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

Jul 6, 2017

$a = 6.52$ ${\text{m/s}}^{2}$

$\phi = {259}^{\text{o}}$

#### Explanation:

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

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

Finding the derivative of velocity component equations, we have

${v}_{x} = {e}^{t} - 2 {t}^{2}$

${v}_{y} = t - t {e}^{2}$

${a}_{x} = \frac{d}{\mathrm{dt}} \left[{e}^{t} - 2 {t}^{2}\right] = {e}^{t} - 4 t$

${a}_{y} = \frac{d}{\mathrm{dt}} \left[t - t {e}^{2}\right] = 1 - {e}^{2}$

At time $t = 1$ $\text{s}$, the acceleration components are thus

${a}_{x} = {e}^{\left(1\right)} - 4 \left(1\right) = - 1.28$ ${\text{m/s}}^{2}$

${a}_{y} = 1 - {e}^{2} = - 6.39$ ${\text{m/s}}^{2}$

The magnitude (rate) of the acceleration is

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

The direction of the acceleration is

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

measured anticlockwise from the positive $x$-axis.

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