^{2}$ $a_y = d/(dt) [t - te^2] = 1 - e^2$

At time $t = 1$ $t = 1$ $s$ $"s"$ , the acceleration components are thus

$a_{x} = e^{(1)} - 4 (1) = - 1.28$ $a_x = e^((1)) - 4(1) = -1.28$ ${m/s}^{2}$ $"m/s"^2$

$a_{y} = 1 - e^{2} = - 6.39$ $a_y = 1 - e^2 = -6.39$ ${m/s}^{2}$ $"m/s"^2$

The magnitude (rate) of the acceleration is

$a = \sqrt{{(- 1.28 l {m/s}^{2})}^{2} + {(- 6.39 l {m/s}^{2})}^{2}} = 6.52$ $a = sqrt((-1.28color(white)(l)"m/s"^2)^2 + (-6.39color(white)(l)"m/s"^2)^2) = color(red)(6.52$ ${m/s}^{2}$ $color(red)("m/s"^2$

The direction of the acceleration is

$ϕ = arctan (\frac{- 6.39 l {m/s}^{2}}{- 1.28 l {m/s}^{2}}) = {78.7}^{o}$ $phi = arctan((-6.39color(white)(l)"m/s"^2)/(-1.28color(white)(l)"m/s"^2)) = 78.7^"o"$ $+ 180^{o} = 259^{o}$ $+ 180^"o" = color(blue)(259^"o"$

measured anticlockwise from the positive $x$ $x$ -axis.

The $180^{o}$ $180^"o"$ was added to fix the calculator angle error.

Science

Math

Humanities

... and beyond

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

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)=(et−2t2,t−te2)v(t) = ( e^t-2t^2 , t-te^2 ). What is the object's rate and direction of acceleration at t=1t=1 ?

1 Answer

Explanation:

Related questions

Impact of this question

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